Chapter 6 of Algebra 1: Concepts and Skills, built for a visual learner. Solution sets drawn on the number line, the flip rule explained as a reflection through zero, compound and and or inequalities as stacked shadings, absolute-value equations as two symmetric solutions, absolute-value inequalities as close-or-far, and shaded half-planes in two variables.
Subject: Algebra 1 · 62 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6
Solution sets instead of single answers, the sign that flips, compound statements, and shaded regions
Objectives
An equation asks which number works. An inequality asks which numbers work — usually infinitely many, and the answer is a picture.
Figure (svg): Two number lines side by side, one with an open circle for greater than and one with a filled circle for greater than or equal to
Section
Section 6.1
Concept
Adding or subtracting the same amount on both sides is just as legal for an inequality as for an equation, and the sign never changes.
Figure (svg): Two number lines side by side, one with an open circle for greater than and one with a filled circle for greater than or equal to
solution set — All the values that make the statement true. For an inequality this is usually a whole stretch of the number line, which is why the answer is drawn rather than written as a single number.
Worked example
Solve the inequality and graph its solution set.
\[ x - 3 < 5 \]
Add 3 to both sides
Why: Three is being subtracted from x, so adding three is the inverse. Adding never affects the direction of the sign.
\[ x < 8 \]
Draw the solution set with an open circle at 8
Why: The sign is strictly less than, so eight itself does not satisfy the statement and the circle stays hollow.
Figure (svg): A number line with an open circle at eight and everything to the left shaded
Verify: test one number from inside and one from outside
Why: Trying 7 gives 4, which is less than 5, so it works. Trying 9 gives 6, which is not less than 5, so it fails. The boundary really is at 8.
Prediction
Commit before drawing.
\[ x + 4 \ge 9 \]
Predict first
What is the solution set?
Correct: x is at least 5, shaded right with a solid circle.
Why: Subtracting four from both sides gives x greater than or equal to 5. The or-equal-to part means 5 itself is a solution, so the circle is filled in. Adding four instead of subtracting would give 13, which is the commonest arithmetic slip here.
Matching
Read the sign and the circle together — they carry different information.
Match the pairs
Why: The direction of shading comes from the inequality sign, and the type of circle comes from whether the phrase includes the boundary. At least and at most both include it; greater than and less than both exclude it. Those two everyday phrases carry the or-equal-to quietly, which is why they are so easy to misread.
Sorting
For the inequality shown, sort each candidate.
\[ x - 3 \le 4 \]
Sort into buckets
Which values satisfy the inequality?
Explain it to yourself
Multiplying by a negative flips the inequality. Adding a negative does not. Say why.
Discussion prompt
Why does adding the same number to both sides leave the direction of an inequality unchanged?
Hint: Picture both numbers sliding along the line by the same amount.
Answer:
Adding the same amount to both sides slides both numbers along the line by the same distance in the same direction. Their relative positions are untouched, so whichever was further left is still further left.
Multiplying by a negative is different in kind: it reflects the whole line through zero, so left and right genuinely swap. That is the real reason for the rule, and it is why only that one operation needs an exception.
Section
Section 6.2
Concept
This is the only rule in inequalities that differs from equations, and forgetting it is the single largest source of wrong answers in the chapter.
Figure (svg): A number line showing three and five, then the same two numbers after multiplying by negative one, with their order reversed
\[ 3 < 5 \quad \text{but} \quad -3 > -5 \]
Worked example
Solve the inequality and graph the solution set.
\[ -2x < 8 \]
Divide both sides by negative 2
Why: The coefficient is negative, so this is exactly the situation where the sign must reverse.
Flip the inequality sign as you divide
Why: Dividing by a negative reflects both sides through zero, which swaps which one is larger. The sign must follow.
\[ x > -4 \]
Figure (svg): A number line with an open circle at minus four and everything to the right shaded
Verify: test a number from the shaded region
Why: Trying 0 gives negative 2 times 0, which is 0, and 0 is less than 8, so 0 is a solution. Trying negative 10 gives 20, which is not less than 8, confirming the shaded side is correct.
Trap
Solve the inequality below.
\[ -3x \ge 12 \]
Divide both sides by negative 3 and keep the sign as it is
Why: Every other solving move so far has left the sign alone, so leaving it alone here feels consistent.
\[ x \ge -4 \]
Test x equal to 0, which this answer claims is a solution: negative 3 times 0 is 0, and 0 is not greater than or equal to 12. The answer set is exactly backwards.
Solve the same inequality, reversing the sign as you divide.
Divide by negative 3 and flip the sign
Why: Dividing by a negative reflects the line, so the direction of the comparison must reverse with it.
\[ x \le -4 \]
Test x equal to negative 10: negative 3 times negative 10 is 30, which is greater than 12. The solution set is the numbers at negative four and below.
Prediction
Do not solve. Just decide whether the sign reverses.
Predict first
In which of these does the inequality sign have to flip?
Correct: Only dividing both sides by negative 5.
Why: The flip is triggered by multiplying or dividing by a negative number, and by nothing else. Adding negative four looks like it should count because there is a minus sign, but adding merely slides both sides along the line without reversing their order.
Error analysis
This solution looks careful and is still wrong. Find where.
Annotate
On: \( \begin{aligned} 5 - 2x &> 11 \\ -2x &> 6 \\ x &> -3 \end{aligned} \)
Always test one number from your final region against the original inequality. It catches a missed flip instantly.
Ranking
For the inequality below, rank the moves.
\[ 4 - 3x \le 19 \]
Put in order
Why: Clear the constant first, giving negative 3x less than or equal to 15. Then divide by negative 3, and the flip happens as part of that same move, giving x greater than or equal to negative 5. Testing a value last is what confirms the flip was applied.
Section
Section 6.3
Concept
Distribute, combine, gather the variable, undo the constant, undo the coefficient. The only addition is checking for a negative divisor at the end.
\[ 2(x - 3) + 4 > 5x + 7 \]
Figure (svg): A checklist of the five solving moves with the flip rule highlighted at the last step
Worked example
Solve the inequality and graph the solution.
\[ 2(x - 3) + 4 > 5x + 7 \]
Distribute and simplify the left side
Why: Two times the bracket gives 2x minus 6, and adding 4 leaves 2x minus 2.
\[ 2x - 2 > 5x + 7 \]
Gather the variable terms
Why: Subtracting 2x from both sides keeps the right coefficient positive and clears the left.
\[ -2 > 3x + 7 \]
Undo the constant, then the coefficient
Why: Subtracting 7 gives negative 9 greater than 3x, and dividing by positive 3 needs no flip.
\[ -3 > x \quad \text{or equivalently} \quad x < -3 \]
Figure (svg): A number line with an open circle at minus three and everything left of it shaded
Verify: test x equal to negative 4 in the original
Why: The left gives 2 times negative 7 plus 4, which is negative 10. The right gives negative 20 plus 7, which is negative 13. Negative 10 is greater than negative 13, so the statement holds.
Fill the middle
Fill each blank.
Fill in the blanks
3x + 5 \le 2x - 4 \;\Longrightarrow\; x + 5 \le -4 \;\Longrightarrow\; x \le -9
Why: Subtracting 2x from both sides leaves x plus 5 on the left and negative 4 on the right. Subtracting 5 then gives x at most negative 9. No division by a negative happened anywhere, so the sign never flipped.
Discrimination
Do not solve. Just say whether a flip will be needed on the final step.
Sort into buckets
Sort each inequality by whether the sign will reverse.
Error analysis
A student solved correctly and then panicked. Diagnose the confusion.
Annotate
On: \( -7 > x \;\overset{?}{\Longrightarrow}\; x > -7 \)
Say it aloud: the sign always opens towards the larger side, whichever way round you write it.
Pattern
Every inequality in this chapter, including the absolute-value ones, ends up here.
The final test is what separates people who get inequalities right from people who mostly do.
Section
Section 6.4
Concept
An and inequality demands both statements hold together. The solution is the overlap, which is always a single band.
Figure (svg): Two number lines: an and-inequality shaded only between two points, and an or-inequality shaded outside them
\[ -2 < x < 3 \]
Worked example
Solve the compound inequality and graph it.
\[ -5 \le 2x + 1 < 7 \]
Do the same move to all three parts
Why: There are three expressions here, not two, so every operation happens in triplicate. Subtract one from each part.
\[ -6 \le 2x < 6 \]
Divide all three parts by 2
Why: The divisor is positive, so no signs reverse. If it had been negative, both signs would flip and the whole statement would be read the other way round.
\[ -3 \le x < 3 \]
Figure (svg): A number line with a solid circle at minus three, an open circle at three, and the band between them shaded
Verify: test both endpoints and one interior value
Why: At x equal to negative 3 the middle is negative 5, which satisfies the or-equal-to. At x equal to 3 the middle is 7, which fails the strict less-than, so 3 is correctly excluded. At x equal to 0 the middle is 1, comfortably inside.
Picture it
An and-inequality is two separate shadings, and the answer is where both are shaded.
Figure (svg): Three stacked number lines showing x greater than minus two, x less than three, and the overlapping band that satisfies both
If the two stripes never overlap, the and-inequality has no solutions at all — and the picture tells you that immediately.
Sorting
Sort each number against the compound inequality shown.
\[ -3 \le x < 3 \]
Sort into buckets
Which values are solutions?
Prediction
Commit before reasoning it through.
\[ x > 5 \;\text{ and }\; x < 2 \]
Predict first
How many numbers satisfy both conditions?
Correct: None — no number is both bigger than 5 and smaller than 2.
Why: Drawing the two shadings shows they never overlap: one covers everything to the right of 5 and the other everything to the left of 2, with a gap between. An and-inequality whose two conditions do not overlap has an empty solution set, which is a perfectly good answer rather than a mistake.
Socratic
One question, no computation.
\[ -2 < x < 3 \]
Discussion prompt
What is gained by writing an and-inequality in this compressed three-part form rather than as two separate statements?
Hint: Could you write an or-statement in this same compressed form?
Answer:
The compressed form shows the answer is a single connected band, which is exactly what and produces. It also lets you operate on all three parts at once instead of solving two inequalities separately and then intersecting them.
It only works for and, though. An or-inequality has two disconnected pieces and cannot be squeezed onto one line — writing it that way is a genuine error rather than a style choice.
Section
Section 6.5
Concept
An or inequality is satisfied when at least one of the two statements holds. The solution set is usually two separate pieces.
\[ x < -2 \quad \text{or} \quad x \ge 3 \]
Figure (svg): Two number lines: an and-inequality shaded only between two points, and an or-inequality shaded outside them
Worked example
Solve and graph the compound inequality below.
\[ 3x + 2 < -4 \quad \text{or} \quad 2x - 1 \ge 7 \]
Solve each piece completely and separately
Why: The two halves share nothing, so solve them as two independent inequalities.
\[ 3x < -6 \;\Longrightarrow\; x < -2 \]
\[ 2x \ge 8 \;\Longrightarrow\; x \ge 4 \]
Graph both solution sets on one line
Why: Shade each piece; anything shaded by either one is a solution.
Figure (svg): A number line shaded to the left of an open circle at minus two and to the right of a solid circle at four
Verify: test one value from each piece and one from the gap
Why: At x equal to negative 3 the first half gives negative 7, which is less than negative 4, so it qualifies. At x equal to 5 the second gives 9, which is at least 7. At x equal to 0 both halves fail, confirming the gap.
Comparison
Fill the blanks. These two words produce opposite-shaped answers.
Comparison matrix
| and | or | |
|---|---|---|
| how many conditions must hold | both | at least one |
| shape of the picture | one connected band | two separate pieces |
| can be written on one line | yes, as a three-part inequality | no, it must stay as two statements |
When the pieces of an or happen to overlap, the answer collapses to a single stretch — and occasionally to the entire number line.
Matching
Everyday phrases carry and or or without saying so.
Match the pairs
Why: Between always means and, and not between always means or. The phrase at least but no more than is the one worth memorising, because both ends are inclusive and the everyday wording never says so explicitly.
Two truths and a lie
Three claims about compound inequalities.
Eliminate the wrong options
Which statement is false?
Survives elimination: C
Why: Keep the false statement, which is C. When the two pieces of an or overlap, they merge into a single stretch — and sometimes into the whole number line. Drawing both shadings before describing the answer is the only reliable way to see which case you are in.
Check
Solve it on paper before you click.
Check your understanding
Solve -1 < 3x + 2 <= 11.
Answer: A
Why: Subtracting 2 from all three parts gives -3 < 3x <= 9. Dividing all three by 3 gives -1 < x <= 3. The divisor is positive, so neither sign flips, and the strict and inclusive ends stay attached to the same sides they started on.
Section
Section 6.6
Concept
An absolute-value equation asks which numbers sit a given distance from a point. There are normally two.
Figure (svg): A number line showing that the absolute value of x equals five has two answers, at minus five and five
\[ |x| = 5 \;\Longrightarrow\; x = 5 \;\text{ or }\; x = -5 \]
Worked example
Solve the equation below.
\[ |2x - 3| = 7 \]
Split into the two cases the bars allow
Why: Whatever is inside the bars is either positive seven or negative seven, because both are seven units from zero.
\[ 2x - 3 = 7 \quad \text{or} \quad 2x - 3 = -7 \]
Solve each ordinary equation separately
Why: From here there is no absolute value left; these are Chapter 3 problems.
\[ 2x = 10 \;\Longrightarrow\; x = 5 \]
\[ 2x = -4 \;\Longrightarrow\; x = -2 \]
Figure (svg): A number line marking the two solutions minus two and five, each seven units from the midpoint at one point five
Verify: substitute both solutions into the original
Why: At x equal to 5 the inside is 7, whose absolute value is 7. At x equal to negative 2 the inside is negative 7, whose absolute value is also 7. Both check out.
Prediction
Commit before working.
\[ |x + 4| = -3 \]
Predict first
How many solutions does this equation have?
Correct: None — an absolute value can never equal a negative number.
Why: The bars produce a distance, and distances are never negative. So no value of x can make the left side equal negative three, and the equation has an empty solution set. Recognising this saves you from splitting into two cases and producing two invalid answers.
Error analysis
A student solved this equation like so. Find the error.
Annotate
On: \( |x - 2| + 5 = 11 \;\overset{?}{\Longrightarrow}\; x - 2 = 6 \;\text{ only} \)
Isolate the bars, then split. Doing those in the other order is the other common failure.
Reverse engineer
Work backwards from a pair of solutions.
Fill in the blanks
|x - 5| = 4 \;\text___ x = 1 \text___ x = 9
Why: The two solutions are symmetric about their midpoint, which is 5, and each sits 4 units away from it. So the equation is the absolute value of x minus 5 equals 4. This is exactly the structure every absolute-value equation has: a centre and a distance.
Edge cases
Usually there are two answers, and occasionally there are none. There is a third possibility.
Discussion prompt
What value on the right-hand side makes an absolute-value equation have exactly one solution, and why?
Hint: How many numbers are zero units away from a given point?
Answer:
Setting it equal to zero. The absolute value of something is zero only when that something is exactly zero, and there is only one way to be zero distance from a point.
\[ |2x - 6| = 0 \;\Longrightarrow\; 2x - 6 = 0 \;\Longrightarrow\; x = 3 \]
So the full picture is: a negative right-hand side gives no solutions, zero gives one, and any positive value gives two. That is the whole classification.
Section
Section 6.7
Concept
An absolute-value inequality asks which numbers are close to or far from a centre. Those two questions have opposite-shaped answers.
Figure (svg): Two number lines contrasting an absolute value less than three, shaded between, with greater than three, shaded outside
\[ |x| < 3 \;\Longleftrightarrow\; -3 < x < 3 \]
Worked example
Solve the inequality and graph it.
\[ |x - 1| \le 4 \]
Recognise the shape: less-than means close to the centre
Why: Being within 4 units of 1 is a single band, so this will become an and-inequality.
Write it as a three-part inequality
Why: The inside must sit between negative four and four.
\[ -4 \le x - 1 \le 4 \]
Add 1 to all three parts
Why: The same move applies to every part, and the divisor is never negative here so nothing flips.
\[ -3 \le x \le 5 \]
Figure (svg): A number line with solid circles at minus three and five and the band between them shaded, centred on one
Verify: test both endpoints and one outside value
Why: At x equal to 5 the inside is 4, whose absolute value is 4, satisfying the or-equal-to. At x equal to 6 the inside is 5, which is too far. The boundary is correct.
Comparison
Fill the blanks. This one table replaces a lot of memorisation.
Comparison matrix
| the inequality | in words | becomes |
|---|---|---|
| absolute value less than a | close to the centre | an and: one band |
| absolute value greater than a | far from the centre | an or: two pieces |
| absolute value less than a negative | impossible | no solutions at all |
The middle column is the one to think in. Once you can say close or far in words, the shape of the answer follows without a rule.
Sorting
Sort each statement by the shape of its solution set. Do not solve.
Sort into buckets
What shape is each answer?
Faded example
Fill every blank.
Fill in the blanks
|3x| > 12 \;\Longrightarrow\; 3x > 12 \text-12 3x < -4 \;\Longrightarrow\; x > 4 \text___ x < ___
Why: Greater-than means far from zero, so it splits into two cases with the second one reversed and negated. Dividing each by positive 3 gives x greater than 4 or x less than negative 4 — two pieces with a gap between them. Note the sign in the second case flipped when the case was written, not when it was divided.
Check
Solve it on paper before you click.
Check your understanding
Solve |x + 2| >= 5.
Answer: A
Why: Greater-than means far from the centre, so it splits into x plus 2 at least 5, giving x at least 3, or x plus 2 at most negative 5, giving x at most negative 7. The answer is two pieces with a gap between them.
Section
Section 6.8
Concept
With two variables the solution set is a region. Draw the boundary line, then shade the side that works.
Figure (svg): A coordinate plane with a dashed boundary line and the region above it shaded, with a test point marked
A strict inequality gets a dashed boundary because the line itself is excluded; an or-equal-to gets a solid one.
Worked example
Graph the inequality below.
\[ y > x + 1 \]
Graph the boundary line as if it were an equation
Why: Treat it as y equals x plus 1 and draw it, starting at the intercept and walking the slope.
Make it dashed, because the sign is strict
Why: Points on the line give y exactly equal to x plus 1, which is not greater than, so the line is not part of the answer.
Test a point not on the line
Why: The origin is easiest whenever it is not on the boundary. Substituting gives 0 greater than 1, which is false.
Shade the other side
Why: Since the origin fails, the solution region is the side that does not contain it — the region above the line.
Figure (svg): A coordinate plane with a dashed boundary line and the region above it shaded, with a test point marked
Verify: test a point from inside the shaded region
Why: The point (0, 3) gives 3 greater than 1, which is true, so the shaded side is correct. Testing a second point from the unshaded side would give a false statement, confirming the split.
Tweak it
The shading always sits on one side of this line. Watch which side as the line moves.
Parameter explorer
Drag the intercept. Does the shaded region ever change which side of the line it is on?
\[ y > x + {b} \]
Prediction
Commit before drawing.
\[ 2x + y \le 6 \]
Predict first
How should the boundary line be drawn, and which side is shaded?
Correct: Solid line, shade the side containing the origin.
Why: The sign includes equality, so points on the line do satisfy the inequality and the line is drawn solid. Testing the origin gives 0 less than or equal to 6, which is true, so the origin's side is the shaded one. The test point decides the side; the sign decides the line style.
Discrimination
Substitute each point. No graphing needed.
\[ y \ge 2x - 3 \]
Sort into buckets
Sort each point by whether it is in the solution region.
Real world
You have 20 dollars. Sandwiches cost 4 dollars and drinks cost 2 dollars.
Discussion prompt
Write an inequality for what you can afford, and say what the shaded region means — including why only part of it is usable.
Hint: Can you buy minus two drinks?
Answer:
\[ 4s + 2d \le 20 \]
The shaded region is every affordable combination of sandwiches and drinks. The boundary line is the combinations that spend exactly 20 dollars.
Only part of the region is usable: you cannot buy a negative or fractional number of sandwiches, so the meaningful answers are the whole-number points in the first quadrant. The algebra gives a region; the situation restricts it.
Commit first
Answer, then rate your confidence.
\[ -4x \ge 20 \]
Predict first
What is the solution set?
Correct: x is at most -5.
\[ -4x \ge 20 \;\Longrightarrow\; x \le -5 \]
Why: Dividing both sides by negative 4 flips the sign, giving x less than or equal to negative 5. Testing x equal to negative 10 in the original gives 40, which is at least 20, so it works. Testing x equal to 0 gives 0, which fails — confirming that the solution set lies to the left.
Exit ticket
Last commitment of the chapter.
Predict first
Which of these is shakiest right now?
Correct: Whatever you picked is the one to drill first.
Why: All four have the same fix, which is testing a value from your final answer against the original statement. That single habit catches a missed flip, a wrong-shaped compound answer, a lost case and a wrongly shaded region — four different errors, one check.
Connect it up
One page, drawn by you.
Draw it
Draw a number line down the middle. Off it, attach: one-step, multi-step, the flip rule, and, or, absolute-value equations, absolute-value inequalities. For each, sketch the shape of its answer — one band, two pieces, two points, or a region. Circle the two places a sign can flip.
If your map does not connect absolute value less than to and, add that arrow — it is the link the chapter is built on.
Recap
You can now answer questions whose answer is a whole set of numbers rather than a single one.
| if you remember one thing | it should be |
|---|---|
| about the flip | only multiplying or dividing by a negative reverses the sign |
| about circles | the sign decides hollow or solid, and that is one whole number of difference |
| about absolute value | less-than means and, greater-than means or |
| about regions | the sign picks the line style, and a test point picks the side |
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