Equations, solutions and equivalent equations; the four properties of equality; isolating a variable on one side and then on both; clearing brackets and fractions; and writing an equation from a verbal model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Solve Linear Equations
Objectives
Five outcomes. The fifth is the one every word problem in this book depends on.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-25 — the lesson these objectives are drawn from
Warm-up
Lesson 1.1 gave you the properties of real numbers. This lesson adds the four that apply to equations.
Discussion prompt
If you know that a equals b, name two different things you could do to both sides that would keep the statement true. Is there anything you could do to both sides that would break it?
Hint: One of the four legal moves carries a restriction. Which, and what is it?
Answer:
\[ a = b \;\Longrightarrow\; a + c = b + c \qquad a - c = b - c \]
\[ a = b, \; c \neq 0 \;\Longrightarrow\; ac = bc \qquad \tfrac{a}{c} = \tfrac{b}{c} \]
Multiplying both sides by zero breaks it — not by making a false statement, but by making a useless one. Zero equals zero is true for every value of x, so the information about which x you wanted is destroyed. That is why the multiplication and division properties both say nonzero.
Concept
An equation claims that two expressions have the same value. A solution is a number that makes the claim true when it is substituted. Solving means rewriting the equation into simpler equations that have exactly the same solutions, until the answer is unavoidable.
equivalent equations — Two equations that have the same solution or solutions. Each of the four properties of equality turns an equation into an equivalent one.
\[ ax + b = 0, \quad a \neq 0 \]
That is the definition of a linear equation in one variable: it can be written in that form, with a not zero.
Figure (svg): A balance scale with four fifths of x and 8 in the left pan and 20 in the right pan, level
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-18
Section
Section 1
Concept
You may add, subtract, multiply or divide — but only if you do it to both sides, and only by a nonzero number in the multiply and divide cases. Every legitimate line of every solution in this course is one of these four moves.
solution — A number that, when substituted for the variable, turns the equation into a true statement.
Because each move produces an equivalent equation, no solution is lost and none is invented along the way. That guarantee is what makes the final line trustworthy.
Figure (svg): The four properties of equality as a table: add, subtract, multiply and divide the same number on each side
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-18 — Transformations That Produce Equivalent Equations
Picture it
The balance is not a metaphor for the answer; it is a metaphor for the rule.
Figure (svg): A balance scale with four fifths of x and 8 in the left pan and 20 in the right pan, level
Taking eight off only the left pan does not find x — it just breaks the scale, and with it the claim you were trying to preserve.
Worked example
Before solving anything, be clear on what solving is looking for.
\[ \text{Is } x = 15 \text{ a solution of } \tfrac{4}{5}x + 8 = 20? \]
Substitute 15 for x in the original equation
Why: Checking always uses the original, never a rewritten line, because a rewritten line could itself contain the error.
Evaluate the left side
Why: Four fifths of fifteen is twelve, and twelve plus eight is twenty.
\[ 12 + 8 = 20 \]
Compare the two sides
Why: Both sides read twenty, so the statement is true.
\[ 20 = 20 \]
Figure (svg): The solution to Worked example check a proposed solution shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{4}{5}(15) + 8 = 20 \quad \checkmark \]
Verify: try a nearby value to see that it fails
Why: At x equal to 14 the left side is 11.2 plus 8, which is 19.2, not 20. A solution is a single value, not a range, so a neighbour failing is exactly what should happen and confirms the check meant something.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-18
Matching
Each line changed the equation in exactly one way.
Match the pairs
Why: Name the property by what was done, not by what disappeared. Nine went away because nine was subtracted; the coefficient four went away because both sides were divided by four. The two moves that undo a subtraction or a division are addition and multiplication respectively, which is why the last two are the reverses of the first two.
Worked example
The same solution, annotated. Being able to name the move is what makes an unfamiliar equation approachable.
\[ \text{Solve } \tfrac{4}{5}x + 8 = 20, \text{ naming each property used.} \]
Subtract 8 from each side
Why: Subtraction property of equality. The 8 is the outermost thing wrapped around x, so it comes off first.
\[ (\frac{4}{5}) x = 12 \]
Multiply each side by 5/4
Why: Multiplication property of equality, using the reciprocal of the coefficient. Multiplying by the reciprocal is cleaner than dividing by a fraction.
\[ x = (\frac{5}{4}) (12) \]
Simplify the right side
Why: Five fourths of twelve is fifteen.
\[ x = 15 \]
Figure (svg): The solution to Worked example name the property behind each move shown as a ladder of expressions, one row per algebraic move
\[ x = 15 \]
Verify: substitute 15 into the original equation
Why: Four fifths of fifteen is twelve, and twelve plus eight is twenty, which matches the right side exactly. Because every step was one of the four properties, this check confirms arithmetic rather than legality.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-18
Trap
\[ 2x + 6 = 14 \]
Move the 6 across and change its sign
Why: The 6 is described as moving, which hides the fact that a move must happen on both sides.
\[ 2x = 14 \quad \text{(the 6 vanished, the 14 was never touched)} \]
\[ x = 7 \quad \text{(wrong)} \]
\[ 2x + 6 = 14 \]
Subtract 6 from BOTH sides
Why: Subtraction property of equality. What looks like moving a term is really subtracting it twice, once on each side.
\[ 2x = 8 \]
\[ x = 4 \]
The shortcut phrase move it across the equals sign and flip the sign is fine once you know what it stands for — and dangerous before then, because it lets you forget the other side.
Sorting
Each of these is applied to the equation 3x = 12. Which preserve the solution?
Sort into buckets
Sort each proposed move.
Notice that legal and useful are separate questions. The properties tell you what you may do; strategy tells you what is worth doing.
Explain it to yourself
Two of the four properties carry a restriction and two do not.
\[ a = b, \; c \neq 0 \;\Longrightarrow\; ac = bc \]
Discussion prompt
Explain why the addition and subtraction properties need no restriction on c, while multiplication and division do. What exactly goes wrong if c is zero in each of those two cases?
Hint: Try multiplying x = 5 by zero and see what you are left with.
Answer:
Adding or subtracting any number, including zero, is reversible: you can always add it back. Multiplying by zero is not reversible, because every equation becomes zero equals zero and there is no way back.
\[ x = 5 \;\xrightarrow{\times 0}\; 0 = 0 \]
Dividing by zero is worse still: it is not an operation at all, since no number multiplied by zero gives a nonzero result. So one case destroys information and the other is undefined, and the restriction rules out both.
Prediction
Commit before you reason.
Predict first
In 4x + 9 = 21, which move should come first and why?
Correct: Subtract 9 first — but the third option is not wrong about legality, only about tidiness.
\[ 4x + 9 = 21 \;\longrightarrow\; 4x = 12 \;\longrightarrow\; x = 3 \]
\[ 4x + 9 = 21 \;\longrightarrow\; x + \tfrac{9}{4} = \tfrac{21}{4} \;\longrightarrow\; x = 3 \]
Why: Both orders are legal and both reach x equal to 3. Subtracting 9 first keeps whole numbers throughout. Dividing by 4 first gives x plus 9 over 4 equals 21 over 4, which is correct but drags fractions through the rest of the work. Undoing the outermost operation first is a strategy, not a law, and knowing it is a strategy is what lets you break it deliberately when fractions are already unavoidable.
Section
Section 2
Concept
Look at what has been done to the variable, in order, then undo those operations from the outside in. It is the same reasoning as taking off a coat before a jumper.
\[ \tfrac{4}{5}x + 8 = 20 \]
When the coefficient is a fraction, multiply by its reciprocal rather than dividing. Dividing by four fifths and multiplying by five fourths are the same move, and one of them is much easier to write.
Figure (svg): Two columns showing the order operations were applied to x and the reverse order used to undo them
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-18
Picture it
Read the left column downwards and the right column downwards; they are mirror images.
Figure (svg): Two columns showing the order operations were applied to x and the reverse order used to undo them
This is why the constant comes off before the coefficient: the coefficient was applied first, so it must be undone last.
Worked example
Example 1. Two ways are shown in the textbook; here is the one that keeps the numbers small.
\[ \text{Solve } \tfrac{4}{5}x + 8 = 20. \]
Subtract 8 from each side
Why: Eight is the last thing that was done to the variable expression, so it is the first thing undone.
\[ (\frac{4}{5}) x = 12 \]
Multiply each side by 5/4
Why: Five fourths is the reciprocal of four fifths, and a number times its reciprocal is 1, leaving x alone.
\[ x = (\frac{5}{4}) (12) \]
Simplify
Why: Twelve divided by four is three, and three times five is fifteen.
\[ x = 15 \]
Figure (svg): The solution to Worked example a fractional coefficient shown as a ladder of expressions, one row per algebraic move
\[ x = 15 \]
Verify: substitute into the original
Why: Four fifths of fifteen is twelve; twelve plus eight is twenty, which is the right side. Note also the alternative route: multiplying the whole original equation by 5 gives 4x plus 40 equals 100, then 4x equals 60, then x equals 15 — the same answer by a different first move, which is the strongest kind of check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-18
Ranking
Solving 3 minus five thirds c equals negative two, from Exercise 17.
Put in order
Why: The constant on the same side as the variable comes off first, then the coefficient is undone by multiplying by its reciprocal — and because the coefficient is negative, the reciprocal is negative too, which is what turns negative five into positive three. The check is a genuine step, not decoration: it is the only line that uses the original equation.
Worked example
Guided Practice 9. Two fractions on one side, and one clean way through.
\[ \text{Solve } \tfrac{1}{4}x + \tfrac{2}{5}x = 39. \]
Notice both terms are like terms
Why: Both are a number times x, so they combine before anything else needs to happen.
\[ (\frac{1}{4} + \frac{2}{5}) x = 39 \]
Add the coefficients over a common denominator
Why: Twenty is the least common denominator, so one quarter is five twentieths and two fifths is eight twentieths.
\[ (\frac{13}{20}) x = 39 \]
Multiply each side by 20/13
Why: The reciprocal of thirteen twentieths, which leaves x alone.
\[ x = (\frac{20}{13}) (39) \]
Simplify
Why: Thirty-nine divided by thirteen is three, and three times twenty is sixty.
\[ x = 60 \]
Figure (svg): The solution to Worked example two fractional terms shown as a ladder of expressions, one row per algebraic move
\[ x = 60 \]
Verify: substitute into the original
Why: One quarter of sixty is fifteen, and two fifths of sixty is twenty-four. Fifteen plus twenty-four is thirty-nine, matching the right side. Choosing sixty was not luck: the least common denominator of 4 and 5 divides it, which is why the arithmetic came out whole.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 21-21
Trap
\[ 4x + 12 = 28 \]
Divide by 4 to remove the coefficient
Why: The division is applied to the term the student was looking at, not to the whole side.
\[ x + 12 = 7 \quad \text{(wrong: the 12 was not divided)} \]
\[ x = -5 \]
\[ 4x + 12 = 28 \]
Divide EVERY term on both sides by 4
Why: The division property applies to the whole side, and a side is a single quantity even when it is written as a sum.
\[ x + 3 = 7 \]
\[ x = 4 \]
Or subtract 12 first and avoid the issue entirely — which is why undoing the outermost operation first is the standard advice.
Fill the middle
Exercise 16. The middle step has been erased.
Fill in the blanks
\tfrac\frac{3}{11}b = 0___b + 5 = 5 \;\longrightarrow\; ___ \;\longrightarrow\; b = 0
Why: Subtracting five from each side leaves three elevenths of b equal to zero. Multiplying by the reciprocal eleven thirds then gives b equal to zero, because any number times zero is zero. This one is worth doing because zero is a perfectly good solution and students often distrust it, expecting solutions to be interesting.
Elimination
Exercise 19, as it appears on the page.
Eliminate the wrong options
What is the solution of 4x - 7 = -15?
Survives elimination: B
Why: Adding 7 to each side gives 4x equal to -8, and dividing by 4 gives x equal to -2. Checking: 4 times -2 is -8, minus 7 is -15, which matches. Every wrong option here can be eliminated by substitution alone in under ten seconds, which is why checking is worth the time on a multiple-choice question.
Edge cases
The division property says you may divide by any nonzero number.
Discussion prompt
Solve ax = 12 for x, in general. Then describe what happens to the solution as a gets closer and closer to zero, and say what happens at a equal to zero itself. Why is this the same restriction you met in the property table?
Hint: Try a equal to 1, then 0.1, then 0.01.
Answer:
\[ ax = 12 \;\Longrightarrow\; x = \tfrac{12}{a}, \quad a \neq 0 \]
As a shrinks toward zero the solution grows without bound: at a equal to one tenth, x is 120; at one hundredth, 1200. At a equal to zero the equation becomes 0 equals 12, which is false for every x, so there is no solution at all.
That is the same restriction from the division property, seen from the other side: division by zero is not merely awkward, it corresponds to an equation with no solution.
Section
Section 3
Concept
When the variable appears on both sides, one extra step comes first: use the addition or subtraction property to gather every variable term on one side and every constant on the other. After that it is an equation you already know how to solve.
\[ 4p + 15 = 7p - 3 \]
Which side you send the variables to is free. Sending them to the side that ends up with a positive coefficient saves you a sign to worry about later.
Figure (svg): A number line style diagram showing the variable terms collected on the left and the constants on the right
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 19-19
Picture it
The equation has variable terms and constant terms scattered across both sides.
Figure (svg): A number line style diagram showing the variable terms collected on the left and the constants on the right
Subtracting 4p sends the variables right and leaves 3p; subtracting 7p sends them left and leaves -3p. Both give p equal to 6, but only one of them avoids dividing by a negative.
Worked example
Example 3, which the textbook presents as a multiple-choice question.
\[ \text{Solve } 4p + 15 = 7p - 3. \]
Subtract 4p from each side
Why: Four p is the smaller variable term, so removing it leaves a positive coefficient on the right.
\[ 15 = 3 p - 3 \]
Add 3 to each side
Why: Now the constants are gathered on the left and the variable term stands alone on the right.
\[ 18 = 3 p \]
Divide each side by 3
Why: The coefficient is undone last, as always.
\[ 6 = p \]
Figure (svg): The solution to Worked example variables on both sides shown as a ladder of expressions, one row per algebraic move
\[ p = 6 \]
Verify: substitute 6 into the original
Why: The left side is 4 times 6 plus 15, which is 24 plus 15, or 39. The right side is 7 times 6 minus 3, which is 42 minus 3, also 39. Both sides agree at 39, so 6 is genuinely a solution and not just the end of a chain of arithmetic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 19-19
Discrimination
Do not solve these. Just say which move avoids a negative coefficient.
Sort into buckets
Sort each equation by the move that leaves a positive coefficient.
Worked example
Guided Practice 5, solved by sending the variables the awkward way first, to see what it costs.
\[ \text{Solve } -2x + 9 = 2x - 7. \]
Subtract 2x from each side
Why: This sends the variables left, where the coefficient will end up negative — deliberately, to see the cost.
\[ -4 x + 9 = -7 \]
Subtract 9 from each side
Why: The constants gather on the right.
\[ -4 x = -16 \]
Divide each side by -4
Why: Dividing a negative by a negative gives a positive, but this is one more sign decision than the other route required.
\[ x = 4 \]
Compare with the other route
Why: Adding 2x to each side instead gives 9 equals 4x minus 7, then 16 equals 4x, then x equals 4 — the same answer with no negative coefficient anywhere.
\[ x = 4\text{ either way} \]
Figure (svg): The solution to Worked example the other direction, on purpose shown as a ladder of expressions, one row per algebraic move
\[ x = 4 \]
Verify: substitute 4 into the original
Why: The left side is -8 plus 9, which is 1. The right side is 8 minus 7, which is also 1. Both routes reached the same value and the check confirms it, so the choice of direction really is only about convenience.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 21-21
Error analysis
Exercise 24 as a student solved it. The answer is wrong and one line is responsible.
Annotate
On: \( \begin{aligned} 3y + 7 &= y - 3 \\ 3y + 7 - y &= y - 3 - y \\ 2y + 7 &= -3 \\ 2y &= 4 \\ y &= 2 \end{aligned} \)
The tell is that the wrong answer is positive while the original equation has a bigger constant on the left, which forces the variable to be negative. A quick sign estimate before solving catches this class of error.
Prediction
Before solving, estimate. This is a habit that catches sign errors for free.
Predict first
In 2z - 3 = 6z + 25, will z be positive or negative?
Correct: Negative — z equals -7.
\[ 2z - 3 = 6z + 25 \;\longrightarrow\; -28 = 4z \;\longrightarrow\; z = -7 \]
Why: The right side starts 28 higher than the left when z is zero, and each unit increase in z raises the right side faster than the left, since 6 is bigger than 2. So increasing z only widens the gap; the sides can only meet by going the other way, which means z is negative. Solving confirms it: subtracting 2z gives -3 = 4z + 25, then -28 = 4z, so z = -7.
Explain it
A classmate insists there is a rule about which side the variables must go to.
Discussion prompt
Explain to them why both directions are legal, why one is usually easier, and how they could convince themselves of this in thirty seconds with a single equation.
Hint: The thirty-second demonstration is to solve one equation twice.
Answer:
Both directions use the same property — subtraction of equals from equals — so both produce equivalent equations and both must give the same solution. There is no rule, only a preference.
\[ 4p + 15 = 7p - 3 \;\longrightarrow\; 18 = 3p \quad \text{or} \quad -3p + 15 = -3 \]
The demonstration is exactly that: solve one equation both ways and watch the same answer appear. The right-hand route needs a division by negative three; the left-hand one does not.
Comparison
Fill the blanks. Both columns must land on the same solution.
Comparison matrix
| Step | Send variables right | Send variables left |
|---|---|---|
| Start | 4p + 15 = 7p - 3 | 4p + 15 = 7p - 3 |
| Collect | 15 = 3p - 3 | -3p + 15 = -3 |
| Constants | 18 = 3p | -3p = -18 |
| Divide | p = 6 | p = 6 |
The left column never divides by a negative. That is the entire advantage, and it is worth having on a long problem.
Section
Section 4
Concept
If there are brackets, distribute. If there are fractions, multiply every term by the least common denominator. Both moves turn an unfamiliar equation into the familiar both-sides case, and both are single applications of properties you already have.
\[ \tfrac{2}{3}x + \tfrac{5}{6} = x - \tfrac{1}{2} \]
Clearing fractions is the multiplication property of equality applied to the whole equation at once, which is why every term is multiplied, including the ones that have no fraction in them.
Figure (svg): An equation with fractions being multiplied through by the least common denominator to remove every fraction at once
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 20-21
Picture it
Denominators of 3, 6 and 2, so the least common denominator is 6.
Figure (svg): An equation with fractions being multiplied through by the least common denominator to remove every fraction at once
Multiplying by 6 is legal because 6 is not zero. The reason to use the LEAST common denominator rather than any common one is simply that it keeps the resulting numbers small.
Worked example
Example 4. Distribute first, then it is an ordinary both-sides equation.
\[ \text{Solve } 3(5x - 8) = -2(-x + 7) - 12x. \]
Distribute on each side
Why: Three across the left bracket; negative two across the right bracket, including across its negative first term.
\[ 15 x - 24 = 2 x - 14 - 12 x \]
Combine like terms on the right
Why: Two x minus twelve x is negative ten x.
\[ 15 x - 24 = -10 x - 14 \]
Add 10x to each side
Why: Sending the variables left leaves a positive coefficient there.
\[ 25 x - 24 = -14 \]
Add 24 to each side
Why: The constants gather on the right.
\[ 25 x = 10 \]
Divide each side by 25 and simplify
Why: Ten over twenty-five reduces to two fifths.
\[ x = \frac{2}{5} \]
Figure (svg): The solution to Worked example brackets on both sides shown as a ladder of expressions, one row per algebraic move
\[ x = \tfrac{2}{5} \]
Verify: substitute two fifths into the original
Why: The left side is 3 times the quantity 2 minus 8, which is 3 times -6, or -18. The right side is -2 times the quantity -2/5 plus 7, minus 24/5, which is -2 times 33/5 minus 24/5, or -66/5 minus 24/5, which is -90/5, or -18. Both sides read -18.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 20-20
Step zero
Look at the equation. Do not solve it.
\[ 3(x + 2) = 5(x + 4) \]
Discussion prompt
In plain English, list the moves you would make and in what order, and say which side you would send the variables to and why. Predict the sign of the answer before you do any arithmetic.
Hint: The right side grows faster as x grows. What does that tell you about where the two sides can be equal?
Answer:
Distribute both brackets, collect variables on the right because 5x is larger than 3x, collect constants on the left, then divide. The right side grows faster and already starts higher at x equal to zero, so the sides can only meet at a negative x.
\[ 3x + 6 = 5x + 20 \;\longrightarrow\; -14 = 2x \;\longrightarrow\; x = -7 \]
Worked example
Guided Practice 10. The alternative is to work with fractions throughout, which is legal and unpleasant.
\[ \text{Solve } \tfrac{2}{3}x + \tfrac{5}{6} = x - \tfrac{1}{2}. \]
Find the least common denominator of 3, 6 and 2
Why: Six is divisible by all three, and nothing smaller is.
\[ LCD = 6 \]
Multiply every term on both sides by 6
Why: The multiplication property of equality, applied to the whole equation. Every term is multiplied, including the bare x.
\[ 4 x + 5 = 6 x - 3 \]
Subtract 4x from each side
Why: Sending the variables right leaves a positive coefficient.
\[ 5 = 2 x - 3 \]
Add 3 to each side
Why: The constants gather on the left.
\[ 8 = 2 x \]
Divide each side by 2
Why: The coefficient comes off last.
\[ 4 = x \]
Figure (svg): The solution to Worked example clear the fractions shown as a ladder of expressions, one row per algebraic move
\[ x = 4 \]
Verify: substitute 4 into the ORIGINAL fractional equation
Why: The left side is two thirds of four plus five sixths, which is eight thirds plus five sixths, or sixteen sixths plus five sixths, which is twenty-one sixths, or 3.5. The right side is 4 minus one half, which is also 3.5. Checking against the original rather than the cleared version is what confirms the clearing step itself.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 21-21
Trap
\[ \tfrac{2}{3}x + \tfrac{5}{6} = x - \tfrac{1}{2} \]
Multiply the terms that have fractions by 6, leaving x alone
Why: The x looks like it has nothing to clear, so it is skipped.
\[ 4x + 5 = x - 3 \quad \text{(wrong: the x was not multiplied)} \]
\[ 3x = -8, \; x = -\tfrac{8}{3} \]
\[ \tfrac{2}{3}x + \tfrac{5}{6} = x - \tfrac{1}{2} \]
Multiply EVERY term on both sides by 6
Why: The multiplication property applies to each side as a whole, so a term with no fraction is multiplied like any other.
\[ 4x + 5 = 6x - 3 \]
\[ 8 = 2x, \; x = 4 \]
Substituting the wrong answer back into the original gives different values on the two sides, which is how this one is caught.
Fill the middle
Denominators 4 and 5. Fill in the cleared equation.
Fill in the blanks
\tfrac5x + 8x = 780___x + \tfrac______x = 39 \;\xrightarrow___\; ___ \;\longrightarrow\; x = 60
Why: Twenty is the least common denominator of 4 and 5. One quarter of x times 20 is 5x, two fifths of x times 20 is 8x, and — the step that gets forgotten — 39 times 20 is 780. Combining gives 13x equal to 780, so x is 60. Forgetting to multiply the constant is the single most common error in clearing fractions.
Error analysis
A student solves an equation with brackets. One line is wrong.
Annotate
On: \( \begin{aligned} 3(5x - 8) &= -2(-x + 7) - 12x \\ 15x - 24 &= 2x + 14 - 12x \\ 15x - 24 &= -10x + 14 \\ 25x &= 38 \end{aligned} \)
When a multiplier is negative, every term inside the bracket changes sign - including the ones that were already positive.
Commit first
Answer, then rate your confidence honestly.
Predict first
When you multiply an equation through by the least common denominator, which terms must be multiplied?
Correct: Every term on both sides, fractions or not.
\[ 6\left(\tfrac{2}{3}x + \tfrac{5}{6}\right) = 6(x) - 6\left(\tfrac{1}{2}\right) \]
\[ 4x + 5 = 6x - 3 \]
Why: The multiplication property of equality multiplies each side as a single quantity, and multiplying a sum means multiplying every term in it — that is the distributive property. A term with no fraction is not exempt; it simply gets larger. Skipping the whole-number terms is the standard way this move goes wrong, and it produces an equation that is no longer equivalent to the original.
Section
Section 5
Concept
Most equations in real problems are not handed to you — you build them. Write the relationship in words with units attached, then replace each phrase with a number or a letter. Percentages become decimals before they enter.
verbal model — A word equation describing the situation, with each quantity labelled by its unit, written before any symbols.
The unit check is what tells you the equation is at least the right shape: dollars on the left, dollars on the right.
Figure (svg): A verbal model showing income equals wages plus percent for tips times food bills, with the equation underneath
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 19-19
Picture it
A waiter earns 30 dollars in wages plus 15 percent of the food bills, and takes home 105 dollars.
Figure (svg): A verbal model showing income equals wages plus percent for tips times food bills, with the equation underneath
Fifteen percent becomes 0.15 in the model, not in the middle of the solving. Converting early is what keeps the arithmetic honest.
Worked example
Example 2. The equation is the hard part; the solving is three lines.
\[ \text{Wages of } \$30 \text{ plus } 15\% \text{ of food bills } x \text{ gives } \$105. \text{ Find } x. \]
Write the verbal model in units
Why: Income equals wages plus tip rate times food bills. Dollars on both sides, so the equation is well formed.
\[ \text{income } =\text{ wages } +\text{ rate } \times\text{ bills} \]
Convert the percent to a decimal and write the equation
Why: Fifteen percent is fifteen hundredths, or 0.15. It has no unit, which is why it can multiply dollars and still give dollars.
\[ 105 = 30 + 0.15 x \]
Subtract 30 from each side
Why: The wages are the constant wrapped around the variable term.
\[ 75 = 0.15 x \]
Divide each side by 0.15
Why: The coefficient comes off last.
\[ 500 = x \]
Figure (svg): The solution to Worked example the waiter's food bills shown as a ladder of expressions, one row per algebraic move
\[ x = 500 \quad \text{dollars in food bills} \]
Verify: substitute back into the situation, not just the equation
Why: Fifteen percent of 500 dollars is 75 dollars in tips, and 75 plus the 30 dollars of wages is 105 dollars — the stated earnings. Checking against the words rather than the symbols also confirms the model was the right one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 19-19
Real world
A gym charges a 40 dollar joining fee plus 25 dollars a month. You have budgeted 340 dollars for the year.
Discussion prompt
Write the equation whose solution is the number of months you can afford, say what unit each part carries, and then say what you would do with the answer if it came out as a decimal.
Hint: Think about what a fractional month would mean in this situation.
Answer:
\[ 340 = 40 + 25m \;\Longrightarrow\; 300 = 25m \;\Longrightarrow\; m = 12 \]
Dollars on both sides: the fee is dollars, and dollars per month times months gives dollars. Here it comes out whole, but had it come out as 12.4 you would round down to 12, because you cannot buy part of a month. Rounding direction is decided by the situation, never by the arithmetic.
Worked example
Guided Practice 4. Same shape, bigger numbers, and a commission instead of tips.
\[ \text{Base salary } \$22\,000 \text{ plus } 4\% \text{ commission on sales } s \text{ gives } \$60\,000. \]
Write the verbal model
Why: Total earnings equals base salary plus commission rate times total sales.
\[ \text{earnings } =\text{ base } +\text{ rate } \times\text{ sales} \]
Write the equation with 4 percent as 0.04
Why: Four percent is four hundredths. Writing 4 instead of 0.04 is the classic error and inflates the answer a hundredfold.
\[ 60000 = 22000 + 0.04 s \]
Subtract 22000 from each side
Why: The base salary is guaranteed regardless of sales, so it comes off first.
\[ 38000 = 0.04 s \]
Divide each side by 0.04
Why: Dividing by four hundredths is the same as multiplying by twenty-five.
\[ 950000 = s \]
Figure (svg): The solution to Worked example the estate agent's target shown as a ladder of expressions, one row per algebraic move
\[ s = 950\,000 \quad \text{dollars of sales} \]
Verify: compute the commission directly
Why: Four percent of 950,000 dollars is 38,000 dollars, and 38,000 plus the 22,000 base is exactly 60,000 dollars. The size is also plausible: a 4 percent commission means sales must be roughly twenty-five times the commission needed, and 25 times 38,000 is 950,000.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 19-19
Error analysis
A student sets up the estate-agent problem and gets an answer that is obviously too small.
Annotate
On: \( 60\,000 = 22\,000 + 4s \;\Longrightarrow\; 38\,000 = 4s \;\Longrightarrow\; s = 9500 \)
Convert every percent to a decimal in the verbal model, before the equation exists. An answer that is off by a factor of a hundred is almost always this.
Matching
Each situation becomes exactly one of these equations.
Match the pairs
Why: Three of the four have the same shape: a fixed amount plus a rate times an unknown. The rates differ in kind — a percent, a percent, and dollars per month — but they enter identically. The fourth has no fixed amount at all, which is what makes it a pure rate problem rather than a start-plus-rate problem.
Estimation
The waiter earns 105 dollars, of which 30 is wages, with tips at 15 percent.
Predict first
Roughly how large must the food bills be?
Correct: About 500 dollars.
\[ 75 = 0.15x \;\Longrightarrow\; x = 500 \]
Why: Tips account for 75 dollars, and 15 percent is roughly one sixth, so the bills must be roughly six times 75, which is about 450 — close enough to identify the right option. The exact answer is 500. Estimating first means a misplaced decimal in the division by 0.15 cannot survive, and dividing by a number smaller than one is exactly where decimals slip.
Two truths and a lie
All three claim to model the waiter's evening.
Eliminate the wrong options
Two of these are correct models. Knock those out and keep the broken one.
Survives elimination: C
Why: The survivor is the broken one. Writing 15 rather than 0.15 claims the waiter receives fifteen dollars of tips for every dollar of food ordered. It gives x equal to 5, meaning five dollars of food bills produced seventy-five dollars in tips, which the situation immediately contradicts.
Comparison
Fill the blanks. The shape of the equation, not its difficulty, decides the opening move.
Comparison matrix
| Equation shape | First move | What it becomes |
|---|---|---|
| 4x + 9 = 21 | subtract the constant | 4x = 12 |
| (4/5)x + 8 = 20 | subtract 8, then multiply by the reciprocal | x = 15 |
| 4p + 15 = 7p - 3 | collect variables one side | 15 = 3p - 3 |
| 3(x + 2) = 5(x + 4) | distribute both brackets | 3x + 6 = 5x + 20 |
| (2/3)x + 5/6 = x - 1/2 | multiply every term by the LCD | 4x + 5 = 6x - 3 |
Every row after the first reduces to the first row within two moves. That is the whole strategy: make it look like something you have already solved.
Pattern
One routine solves every linear equation in this chapter.
The check is not optional politeness. It is the only step that uses the original equation, so it is the only step capable of catching an error made in step two.
OpenStax Algebra and Trigonometry 2e, §2.2 Linear Equations in One Variable §2.2
Check
Variable on both sides. Solve it on paper first.
Check your understanding
What is the solution of 7t - 5 = 3t + 11?
Answer: A
Why: Subtracting 3t from each side gives 4t - 5 = 11, adding 5 gives 4t = 16, and dividing by 4 gives t = 4. Checking: 28 - 5 is 23, and 12 + 11 is 23.
Check
Brackets on both sides. Distribute before anything else.
Check your understanding
What is the solution of 3(x + 2) = 5(x + 4)?
Answer: A
Why: Distributing gives 3x + 6 = 5x + 20. Subtracting 3x gives 6 = 2x + 20, subtracting 20 gives -14 = 2x, and dividing gives x = -7. Both sides then read -15.
Check
A word problem. Build the equation before you solve it.
Check your understanding
A real estate agent earns a base salary of 22,000 dollars plus 4 percent commission on total sales. How much must the agent sell to earn 60,000 dollars in a year?
Answer: A
Why: The equation is 60,000 = 22,000 + 0.04s. Subtracting gives 38,000 = 0.04s, and dividing by 0.04 gives 950,000. Four percent of 950,000 really is 38,000.
Real world
Two phone plans. Plan A charges 20 dollars a month plus 10 cents a minute. Plan B charges 35 dollars a month plus 4 cents a minute.
Discussion prompt
Write the equation that finds the number of minutes at which the two plans cost the same, solve it, and then say which plan is cheaper below that number and which is cheaper above it. How did you decide which is which without solving anything twice?
Hint: Compare the two monthly fees at zero minutes, then think about which cost rises faster.
Answer:
\[ 20 + 0.10m = 35 + 0.04m \;\Longrightarrow\; 0.06m = 15 \;\Longrightarrow\; m = 250 \]
At zero minutes Plan A is 15 dollars cheaper, so Plan A wins below 250 minutes. Plan A's cost rises faster — ten cents a minute against four — so it must lose above the crossing point. The starting gap and the rate of increase settle the direction without any second calculation.
This is exactly the reasoning you will formalise in Chapter 3 as solving a system, and again in Chapter 2 as comparing two lines by their slopes and intercepts.
Commit first
Answer, then rate your confidence honestly.
Predict first
You multiply both sides of an equation by 3 and then by 5. Could you have got the same result in one step, and does the order matter?
Correct: Yes, multiply by 15 — and the order does not matter.
\[ 5 \cdot (3 \cdot a) = (5 \cdot 3) \cdot a = 15a \]
Why: Multiplying by 3 and then by 5 multiplies each side by 15, because multiplication is associative and commutative. Those are exactly the properties from Lesson 1.1, now doing real work: they are what guarantee that combining two legal moves gives another legal move. Adding the multipliers to get 8 confuses repeated multiplication with repeated addition.
Explain it
They can solve 2x equals 10 but freeze when the variable appears twice.
Discussion prompt
In four sentences or fewer, explain what to do when the variable is on both sides, why it is allowed, and how they would know their answer is right without asking anyone.
Hint: The last part is one sentence about substitution.
Answer:
A usable answer: subtract the smaller variable term from both sides, which leaves the variable on one side only. It is allowed because subtracting the same thing from both sides is one of the four properties of equality, so the balance is preserved.
Then they check by putting their answer back into the original equation and confirming both sides give the same number. That check needs no teacher and no answer key, which is the point of it.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold on a quiz?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For a fractional coefficient, multiply by the reciprocal rather than dividing. For both sides, send the variables toward whichever side already has more of them. For a negative in front of a bracket, write the multiplier with its sign attached before you distribute. For a word problem, write the verbal model in units before any symbols, converting percents to decimals first. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
On paper. Fifteen minutes, and worth more than rereading.
Draw it
At the top of a page draw a balance scale and write the four properties of equality beside it, marking with a star the two that carry the nonzero restriction. Down the left margin write five equation shapes: variable on one side, fractional coefficient, variable on both sides, brackets, and fractions. Beside each write its first move in five words or fewer. In the middle of the page solve one equation of your own that needs at least three of those five moves, writing the property name beside every line. At the bottom write a short real situation with a fixed amount and a rate, turn it into an equation, solve it, and then write one sentence saying what the answer means in the situation rather than what the number is.
If that last sentence is hard to write, the algebra was fine but the model was never really yours. Go back to Section 5.
Recap
Five things, and the fifth is what the rest of this book keeps asking for.
| If you see | The first move is |
|---|---|
| A bracket | Distribute, sign included |
| A fraction anywhere | Multiply every term by the LCD |
| The variable twice | Collect toward the bigger coefficient |
| A fractional coefficient | Multiply by its reciprocal |
| A percent in the words | Convert it to a decimal first |
Lesson 1.4 takes the same four properties and applies them to a formula with several letters, where the goal is not a number but a rearranged equation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.3 Solve Linear Equations §1.3, pp. 18-25 — everything on these slides traces back here
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