This deck solves linear equations end to end. It starts with the balance-scale model, then covers clearing fractions and decimals, telling identities from contradictions, literal equations, applied linear models, and inequalities on the number line written in interval notation. It targets the forgotten inequality flip, multiplying only part of the equation by the LCD, calling a true collapsed statement "no solution", and writing an "or" union as an "and" intersection.
Subject: College Algebra · 147 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 04
Keep the scale balanced. Then learn the one move that flips the sign.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open Linear Equations and Inequalities: without looking back, what was the main idea of Rational Expressions, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers domain restrictions, simplifying by factoring, multiplying and dividing, the least common denominator, adding and subtracting, complex fractions, and the difference quotient. It targets the biggest error in the course - cancelling added terms instead of factors - along with restrictions lost after a cancellation, the dropped minus sign when subtracting, and a common denominator that is either wrong or needlessly huge.
Section
Part 1
Concept
An equation is a claim: it says the thing on the left and the thing on the right name the same number.
\[ 3x - 4 = 11 \]
The claim is not automatically true. It is true for some numbers put in place of the variable and false for others.
solution — A number you can substitute for the variable that makes the left side and the right side come out equal.
Solving means finding every number that makes the claim true. That collection is called the solution set.
Counterexample
Discussion prompt
An equation is a claim: it says the thing on the left and the thing on the right name the same number.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The claim is not automatically true. It is true for some numbers put in place of the variable and false for others.
Picture it
Animation
Shows: Solving keeps the equation balanced — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every line does the same thing to both sides.
Ranking
Put in order
Put the moves of Testing a candidate solution into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Substitution is the definition of a solution, so testing is not a shortcut - it is the meaning.
Worked example
A candidate is only a solution if it survives substitution. Nothing else counts as proof.
\[ 5x - 3 = 17 \]
Try the candidate 4: replace the variable with 4 everywhere it appears
Why: Substitution is the definition of a solution, so testing is not a shortcut - it is the meaning.
\[ 5(4) - 3 = 20 - 3 = 17 \]
Both sides equal 17, so 4 is a solution
Why: The left side evaluated to 17 and the right side was already 17. The claim is true at this number.
Now try the candidate 2 the same way
Why: One success does not mean every number works. Testing a second candidate shows the claim is selective.
\[ 5(2) - 3 = 10 - 3 = 7 \]
Check both candidates side by side against the original equation
Why: At 4 the two sides agree, at 2 they do not. So 4 is in the solution set and 2 is not.
| candidate | left side | right side | solution? |
|---|---|---|---|
| 4 | 17 | 17 | yes |
| 2 | 7 | 17 | no |
Picture it
Animation
Shows: Each line of the worked example "Testing a candidate solution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 4 the two sides agree, at 2 they do not. So 4 is in the solution set and 2 is not.
Intuition
Figure (svg): A balance scale with the expression two x plus six on the left pan and fourteen on the right pan, hanging level
The equals sign is a scale that is currently hanging level. Whatever sits in the left pan weighs exactly what sits in the right pan.
You are allowed to change the scale as long as you do the same thing to both pans. Take four ounces off one side, take four off the other, and it is still level.
Touch only one pan and the scale tips. The new equation is a different claim, and its solution is not the one you were hunting.
The whole game is to strip the left pan down until the variable is sitting there alone. Then the right pan is telling you its weight.
Analogy
Discussion prompt
Explain Picture a balance scale by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
The equals sign is a scale that is currently hanging level. Whatever sits in the left pan weighs exactly what sits in the right pan.
Concept
The legal moves have official names. All four say the same thing: do it to both sides.
| property | what you may do | restriction |
|---|---|---|
| addition | add the same quantity to both sides | none |
| subtraction | subtract the same quantity from both sides | none |
| multiplication | multiply both sides by the same number | none |
| division | divide both sides by the same number | the number cannot be zero |
Division is the only one with a restriction, because dividing by zero is undefined. Everything else is unconditionally safe.
There is also the symmetric property: you may swap the two sides whenever the variable ends up on the right and you would rather read it on the left.
\[ \text{if } 7 = x, \text{ then } x = 7 \]
Discrimination
Sort into buckets
Sort these by restriction, from memory, without looking back at The properties of equality. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Starting equation, and the urge to just slide the six out of the way.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It feels like 'moving' the six, but nothing was subtracted from the right pan.
Same starting equation, one honest move.
Why: It feels like 'moving' the six, but nothing was subtracted from the right pan. The scale just tipped.
Trap
Starting equation, and the urge to just slide the six out of the way.
\[ 2x + 6 = 14 \]
Erase the 6 from the left and leave the right alone
Why: It feels like 'moving' the six, but nothing was subtracted from the right pan. The scale just tipped.
\[ 2x = 14 \quad \Rightarrow \quad x = 7 \]
Test the answer 7 in the original equation
Why: The left side becomes twenty, not fourteen. The candidate fails, so it was never a solution.
\[ 2(7) + 6 = 14 + 6 = 20 \ne 14 \]
Same starting equation, one honest move.
\[ 2x + 6 = 14 \]
Subtract 6 from BOTH sides
Why: Removing the same weight from each pan keeps the scale level, so the new equation has the same solution as the old one.
\[ 2x + 6 - 6 = 14 - 6 \quad \Rightarrow \quad 2x = 8 \]
Divide both sides by 2, then test the answer
Why: Four checks out in the original equation, so it really is the solution.
\[ x = 4, \qquad 2(4) + 6 = 8 + 6 = 14 \ \checkmark \]
Notation
Annotate
From Trap: changing only one side — read this one piece at a time. What is each part doing?
On: \( 2x = 14 \quad \Rightarrow \quad x = 7 \)
Section
Part 2
Concept
An equation is linear in a variable when that variable appears only to the first power - never squared, never under a radical, never in a denominator.
\[ \text{linear:} \quad 4x - 9 = 2x + 1, \qquad \frac{x}{3} + 5 = 7 \]
\[ \text{not linear:} \quad x^{2} - 4 = 0, \qquad \sqrt{x} = 3, \qquad \frac{6}{x} = 2 \]
This matters because a linear equation has at most one solution. The moment you see a square, expect the possibility of two.
Explain it
Discussion prompt
Explain What makes an equation linear to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
An equation is linear in a variable when that variable appears only to the first power - never squared, never under a radical, never in a denominator.
Picture it
Animation
Shows: Where two lines meet — a rendered Manim animation.
Rendered with Manim.
Takeaway: The solution of a system is the point both equations agree on.
Step zero
Discussion prompt
One step: undo a fractional coefficient — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Multiply both sides by the reciprocal of the coefficient
Answer:
Worked example
The variable is already alone except for the number multiplying it.
\[ \frac{2}{3}x = 10 \]
Multiply both sides by the reciprocal of the coefficient
Why: A fraction times its own reciprocal is one, which is exactly what strips the coefficient off the variable.
\[ \frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 10 \]
Simplify each side
Why: On the left the fractions cancel to one. On the right, three halves of ten is fifteen.
\[ x = 15 \]
Verify by substituting 15 into the original equation
Why: Two thirds of fifteen is ten, which matches the right side exactly, so fifteen is the solution.
\[ \frac{2}{3}(15) = \frac{30}{3} = 10 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "One step: undo a fractional coefficient", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two thirds of fifteen is ten, which matches the right side exactly, so fifteen is the solution.
Estimation
Predict first
Two things are wrapped around the variable: a plus nine, and a multiply by negative four.
Commit before you compute: what does Two steps, with a negative coefficient come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by substituting negative six into the original equation
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Negative four times negative six is positive twenty-four, and twenty-four plus nine is thirty-three.
Worked example
Two things are wrapped around the variable: a plus nine, and a multiply by negative four.
\[ -4x + 9 = 33 \]
Subtract 9 from both sides
Why: Peel off the outermost layer first. Addition is the loosest wrapper, so it comes off before the multiplication.
\[ -4x = 33 - 9 = 24 \]
Divide both sides by negative four
Why: The coefficient is negative four, not four. Dividing by the whole coefficient, sign included, is what leaves the variable alone.
\[ x = \frac{24}{-4} = -6 \]
Verify by substituting negative six into the original equation
Why: Negative four times negative six is positive twenty-four, and twenty-four plus nine is thirty-three. Both sides agree.
\[ -4(-6) + 9 = 24 + 9 = 33 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Two steps, with a negative coefficient", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Negative four times negative six is positive twenty-four, and twenty-four plus nine is thirty-three. Both sides agree.
Intuition
Think about how the expression was built around the variable. Someone took the variable, multiplied it, then added something.
To undo it you run the film backwards: undo the last thing that was done first. Coat off before shirt off.
That is why addition and subtraction come off before multiplication and division - they were applied last, so they come off first.
It is the order of operations in reverse. You are not memorizing a new rule; you are reading the old one right to left.
Concept
Before any term crosses the equals sign, clean up each side on its own: distribute, then combine like terms.
Distributing means multiplying the outside factor by every term inside the parentheses, sign and all.
\[ -3(2x - 5) = -6x + 15 \]
Notice the second sign flipped. Negative three times negative five is positive fifteen. That single sign is the most-dropped detail in the course.
Missing information
Discussion prompt
Parentheses on both sides. Clean each side first, then move terms.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Three times two x is six x, three times negative five is negative fifteen, and negative fifteen plus four is negative eleven.
Worked example
Parentheses on both sides. Clean each side first, then move terms.
\[ 3(2x - 5) + 4 = 2(x + 3) - 1 \]
Distribute on the left, then combine the constants
Why: Three times two x is six x, three times negative five is negative fifteen, and negative fifteen plus four is negative eleven.
\[ 6x - 15 + 4 = 6x - 11 \]
Distribute on the right, then combine the constants
Why: Two times x is two x, two times three is six, and six minus one is five.
\[ 2x + 6 - 1 = 2x + 5 \]
Rewrite the cleaned-up equation
Why: Both sides are now as simple as they can get, so the moving of terms can begin.
\[ 6x - 11 = 2x + 5 \]
Subtract 2x from both sides and add 11 to both sides
Why: Collect the variable terms on the left and the constants on the right. Both moves are applied to both sides, so the balance holds.
\[ 4x = 16 \]
Divide both sides by 4
Why: Four is the coefficient of the variable, so dividing by it leaves the variable alone.
\[ x = 4 \]
Verify by substituting 4 into the ORIGINAL equation
Why: Left side: three times three plus four is thirteen. Right side: two times seven minus one is thirteen. They agree, so four is the solution.
\[ 3(8 - 5) + 4 = 13, \qquad 2(7) - 1 = 13 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A multi-step equation with parentheses", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Left side: three times three plus four is thirteen. Right side: two times seven minus one is thirteen. They agree, so four is the solution.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Simplify this expression, which shows up inside a much bigger equation.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The negative two only got applied to the first term.
Same expression, with the sign treated as part of the multiplier.
Why: The negative two only got applied to the first term. The second term kept its own minus and became negative two.
Trap
Simplify this expression, which shows up inside a much bigger equation.
\[ 7 - 2(3x - 1) \]
Multiply 2 by each inside term and keep the inside signs as written
Why: The negative two only got applied to the first term. The second term kept its own minus and became negative two.
\[ 7 - 6x - 2 = 5 - 6x \]
Test the result at the input 1
Why: The true value of the original expression at one is three, but this version gives negative one. The simplification changed the expression.
\[ 5 - 6(1) = -1 \ne 3 \]
Same expression, with the sign treated as part of the multiplier.
\[ 7 - 2(3x - 1) \]
Distribute negative two across BOTH inside terms
Why: Negative two times three x is negative six x, and negative two times negative one is positive two. A negative times a negative is positive.
\[ 7 - 6x + 2 = 9 - 6x \]
Test the result at the input 1
Why: The original expression at one is seven minus two times two, which is three, and this version gives three as well.
\[ 9 - 6(1) = 3, \qquad 7 - 2(3 - 1) = 3 \ \checkmark \]
Concept
When the variable appears on both sides, you cannot isolate it until all of it lives on one side.
Pick a side and subtract the other side's variable term from both sides. It is the same legal move as before.
A useful habit: move the variable to whichever side keeps its coefficient positive. Fewer negatives means fewer sign slips.
\[ 9 - 6x = 4x - 11 \quad \longrightarrow \quad 9 = 10x - 11 \]
Adding six x to both sides put the variable on the right with a positive coefficient. Ending with the variable on the right is fine.
Fill the middle
Fill in the blanks
From Variables on both sides, with distribution — finish the line. Write what belongs on the right of the equals sign before you look.
7 - 2(3x - 1) = 4x - 11
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Seven plus two is nine, and the variable term is negative six x.
Worked example
There are parentheses on the left and a variable term on the right.
\[ 7 - 2(3x - 1) = 4x - 11 \]
Distribute negative two, then combine like terms on the left
Why: Seven plus two is nine, and the variable term is negative six x. The right side is already simple.
\[ 9 - 6x = 4x - 11 \]
Add 6x to both sides
Why: This clears the variable off the left and lands it on the right with a positive coefficient, which is easier to work with.
\[ 9 = 10x - 11 \]
Add 11 to both sides
Why: That isolates the variable term by removing the constant that sits with it.
\[ 20 = 10x \]
Divide both sides by 10
Why: Ten is the coefficient. Dividing gives the value of the variable, and the symmetric property lets you write it on the left.
\[ x = 2 \]
Verify by substituting 2 into the ORIGINAL equation
Why: Left side: seven minus two times five is negative three. Right side: eight minus eleven is negative three. Both sides agree.
\[ 7 - 2(6 - 1) = -3, \qquad 4(2) - 11 = -3 \ \checkmark \]
Pattern
Every linear equation, however cluttered, gives way to the same short list of moves in the same order.
Steps one through three are cleanup - nothing crosses the equals sign yet. Steps four and five are the actual solving.
Step six is not optional. It is the only step that can tell you whether steps one through five went wrong.
Picture it
Animation
Shows: The slope, seen — a rendered Manim animation.
Rendered with Manim.
Takeaway: The coefficient tilts the line; the constant slides it.
Check
Distribute first, then collect. Work it on paper before you choose.
Check your understanding
Solve 5(x - 2) = 3x + 8.
Answer: A
Why: Distributing gives 5x - 10 = 3x + 8. Subtract 3x from both sides for 2x - 10 = 8, add 10 for 2x = 18, then divide by 2 to get x = 9. Verify in the original: 5(9 - 2) = 35 and 3(9) + 8 = 35.
Section
Part 3
Concept
Fractions are not harder to solve with. They are just harder to look at. So get rid of them on move one.
Multiply both sides by the least common denominator of every fraction in the equation.
\[ \frac{x}{3} + \frac{1}{2} = \frac{x}{4} + 2 \qquad \text{LCD} = 12 \]
Because both sides were multiplied by the same number, the scale stays level and the solution set does not change.
The one rule: every term gets multiplied, including the terms that were never fractions in the first place.
Step zero
Discussion prompt
Clearing fractions with the LCD — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the LCD of 3, 2, and 4
Answer:
Worked example
Four terms, three of them fractions. The denominators are three, two, and four.
\[ \frac{x}{3} + \frac{1}{2} = \frac{x}{4} + 2 \]
Find the LCD of 3, 2, and 4
Why: The smallest number all three divide into is twelve, so twelve clears every denominator at once.
\[ \text{LCD} = 12 \]
Multiply EVERY term on both sides by 12
Why: Multiplying a side means multiplying each of its terms - including the plain 2, which is a term too.
\[ 12 \cdot \frac{x}{3} + 12 \cdot \frac{1}{2} = 12 \cdot \frac{x}{4} + 12 \cdot 2 \]
Simplify each product
Why: Twelve thirds is four, twelve halves is six, twelve fourths is three, and twelve times two is twenty-four. Every denominator is gone.
\[ 4x + 6 = 3x + 24 \]
Subtract 3x from both sides, then subtract 6 from both sides
Why: Collect the variable on the left and the constants on the right, one legal move at a time.
\[ x = 18 \]
Verify by substituting 18 into the ORIGINAL equation
Why: Left side: eighteen thirds is six, plus one half. Right side: eighteen fourths is four and a half, plus two. Both are thirteen halves.
\[ \frac{18}{3} + \frac{1}{2} = \frac{13}{2}, \qquad \frac{18}{4} + 2 = \frac{13}{2} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Clearing fractions with the LCD", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Left side: eighteen thirds is six, plus one half. Right side: eighteen fourths is four and a half, plus two. Both are thirteen halves.
Trap
Same equation, and the temptation to touch only the terms that look like fractions.
\[ \frac{x}{3} + \frac{1}{2} = \frac{x}{4} + 2 \]
Multiply the three fraction terms by 12 and leave the 2 alone
Why: The plain 2 seemed to have nothing to clear, so it was skipped - and the right side quietly shrank.
\[ 4x + 6 = 3x + 2 \quad \Rightarrow \quad x = -4 \]
Test negative four in the original equation
Why: The left side is negative five sixths and the right side is one. They do not match, so negative four was never a solution.
\[ \frac{-4}{3} + \frac{1}{2} = -\frac{5}{6}, \qquad \frac{-4}{4} + 2 = 1, \qquad -\frac{5}{6} \ne 1 \]
Same equation, with the multiplication applied to every term on both sides.
\[ \frac{x}{3} + \frac{1}{2} = \frac{x}{4} + 2 \]
Multiply all FOUR terms by 12
Why: Twelve times two is twenty-four. A whole-number term still has to be multiplied, or the two sides are no longer equal.
\[ 4x + 6 = 3x + 24 \quad \Rightarrow \quad x = 18 \]
Test eighteen in the original equation
Why: Both sides come out to thirteen halves, so eighteen really is the solution.
\[ \frac{18}{3} + \frac{1}{2} = \frac{13}{2} = \frac{18}{4} + 2 \ \checkmark \]
Translation
\( 4x + 6 = 3x + 24 \quad \Rightarrow \quad x = 18 \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Fill the middle
Fill in the blanks
From Clearing decimals with a power of ten — finish the line. Write what belongs on the right of the equals sign before you look.
0.4x + 0.15 = 0.25x + 0.6
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. One hundred shifts every decimal point two places to the right, which is exactly enough to make all four coefficients whole numbers.
Worked example
Decimals clear the same way. Count the most decimal places on any coefficient, then multiply by that power of ten.
\[ 0.4x + 0.15 = 0.25x + 0.6 \]
The most decimal places on any coefficient is two, so multiply every term by 100
Why: One hundred shifts every decimal point two places to the right, which is exactly enough to make all four coefficients whole numbers.
\[ 100(0.4x) + 100(0.15) = 100(0.25x) + 100(0.6) \]
Simplify each product
Why: Forty x, fifteen, twenty-five x, sixty. No decimals are left to fumble.
\[ 40x + 15 = 25x + 60 \]
Subtract 25x from both sides, then subtract 15 from both sides
Why: Variables to the left, constants to the right, each move applied to both sides.
\[ 15x = 45 \]
Divide both sides by 15
Why: Fifteen is the coefficient of the variable, so dividing by it leaves the variable alone.
\[ x = 3 \]
Verify by substituting 3 into the ORIGINAL equation
Why: Left side: four tenths of three is one and two tenths, plus fifteen hundredths. Right side: a quarter of three is three quarters, plus six tenths. Both are 1.35.
\[ 0.4(3) + 0.15 = 1.35, \qquad 0.25(3) + 0.6 = 1.35 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Clearing decimals with a power of ten", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Left side: four tenths of three is one and two tenths, plus fifteen hundredths. Right side: a quarter of three is three quarters, plus six tenths. Both are 1.35.
Check
Find the least common denominator first, and remember which terms have to be multiplied.
Check your understanding
Solve x/2 - x/5 = 3.
Answer: A
Why: The LCD of 2 and 5 is 10. Multiplying every term by 10 gives 5x - 2x = 30, so 3x = 30 and x = 10. Verify in the original: 10/2 - 10/5 = 5 - 2 = 3.
Concept
So far every equation had exactly one solution. Most linear equations do. Two other endings exist, and both are normal.
| what the equation collapses to | meaning | how to report it |
|---|---|---|
| a single value for the variable | exactly one number works | state the number |
| a true statement with no variable | every number works | all real numbers |
| a false statement with no variable | no number works | no solution |
The variable terms cancelling is not an error and not a dead end. It is the equation telling you which of the last two endings you are in.
What decides between them is whether the statement left behind is true or false.
Comparison
Comparison matrix
From Three things can happen: refill the meaning column from what you know. The rest of the table is as it appeared.
| what the equation collapses to | meaning | how to report it |
|---|---|---|
| a single value for the variable | exactly one number works | state the number |
| a true statement with no variable | every number works | all real numbers |
| a false statement with no variable | no number works | no solution |
Intuition
When the variable disappears, students panic and guess. Do not guess. Read what is actually written on the page.
\[ 6 = 6 \quad \text{is true}, \qquad -4 = 5 \quad \text{is false} \]
If the leftover statement is true, it stayed true no matter what you had substituted - so every real number is a solution.
If the leftover statement is false, it was false no matter what you had substituted - so no number is a solution.
identity — An equation that is true for every value of the variable. The two sides are secretly the same expression written differently.
contradiction — An equation that is false for every value of the variable. The two sides can never be made equal.
Definition probe
Sort into buckets
Every line below is part of the definition of identity or of contradiction — one or the other, never both. Put each where it belongs.
Hypothesis
Predict first
An identity: every number works is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Distribute negative three on the left
Why: Negative three times x is negative three x, and negative three times negative two is positive six.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
It looks like an ordinary equation with the variable on both sides.
\[ 5x - 3(x - 2) = 2x + 6 \]
Distribute negative three on the left
Why: Negative three times x is negative three x, and negative three times negative two is positive six.
\[ 5x - 3x + 6 = 2x + 6 \]
Combine like terms on the left
Why: Five x minus three x is two x, so the left side is two x plus six - which is exactly the right side.
\[ 2x + 6 = 2x + 6 \]
Subtract 2x from both sides
Why: The variable terms are identical, so they cancel completely and carry the variable out of the equation with them.
\[ 6 = 6 \]
Read the leftover statement
Why: Six equals six is true, and it holds no variable, so nothing you substitute could break it. The solution set is every real number.
\[ \text{solution set: all real numbers} \]
Verify by testing three very different numbers in the ORIGINAL equation
Why: At zero both sides are six, at four both sides are fourteen, at negative one both sides are four. Three hits is strong evidence the two sides are the same expression.
| x | left side | right side | equal? |
|---|---|---|---|
| 0 | 6 | 6 | yes |
| 4 | 14 | 14 | yes |
| -1 | 4 | 4 | yes |
Picture it
Animation
Shows: When the variable disappears — a rendered Manim animation.
Rendered with Manim.
Takeaway: A false statement means empty; a true one means everything.
Ranking
Put in order
Put the moves of A contradiction: no number works into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Four times two x is eight x, and four times negative one is negative four.
Worked example
Same shape as the identity, different constants.
\[ 4(2x - 1) = 8x + 5 \]
Distribute the 4 on the left
Why: Four times two x is eight x, and four times negative one is negative four.
\[ 8x - 4 = 8x + 5 \]
Subtract 8x from both sides
Why: Both sides carry exactly eight x, so the variable cancels out entirely and never comes back.
\[ -4 = 5 \]
Read the leftover statement
Why: Negative four equals five is false, and no choice of the variable can repair it. There is no solution.
\[ \text{solution set: no solution} \]
Verify by testing numbers in the ORIGINAL equation
Why: The right side beats the left side by exactly nine at every input. Two parallel expressions never meet, which is what no solution looks like.
| x | left side | right side | gap |
|---|---|---|---|
| 0 | -4 | 5 | 9 |
| 2 | 12 | 21 | 9 |
| -3 | -28 | -19 | 9 |
Picture it
Animation
Shows: Each line of the worked example "A contradiction: no number works", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The right side beats the left side by exactly nine at every input. Two parallel expressions never meet, which is what no solution looks like.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The variable terms cancel, and the reflex is to write down no solution.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Both sides had two x, so the variable vanished.
The same cancellation, but this time the leftover statement gets read.
Why: Both sides had two x, so the variable vanished. The conclusion drawn was that nothing can satisfy the equation.
Trap
The variable terms cancel, and the reflex is to write down no solution.
\[ 5x - 3(x - 2) = 2x + 6 \]
Cancel the variable terms and stop reading
Why: Both sides had two x, so the variable vanished. The conclusion drawn was that nothing can satisfy the equation.
\[ 6 = 6 \quad \Rightarrow \quad \text{no solution (wrong)} \]
Test the number 1 in the original equation
Why: The left side is five minus three times negative one, which is eight, and the right side is eight. A solution exists, so no solution was the wrong call.
\[ 5(1) - 3(1 - 2) = 8, \qquad 2(1) + 6 = 8 \]
The same cancellation, but this time the leftover statement gets read.
\[ 5x - 3(x - 2) = 2x + 6 \]
Cancel the variable terms, then read what is left
Why: Six equals six is a TRUE statement. True means the equation held for whatever was substituted, so every real number is a solution.
\[ 6 = 6 \quad \Rightarrow \quad \text{all real numbers} \]
Test two different numbers in the original equation
Why: At one both sides are eight, and at four both sides are fourteen. That is exactly what an identity looks like from the outside.
\[ x = 1: \ 8 = 8; \qquad x = 4: \ 14 = 14 \ \checkmark \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Six equals six is a TRUE statement. True means the equation held for whatever was substituted, so every real number is a solution.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Both sides had two x, so the variable vanished. The conclusion drawn was that nothing can satisfy the equation.
Ranking
Put in order
These are the steps of Pattern: reading the collapsed equation, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
When the algebra is finished, look at what is written and answer one question: is there still a variable?
Say the leftover statement out loud. Six equals six: true. Negative four equals five: false. That one sentence decides the answer.
Never write the empty answer just because the variable disappeared. Disappearing is neutral. Truth or falsehood is the verdict.
Elimination
Eliminate the wrong options
Solve 2(3x - 4) = 6x - 8. What is the solution set?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Distributing gives 6x - 8 = 6x - 8, the same expression on both sides. Subtracting 6x leaves -8 = -8, a true statement with no variable in it, so every real number is a solution.
Check
Distribute, then compare the two sides carefully before you decide.
Check your understanding
Solve 2(3x - 4) = 6x - 8. What is the solution set?
Answer: A
Why: Distributing gives 6x - 8 = 6x - 8, the same expression on both sides. Subtracting 6x leaves -8 = -8, a true statement with no variable in it, so every real number is a solution.
Section
Part 4
Concept
literal equation — An equation containing two or more variables, usually a formula. Solving it means isolating one chosen variable in terms of all the others.
Nothing new happens here. The same legal moves apply, and the other letters just ride along as if they were numbers you have not been shown yet.
\[ A = \frac{1}{2}bh \qquad \longrightarrow \qquad h = \frac{2A}{b} \]
The answer is an expression, not a number. That feels unfinished the first few times. It is finished.
This is how formulas get rearranged in every science class: the same formula, solved for whichever quantity you are actually missing.
Intuition
Imagine someone will hand you the values of all the other letters tomorrow. You want the recipe that turns them into the letter you care about.
If a plain seven sat where the letter b sits, you would divide both sides by seven without a second thought. So divide by b.
The only extra caution: you may divide by a letter only when it is not zero. For real formulas - a length, a base, a rate - it never is.
Whatever is being done to your target letter, undo it, in reverse order, exactly as you would with numbers.
Picture it
Animation
Shows: Solving for a letter, not a number — a rendered Manim animation.
Rendered with Manim.
Takeaway: Identical moves; the answer just happens to contain letters.
Estimation
Predict first
Simple interest: the interest earned equals the principal times the annual rate times the time in years. Solve it for the rate.
Commit before you compute: what does Solving the simple-interest formula for the rate come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with real numbers in the ORIGINAL formula
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Two thousand dollars held three years earning two hundred seventy dollars gives a rate of 0.045, and putting that rate back into the original formula reproduces two hundred seventy.
Worked example
Simple interest: the interest earned equals the principal times the annual rate times the time in years. Solve it for the rate.
\[ I = Prt \]
Identify what is attached to the target letter
Why: The rate is multiplied by the principal and by the time. Multiplication is undone by division, so one division will do it.
Divide both sides by the product of the principal and the time
Why: Dividing both sides by the same nonzero quantity keeps the equation balanced and cancels both unwanted factors off the rate.
\[ \frac{I}{Pt} = \frac{Prt}{Pt} = r \]
Write the result with the target on the left
Why: The symmetric property lets you swap the sides so the formula reads the way you would actually use it.
\[ r = \frac{I}{Pt} \]
Verify with real numbers in the ORIGINAL formula
Why: Two thousand dollars held three years earning two hundred seventy dollars gives a rate of 0.045, and putting that rate back into the original formula reproduces two hundred seventy.
\[ r = \frac{270}{2000 \cdot 3} = 0.045; \qquad 2000(0.045)(3) = 270 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Solving the simple-interest formula for the rate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two thousand dollars held three years earning two hundred seventy dollars gives a rate of 0.045, and putting that rate back into the original formula reproduces two hundred seventy.
Worked example
The perimeter of a rectangle is twice the length plus twice the width. Solve it for the width.
\[ P = 2L + 2W \]
Subtract twice the length from both sides
Why: The width term is buried inside a sum, so peel off the added term first - the same order you would use with numbers.
\[ P - 2L = 2W \]
Divide both sides by 2
Why: Two is the coefficient of the width. Note that the ENTIRE left side gets divided by two, not just one of its terms.
\[ \frac{P - 2L}{2} = W \]
State the formula for the width
Why: You may also split the fraction term by term, which some books prefer. Both forms are the same number.
\[ W = \frac{P - 2L}{2} = \frac{P}{2} - L \]
Verify with real numbers in the ORIGINAL formula
Why: A perimeter of forty feet with a length of twelve feet gives a width of eight feet, and twice twelve plus twice eight is forty feet.
\[ W = \frac{40 - 2(12)}{2} = \frac{16}{2} = 8; \qquad 2(12) + 2(8) = 40 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Solving the perimeter formula for the width", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A perimeter of forty feet with a length of twelve feet gives a width of eight feet, and twice twelve plus twice eight is forty feet.
Step zero
Discussion prompt
When the target variable appears twice — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Notice the target letter appears in two separate terms
Answer:
Worked example
The amount in an account after simple interest. The principal shows up in both terms on the right.
\[ A = P + Prt \]
Notice the target letter appears in two separate terms
Why: You cannot divide it out yet. Dividing the whole side by the principal would leave a stray principal behind in the second term.
Factor the target letter out of both terms
Why: The principal is a common factor of both terms, so factoring collects it into a single appearance - which is the only way to isolate it.
\[ A = P(1 + rt) \]
Divide both sides by the parenthesized factor
Why: Now the principal is multiplied by exactly one thing, so one division finishes the job.
\[ P = \frac{A}{1 + rt} \]
Verify with real numbers in the ORIGINAL formula
Why: Five hundred dollars at six percent for two years grows to five hundred sixty dollars, and five hundred sixty divided by one and twelve hundredths is five hundred.
\[ A = 500 + 500(0.06)(2) = 560; \qquad \frac{560}{1 + 0.12} = \frac{560}{1.12} = 500 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "When the target variable appears twice", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Five hundred dollars at six percent for two years grows to five hundred sixty dollars, and five hundred sixty divided by one and twelve hundredths is five hundred.
Check
The area of a triangle is one half the base times the height. Solve it for the height.
Check your understanding
Solve A = (1/2)bh for h.
Answer: A
Why: Multiply both sides by 2 to clear the one half, giving 2A = bh, then divide both sides by b, so h = 2A/b. Check with a triangle of area 12 and base 4: h = 24/4 = 6, and (1/2)(4)(6) = 12.
Section
Part 5
Concept
The hard part of a word problem is not the algebra. It is deciding what the letter stands for.
Write a complete sentence with units: let the letter stand for the number of liters of the stronger solution added.
Every other quantity in the story then gets written in terms of that letter. That is where the equation comes from.
The final answer is also a sentence with units, not a bare number. A number without units answers nothing.
Constraint
Discussion prompt
Run Pattern: turning words into an equation with this step confiscated:
Find the sentence that says two things are equal - that phrase is your equals sign.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Step four is the one people skip. Hunt for words like total, altogether, the same as, or is - those words are the equals sign.
Step six catches the classic error of solving correctly but reporting the wrong quantity, like giving the smaller of two amounts when the larger was asked for.
A table of the quantities is worth five minutes of staring. Rate problems, mixture problems, and interest problems all fill in the same three-column shape.
Edge cases
Discussion prompt
Pattern: turning words into an equation works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Step four is the one people skip. Hunt for words like total, altogether, the same as, or is - those words are the equals sign.
Picture it
Animation
Shows: Turning words into an equation — a rendered Manim animation.
Rendered with Manim.
Takeaway: The final step is the one that gets skipped under time pressure.
Fill the middle
Fill in the blanks
From Consecutive integers — finish the line. Write what belongs on the right of the equals sign before you look.
27 + 28 + 29 = 84 \ \checkmark
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Let the letter stand for the smallest of the three integers.
Worked example
Three consecutive integers add up to eighty-four. Find all three.
Define the variable
Why: Let the letter stand for the smallest of the three integers. Consecutive means each is one more than the last, so the other two are then forced.
\[ n, \qquad n + 1, \qquad n + 2 \]
Write the equation from the phrase add up to
Why: Adding the three expressions has to give eighty-four. That phrase is the equals sign.
\[ n + (n + 1) + (n + 2) = 84 \]
Combine like terms
Why: Three copies of the variable, and one plus two is three.
\[ 3n + 3 = 84 \]
Subtract 3 from both sides, then divide both sides by 3
Why: Peel off the constant, then remove the coefficient. Two familiar moves.
\[ 3n = 81 \quad \Rightarrow \quad n = 27 \]
Answer the question that was asked
Why: The question wanted all three integers, not just the smallest one. This is where careless answers lose points.
\[ 27, \ 28, \ 29 \]
Verify against the ORIGINAL story
Why: They are consecutive, and twenty-seven plus twenty-eight is fifty-five, plus twenty-nine is eighty-four. Both conditions of the story hold.
\[ 27 + 28 + 29 = 84 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Consecutive integers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: They are consecutive, and twenty-seven plus twenty-eight is fifty-five, plus twenty-nine is eighty-four. Both conditions of the story hold.
Missing information
Discussion prompt
A store marks up every item by thirty-five percent of what it paid. A jacket sells for sixty-seven dollars and fifty cents. What did the store pay?
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Let the letter stand for the store's cost in dollars. The markup is a percent OF that cost, so the cost is the thing the letter must name.
Worked example
A store marks up every item by thirty-five percent of what it paid. A jacket sells for sixty-seven dollars and fifty cents. What did the store pay?
Define the variable
Why: Let the letter stand for the store's cost in dollars. The markup is a percent OF that cost, so the cost is the thing the letter must name.
\[ c = \text{store cost, in dollars} \]
Write the selling price in terms of the cost
Why: The selling price is the cost plus thirty-five hundredths of that same cost.
\[ c + 0.35c = 67.50 \]
Combine like terms
Why: One whole c plus thirty-five hundredths of a c is one and thirty-five hundredths of a c. Forgetting the original whole c is the most common slip here.
\[ 1.35c = 67.50 \]
Divide both sides by 1.35
Why: That is the coefficient of the variable, so dividing by it leaves the cost alone.
\[ c = \frac{67.50}{1.35} = 50 \]
Verify by rebuilding the price from the ORIGINAL story
Why: Thirty-five percent of fifty dollars is seventeen dollars and fifty cents, and fifty plus that is sixty-seven fifty. The store paid fifty dollars.
\[ 50 + 0.35(50) = 50 + 17.50 = 67.50 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Percent markup: recovering the original cost", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Thirty-five percent of fifty dollars is seventeen dollars and fifty cents, and fifty plus that is sixty-seven fifty. The store paid fifty dollars.
Estimation
Predict first
Eight thousand dollars is split between an account paying four percent and one paying six percent. After one year the two accounts together earn three hundred eighty dollars. How much went into each?
Commit before you compute: what does Splitting an investment come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify both interest amounts against the ORIGINAL story
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Five thousand dollars at four percent earns two hundred dollars, three thousand at six percent earns one hundred eighty, and together that is exactly three hundred eighty dollars.
Worked example
Eight thousand dollars is split between an account paying four percent and one paying six percent. After one year the two accounts together earn three hundred eighty dollars. How much went into each?
Define the variable and express the rest
Why: Let the letter stand for the dollars placed at four percent. Whatever is left of the eight thousand went into the other account, so it is eight thousand minus that letter.
\[ x, \qquad 8000 - x \]
Write each account's one-year interest
Why: Simple interest for one year is the amount times the rate. Laying it out in a table keeps the two accounts from getting mixed up.
| account | amount (dollars) | rate | interest (dollars) |
|---|---|---|---|
| four percent | x | 0.04 | 0.04x |
| six percent | 8000 - x | 0.06 | 0.06(8000 - x) |
Set the total interest equal to 380
Why: The word together is the equals sign: the two interests add to the given total.
\[ 0.04x + 0.06(8000 - x) = 380 \]
Distribute, then combine like terms
Why: Six hundredths times eight thousand is four hundred eighty, and four hundredths minus six hundredths is negative two hundredths.
\[ 0.04x + 480 - 0.06x = 380 \quad \Rightarrow \quad -0.02x = -100 \]
Divide both sides by negative two hundredths
Why: A negative divided by a negative is positive, so the amount comes out positive as it must.
\[ x = \frac{-100}{-0.02} = 5000 \]
Verify both interest amounts against the ORIGINAL story
Why: Five thousand dollars at four percent earns two hundred dollars, three thousand at six percent earns one hundred eighty, and together that is exactly three hundred eighty dollars.
\[ 0.04(5000) + 0.06(3000) = 200 + 180 = 380 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Splitting an investment", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Five thousand dollars at four percent earns two hundred dollars, three thousand at six percent earns one hundred eighty, and together that is exactly three hundred eighty dollars.
Step zero
Discussion prompt
Distance, rate, and time: moving apart — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Define the variable
Answer:
Worked example
Two cyclists leave the same corner at the same moment, riding in opposite directions at twelve and sixteen miles per hour. When are they seventy miles apart?
Define the variable
Why: Let the letter stand for the hours since they left. Both riders have been riding for that same amount of time, which is what makes one letter enough.
\[ t = \text{hours ridden} \]
Write each distance as rate times time
Why: Distance equals rate times time. Miles per hour times hours gives miles, so the units of the answer take care of themselves.
| rider | rate (miles per hour) | time (hours) | distance (miles) |
|---|---|---|---|
| slower | 12 | t | 12t |
| faster | 16 | t | 16t |
Add the two distances, because they ride in opposite directions
Why: Moving apart means the gap between them is the sum of how far each one has travelled. If they rode the same way you would subtract instead.
\[ 12t + 16t = 70 \]
Combine like terms, then divide both sides by 28
Why: Twelve plus sixteen is twenty-eight, which is how many miles of separation open up per hour.
\[ 28t = 70 \quad \Rightarrow \quad t = 2.5 \]
Verify the distances against the ORIGINAL story
Why: In two and a half hours the slower rider covers thirty miles and the faster covers forty. Thirty plus forty is seventy miles apart, after two and a half hours.
\[ 12(2.5) + 16(2.5) = 30 + 40 = 70 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Distance, rate, and time: moving apart", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: In two and a half hours the slower rider covers thirty miles and the faster covers forty. Thirty plus forty is seventy miles apart, after two and a half hours.
Ranking
Put in order
Put the moves of A mixture problem into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Let the letter stand for the liters of fifty percent solution poured in.
Worked example
A lab has ten liters of a twenty percent saline solution. How many liters of fifty percent solution must be added to make the mixture thirty percent?
Define the variable and the resulting total
Why: Let the letter stand for the liters of fifty percent solution poured in. The final volume is that plus the ten liters already in the beaker.
\[ x \ \text{liters added}, \qquad 10 + x \ \text{liters of mixture} \]
Track the salt, not the liquid
Why: Concentration times volume gives the actual amount of salt. Salt is what is conserved when you pour two solutions together.
| source | volume (liters) | concentration | salt (liters) |
|---|---|---|---|
| existing | 10 | 0.20 | 2 |
| added | x | 0.50 | 0.50x |
| mixture | 10 + x | 0.30 | 0.30(10 + x) |
Set the salt in equal to the salt out
Why: The salt from the two sources has to add up to the salt in the mixture, because pouring neither creates nor destroys salt.
\[ 2 + 0.50x = 0.30(10 + x) \]
Distribute on the right, then collect like terms
Why: Three tenths of ten is three. Subtracting three tenths of x and two from both sides leaves two tenths of x on the left.
\[ 2 + 0.50x = 3 + 0.30x \quad \Rightarrow \quad 0.20x = 1 \]
Divide both sides by two tenths
Why: One divided by two tenths is five, so five liters of the stronger solution are needed.
\[ x = \frac{1}{0.20} = 5 \]
Verify the concentration against the ORIGINAL story
Why: Five liters of fifty percent solution carries two and a half liters of salt. Total salt is four and a half liters in fifteen liters of mixture, which is exactly thirty percent.
\[ \frac{2 + 0.50(5)}{10 + 5} = \frac{4.5}{15} = 0.30 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A mixture problem", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Five liters of fifty percent solution carries two and a half liters of salt. Total salt is four and a half liters in fifteen liters of mixture, which is exactly thirty percent.
Prediction
Predict first
After a 15 percent discount, a jacket costs 68 dollars. What was the original price?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 80 dollars
Why: The sale price is 85 percent of the original, so 0.85p = 68 and p = 68/0.85 = 80. Verify against the story: 15 percent of 80 dollars is 12 dollars, and 80 minus 12 is 68 dollars.
Check
Define the variable as the thing the percent is taken of, then write the equation before you compute anything.
Check your understanding
After a 15 percent discount, a jacket costs 68 dollars. What was the original price?
Answer: A
Why: The sale price is 85 percent of the original, so 0.85p = 68 and p = 68/0.85 = 80. Verify against the story: 15 percent of 80 dollars is 12 dollars, and 80 minus 12 is 68 dollars.
Section
Part 6
Concept
An equation asks which number makes the two sides equal. An inequality asks which numbers make one side smaller than the other.
\[ 3x - 5 \le 7 \]
There is almost never a single answer. The solution set is usually an entire stretch of the number line.
So the answer gets drawn as a shaded ray or segment, and written in interval notation, instead of as one number.
The good news: the solving moves are the same ones you already know, plus exactly one new rule.
Intuition
Figure (svg): Number line shaded from the far left up to a filled dot at four, showing all numbers less than or equal to four
Pick a number, test it, and if the inequality comes out true, shade it. Do that for every number and the shaded part is the solution set.
You obviously cannot test every number one at a time. Algebra finds the boundary for you, and then you shade the correct side of it.
A filled dot means the boundary number itself is a solution. An open dot means the shading stops just short of it.
This picture is the answer. Interval notation is just a compact way of writing the same picture on one line.
Concept
Interval notation names the shaded piece with its two ends: left end, comma, right end.
\[ \{\, x \mid x \le 4 \,\} \quad \longleftrightarrow \quad (-\infty, 4] \]
A square bracket means the endpoint is included. A round parenthesis means it is excluded.
Infinity always takes a parenthesis, because infinity is not a number you can reach, so it can never be included.
\[ (-\infty, 4], \qquad (-5, \infty), \qquad (-2, 4], \qquad [0, 3] \]
interval notation — A pair of endpoints naming every real number between them. A square bracket includes that endpoint; a parenthesis excludes it; infinity always gets a parenthesis.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of solution, identity, contradiction, literal equation, interval notation as Linear Equations and Inequalities uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: Interval notation, read once — a rendered Manim animation.
Rendered with Manim.
Takeaway: Infinity is not a number, so it can never be included.
Picture it
Figure (svg): Number line shaded from the far left up to a filled dot at four
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Solve it exactly the way you would solve the matching equation.
Worked example
Solve it exactly the way you would solve the matching equation.
\[ 3x - 5 \le 7 \]
Add 5 to both sides
Why: Adding the same number to both sides shifts both by the same amount, so it never changes which side is larger. The symbol is untouched.
\[ 3x \le 12 \]
Divide both sides by 3
Why: Three is positive, so the direction of the symbol still does not change. Only a negative divisor would flip it.
\[ x \le 4 \]
Write the interval and graph the solution set
Why: Every number four or smaller works, so shade to the left from a filled dot at four - filled, because the symbol allows equality.
\[ (-\infty, 4] \]
Figure (svg): Number line shaded from the far left up to a filled dot at four
Verify with a test point inside and one outside, in the ORIGINAL inequality
Why: At zero the left side is negative five, which is at most seven, so zero belongs. At five the left side is ten, which is not, so five is correctly excluded.
\[ x = 0: \ -5 \le 7 \ \checkmark \qquad x = 5: \ 10 \le 7 \ \text{false} \]
Reverse engineer
Discussion prompt
Work backwards. The example finished here:
Verify with a test point inside and one outside, in the ORIGINAL inequality
What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.
Hint: Every quantity in the result had to enter somewhere. Account for each one.
Answer:
Solve it exactly the way you would solve the matching equation.
Concept
Adding and subtracting never change the direction of an inequality. Multiplying or dividing by a positive number does not either.
Multiplying or dividing both sides by a negative number reverses the direction of the symbol. Every time. No exceptions.
\[ -3x < 15 \quad \Longrightarrow \quad x > -5 \]
Notice the numbers came out the same as they would have. Only the symbol turned around.
Miss that flip and every single number in your answer is on the wrong half of the line - which is why it is worth its own rule.
Intuition
Figure (svg): Number line with dots at two and five on the right of zero and at negative two and negative five on the left, showing that negative two lies to the right of negative five
Start with something obviously true: two is less than five.
\[ 2 < 5 \]
Multiply both sides by negative one. Two becomes negative two, and five becomes negative five.
On the number line negative two sits to the right of negative five, so now the larger one is negative two.
\[ -2 > -5 \]
Multiplying by a negative mirrors the whole number line about zero. Mirroring swaps left and right, so it must swap the direction of the comparison.
Explain it
Discussion prompt
Explain Why the flip has to happen to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Start with something obviously true: two is less than five.
Picture it
Figure (svg): Number line with an open circle at negative five and shading extending to the right, showing all numbers greater than negative five
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The variable term is negative, which is your warning that a flip is coming at the last step.
Worked example
The variable term is negative, which is your warning that a flip is coming at the last step.
\[ 4 - 3x < 19 \]
Subtract 4 from both sides
Why: Subtraction never touches the direction of the symbol, so it still points the same way it did.
\[ -3x < 15 \]
Divide both sides by negative three and REVERSE the symbol
Why: Dividing by a negative mirrors both sides across zero, which swaps which one is larger. The less-than becomes a greater-than.
\[ x > -5 \]
Write the interval and graph the solution set
Why: Every number strictly greater than negative five works. Open dot at negative five, because it is strictly greater, and shade to the right.
\[ (-5, \infty) \]
Figure (svg): Number line with an open circle at negative five and shading extending to the right, showing all numbers greater than negative five
Verify with a test point inside and one outside, in the ORIGINAL inequality
Why: At zero the left side is four, which is less than nineteen, so zero belongs. At negative six the left side is twenty-two, which is not less than nineteen, so it is correctly excluded.
\[ x = 0: \ 4 < 19 \ \checkmark \qquad x = -6: \ 4 + 18 = 22 < 19 \ \text{false} \]
Reverse engineer
Discussion prompt
Work backwards. The example finished here:
Verify with a test point inside and one outside, in the ORIGINAL inequality
What was it asked to do, and what must it have been given? Reconstruct the problem from its answer.
Hint: Every quantity in the result had to enter somewhere. Account for each one.
Answer:
The variable term is negative, which is your warning that a flip is coming at the last step.
Picture it
Animation
Shows: The one rule that surprises people — a rendered Manim animation.
Rendered with Manim.
Takeaway: Multiplying by a negative reverses order on the number line.
Anomaly
Predict first
A student writes this, and it looks reasonable:
One clean division, and the symbol left exactly as it was written.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The arithmetic is right - twenty divided by negative five is negative four - but nothing was done about the direction.
The same division, with the direction reversed because the divisor is negative.
Why: The arithmetic is right - twenty divided by negative five is negative four - but nothing was done about the direction.
Trap
One clean division, and the symbol left exactly as it was written.
\[ -5x \ge 20 \]
Divide both sides by negative five and keep the symbol as it is
Why: The arithmetic is right - twenty divided by negative five is negative four - but nothing was done about the direction.
\[ x \ge -4 \qquad (\text{wrong}) \]
Test zero, which this answer claims is a solution
Why: Zero is greater than negative four, so this answer includes it. But negative five times zero is zero, and zero is not at least twenty. The answer is the wrong half of the line.
\[ -5(0) = 0 \ge 20 \ \text{is false} \]
The same division, with the direction reversed because the divisor is negative.
\[ -5x \ge 20 \]
Divide both sides by negative five and REVERSE the symbol
Why: Dividing by a negative mirrors the line, so at-least becomes at-most. The number is the same; the direction is not.
\[ x \le -4, \qquad (-\infty, -4] \]
Test a point inside and the boundary itself
Why: At negative five, negative five times negative five is twenty-five, which is at least twenty. At the boundary the left side is exactly twenty, so the filled dot is right.
\[ -5(-5) = 25 \ge 20 \ \checkmark, \qquad -5(-4) = 20 \ge 20 \ \checkmark \]
Notation
Annotate
From Trap: forgetting to flip the symbol — read this one piece at a time. What is each part doing?
On: \( -5(-5) = 25 \ge 20 \ \checkmark, \qquad -5(-4) = 20 \ge 20 \ \checkmark \)
Prediction
Predict first
Solve -2x + 5 > 11 and give the solution in interval notation.
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: (-infinity, -3)
Why: Subtract 5 from both sides to get -2x > 6, then divide by -2 and reverse the symbol, giving x < -3, which is (-infinity, -3). Verify at x = -4: -2(-4) + 5 = 13, and 13 is greater than 11.
Check
Watch the last division carefully, and write the answer in interval notation.
Check your understanding
Solve -2x + 5 > 11 and give the solution in interval notation.
Answer: A
Why: Subtract 5 from both sides to get -2x > 6, then divide by -2 and reverse the symbol, giving x < -3, which is (-infinity, -3). Verify at x = -4: -2(-4) + 5 = 13, and 13 is greater than 11.
Section
Part 7
Concept
Figure (svg): Two shaded rays stacked above a number line, one starting at an open circle at negative two going right and one ending at a filled dot at four, with their overlap from negative two to four highlighted on the axis
A compound inequality joins two inequalities with the word and or the word or. That little word decides the whole answer.
With and, a number must satisfy both parts to belong. On the picture that is the overlap of the two shadings.
\[ x > -2 \quad \text{and} \quad x \le 4 \qquad \longleftrightarrow \qquad (-2, 4] \]
If the two shadings never overlap, the and answer is the empty set. No number can be in two places that do not touch.
Analogy
Discussion prompt
Explain And means both at once by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
A compound inequality joins two inequalities with the word and or the word or. That little word decides the whole answer.
Concept
Figure (svg): Number line with shading left of an open circle at negative two and shading right of a filled dot at three, with an unshaded gap between them
With or, a number only has to satisfy one of the two parts. On the picture that is everything either shading covers.
\[ x < -2 \quad \text{or} \quad x \ge 3 \qquad \longleftrightarrow \qquad (-\infty, -2) \cup [3, \infty) \]
The cup-shaped symbol between the two intervals is the union. It says take both pieces, and it is what you write when the pieces do not touch.
An or answer is usually bigger than either piece alone. An and answer is usually smaller. That is the whole difference between them.
Counterexample
Discussion prompt
With or, a number only has to satisfy one of the two parts. On the picture that is everything either shading covers.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The cup-shaped symbol between the two intervals is the union. It says take both pieces, and it is what you write when the pieces do not touch.
Picture it
Figure (svg): Number line shaded between an open circle at negative two and a filled dot at four
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
When one expression is trapped between two numbers, write it as a single three-part inequality and work on all three parts at once.
Worked example
When one expression is trapped between two numbers, write it as a single three-part inequality and work on all three parts at once.
\[ -7 < 2x - 3 \le 5 \]
Add 3 to all THREE parts
Why: Whatever you do to the middle you must do to both outer parts, or the middle is no longer trapped between the same two bounds.
\[ -4 < 2x \le 8 \]
Divide all three parts by 2
Why: Two is positive, so neither symbol flips. The middle is now the bare variable, which is the goal.
\[ -2 < x \le 4 \]
Write the interval and graph the solution set
Why: Open dot at negative two because that end is strict, filled dot at four because that end allows equality. This is an and: only the overlap.
\[ (-2, 4] \]
Figure (svg): Number line shaded between an open circle at negative two and a filled dot at four
Verify one interior point and both endpoints in the ORIGINAL inequality
Why: At zero the middle is negative three, which sits between negative seven and five. At four the middle is exactly five, which is allowed. At negative two the middle is negative seven, which is not strictly greater than negative seven, so that end is correctly open.
| x | middle value | between -7 and 5? | in the answer? |
|---|---|---|---|
| 0 | -3 | yes | yes |
| 4 | 5 | yes, since 5 is allowed | yes |
| -2 | -7 | no, -7 is not greater than -7 | no |
Comparison
Comparison matrix
From The three-part form: refill the in the answer? column from what you know. The rest of the table is as it appeared.
| x | middle value | between -7 and 5? | in the answer? |
|---|---|---|---|
| 0 | -3 | yes | yes |
| 4 | 5 | yes, since 5 is allowed | yes |
| -2 | -7 | no, -7 is not greater than -7 | no |
Picture it
Figure (svg): Number line with shading left of an open circle at negative two and shading right of a filled dot at three
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Two separate inequalities joined by the word or. Solve each one on its own, then combine them at the very end.
Worked example
Two separate inequalities joined by the word or. Solve each one on its own, then combine them at the very end.
\[ 2x + 1 < -3 \qquad \text{or} \qquad 3x - 2 \ge 7 \]
Solve the left inequality
Why: Subtract one from both sides, then divide both sides by two. Two is positive, so nothing flips.
\[ 2x < -4 \quad \Rightarrow \quad x < -2 \]
Solve the right inequality
Why: Add two to both sides, then divide both sides by three - again a positive divisor, so the symbol keeps its direction.
\[ 3x \ge 9 \quad \Rightarrow \quad x \ge 3 \]
Join the two pieces with a union, because the word was or
Why: The pieces do not overlap, so there is no single interval that names them. Two rays joined by the union symbol is the answer.
\[ (-\infty, -2) \cup [3, \infty) \]
Figure (svg): Number line with shading left of an open circle at negative two and shading right of a filled dot at three
Verify one point from each piece and one from the gap, in the ORIGINAL inequalities
Why: At negative five the first part gives negative nine, which is less than negative three, so it qualifies. At four the second part gives ten, which is at least seven. At zero the parts give one and negative two, and neither qualifies, so zero is correctly left out.
| x | left part | right part | in the solution? |
|---|---|---|---|
| -5 | -9 is less than -3: true | -17 is at least 7: false | yes, one part is enough |
| 4 | 9 is less than -3: false | 10 is at least 7: true | yes, one part is enough |
| 0 | 1 is less than -3: false | -2 is at least 7: false | no, neither part holds |
Comparison
Comparison matrix
From An or compound inequality: refill the left part column from what you know. The rest of the table is as it appeared.
| x | left part | right part | in the solution? |
|---|---|---|---|
| -5 | -9 is less than -3: true | -17 is at least 7: false | yes, one part is enough |
| 4 | 9 is less than -3: false | 10 is at least 7: true | yes, one part is enough |
| 0 | 1 is less than -3: false | -2 is at least 7: false | no, neither part holds |
Picture it
Animation
Shows: And versus or — a rendered Manim animation.
Rendered with Manim.
Takeaway: The connective decides the shape of the answer set.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The two solved pieces, and the urge to write them as one tidy sandwich.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The compact form used for and problems looks tidier, so it gets reused here where it does not belong.
The same two pieces, joined the way the word or requires.
Why: The compact form used for and problems looks tidier, so it gets reused here where it does not belong.
Trap
The two solved pieces, and the urge to write them as one tidy sandwich.
\[ x < -2 \qquad \text{or} \qquad x \ge 3 \]
Write it as a single three-part inequality
Why: The compact form used for and problems looks tidier, so it gets reused here where it does not belong.
\[ 3 \le x < -2 \qquad (\text{wrong}) \]
Read what that statement actually claims
Why: It demands a number that is at least three AND less than negative two at the same moment. No number is. The written answer is the empty set, while the real answer is most of the number line.
\[ \{\, x \mid x \ge 3 \text{ and } x < -2 \,\} = \varnothing \]
The same two pieces, joined the way the word or requires.
\[ x < -2 \qquad \text{or} \qquad x \ge 3 \]
Write the two intervals joined by the union symbol
Why: The three-part form is legal only when one expression is trapped between two bounds, which is an and. Two disjoint pieces need a union.
\[ (-\infty, -2) \cup [3, \infty) \]
Test one number from each piece
Why: Negative five is less than negative two, and four is at least three, so both belong. The answer covers two whole rays, not nothing.
\[ x = -5 \ \checkmark, \qquad x = 4 \ \checkmark \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Ranking
Put in order
These are the steps of Pattern: solving any inequality, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Only step two is genuinely new compared with equations. Everything else you already knew before this deck started.
The test points at the end are the inequality version of substituting back, and they catch a missed flip instantly.
Real world
Discussion prompt
Outside this lesson: where does Linear Equations and Inequalities actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: solving any inequality is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck solves linear equations end to end. It starts with the balance-scale model, then covers clearing fractions and decimals, telling identities from contradictions, literal equations, applied linear models, and inequalities on the number line written in interval notation. It targets the forgotten inequality flip, multiplying only part of the equation by the LCD, calling a true collapsed statement "no solution", and writing an "or" union as an "and" intersection.
Elimination
Eliminate the wrong options
Solve the compound inequality -1 <= 3x + 2 < 8 and write the solution in interval notation.
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Subtract 2 from all three parts to get -3 <= 3x < 6, then divide all three by 3 to get -1 <= x < 2, which is [-1, 2). Verify at x = -1: 3(-1) + 2 = -1, and the left end allows equality, so -1 belongs.
Check
Apply each move to all three parts, then be careful about which endpoint gets a bracket.
Check your understanding
Solve the compound inequality -1 <= 3x + 2 < 8 and write the solution in interval notation.
Answer: A
Why: Subtract 2 from all three parts to get -3 <= 3x < 6, then divide all three by 3 to get -1 <= x < 2, which is [-1, 2). Verify at x = -1: 3(-1) + 2 = -1, and the left end allows equality, so -1 belongs.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — What Solving Actually Means · Solving Linear Equations · Fractions, Decimals, and Special Cases · Literal Equations · Linear Models in the Wild · Inequalities on the Number Line. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
One toolkit handles all of it: clean up each side, keep the scale balanced, then read what the equation tells you at the end.
| what you see at the end | what to write |
|---|---|
| a variable survives | the single number |
| variables cancel, statement true | all real numbers |
| variables cancel, statement false | no solution |
| you divided by a negative | reverse the inequality symbol |
| the pieces are joined by and | the overlap of the two shadings |
| the pieces are joined by or | the union of the two pieces |
Whatever you solved, finish by substituting back into the original problem. That one habit is worth more than every shortcut in this deck.
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