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4.6 — Taking Sides
Human review completed .
Explain why an inequality solution is correct, recognize common mistakes, and describe the whole solution set.
1. In class: A Practice Understanding Task
The packet asks you to evaluate Joaquin’s and Serena’s reasoning. For a refresher on solving inequalities, see Section 4.4: Keep the comparison true.
Read the words and symbols carefully — Main 1–4
“6 less than a number” means subtract 6 from the number: \(n-6\). Similarly, “4 greater than x” means \(x+4\). An expression describes a quantity; an inequality compares quantities.
| Main problem | Answer and reason |
|---|---|
| 1 | Serena is correct. \(x\ge-7\) includes −7, so use a closed dot at −7 and shade to the right. |
| 2 | Serena is not correct. For example, \(x=-1\) gives \(3(-1)+1=-2\), which is not greater than zero. Solving \(3x+1>0\) gives \(x>-\frac13\). |
| 3 | Serena is correct: “4 greater than x” is \(4+x\). Joaquin’s \(4>x\) says that 4 is greater than x, a different statement. |
| 4 | Joaquin is not correct. \(45+47<t\) means \(92<t\), or \(t>92\). When you exchange the sides of an inequality, reverse the symbol to keep the same comparison. |
Worked example 1 — Compare an equation with an inequality
Original problem — Main 5: Compare solving \(-\frac{x}{3}=4\) with solving \(-\frac{x}{3}>4\).
Worked solution: Multiply both sides by −3 in each problem. In the inequality, reverse the symbol:
| Equation | Inequality |
|---|---|
| \(-\frac{x}{3}=4\) | \(-\frac{x}{3}>4\) |
| \(x=-12\) | \(x<-12\) |
The equation has one solution, −12. The inequality has all numbers less than −12 as solutions; −12 itself is excluded.
Checking one value does not check a whole solution set — Main 6
For \(-15q\le135\), divide both sides by −15 and reverse the symbol to get \(q\ge-9\). Serena is correct: adding 15 does not undo multiplication by −15.
Joaquin’s value \(q=150\) does satisfy the original inequality, but that does not prove that his proposed answer \(q\le150\) is correct. For example, −10 satisfies his answer, yet \(-15(-10)=150\) is not less than or equal to 135. One working value confirms only that value; one counterexample disproves a proposed solution set.
Keep every variable term — Main 7
Serena’s last step loses a variable term. Here are the correct steps:
| Inequality | Reason |
|---|---|
| \(3x+3\le-2x+5\) | Original inequality |
| \(3x\le-2x+2\) | Subtract 3 from both sides. |
| \(5x\le2\) | Add 2x to both sides. |
| \(x\le\frac25\) | Divide both sides by 5; keep the symbol because 5 is positive. |
Worked example 2 — Distribute, collect terms, and solve
Original problem — Main 8: Solve \(-4(3m-1)\ge2(m+3)\), explaining the steps.
Worked solution:
| Inequality | Reason |
|---|---|
| \(-4(3m-1)\ge2(m+3)\) | Original inequality |
| \(-12m+4\ge2m+6\) | Distribute on both sides. |
| \(-14m+4\ge6\) | Subtract 2m from both sides. |
| \(-14m\ge2\) | Subtract 4 from both sides. |
| \(m\le-\frac17\) | Divide both sides by −14 and reverse the symbol. |
Note: Distributing −4 within the left expression does not reverse the symbol. Dividing both sides by −14 does.
All real numbers or no solutions — Main 9
Sometimes the variable terms cancel. If the remaining statement is true, every real value works. If it is false, none work.
| Part | Simplification | Solution and number-line graph |
|---|---|---|
| (a) | \(4s+6\ge6+4s\) becomes \(6\ge6\) | All real numbers. Shade the entire number line. |
| (b) | \(3r+5>3r-2\) becomes \(5>-2\) | All real numbers. Shade the entire number line. |
| (c) | \(4(n+1)<4n-3\) becomes \(4< -3\) | No solutions. Leave the number line unshaded. |
Worked example 3 — A coefficient whose sign is unknown
Original problem — Main 10: Solve \(ax+b>c\) for x.
Worked solution: Subtract b from both sides to get \(ax>c-b\). Serena is right: the sign of a determines what happens when you divide.
| Condition | Solution |
|---|---|
| \(a>0\) | \(x>\frac{c-b}{a}\) — divide by a positive number; keep the symbol. |
| \(a<0\) | \(x<\frac{c-b}{a}\) — divide by a negative number; reverse the symbol. |
| \(a=0\) | The original inequality becomes \(b>c\). All real x work if \(b>c\); there are no solutions if \(b\le c\). |
When \(a=0\), check \(b>c\) directly; you cannot divide by zero.
2. Ready — Amusement-park costs and test scores
These are applications of familiar equations and inequalities. Include the membership fee in total cost, and remember that visits and correct answers are whole-number counts. Try each problem before opening the key.
Check your work — Ready 1–2 answer key
1 — Amusement-park visits
Let v be the number of visits. Each answer below gives the maximum number of visits within the budget.
\(50+15v\le100\), so \(v\le\frac{10}{3}\). 3 visits
\(25v\le100\), so \(v\le4\). 4 visits
\(50+15v\le160\), so \(v\le\frac{22}{3}\). 7 visits
\(25v\le160\), so \(v\le\frac{32}{5}\). 6 visits
2 — Test scores
Let q be the number of correct answers and S the score in points. Each question is worth \(100\div20=5\) points.
\(S=5q\)
\(5q\ge70\), so \(q\ge14\). 14 or more correct answers
\(5q\ge83\), so \(q\ge16.6\). 17 or more correct answers
Initially 12 questions were correct and 8 were wrong. Half of those 8 become correct, giving \(12+4=16\) correct answers: \(\frac{16}{20}\times100\%=80\%\).
3. Set — Inequalities and literal equations
For inequality methods, use the main-task explanations above or Section 4.4. For rearranging formulas, see Section 4.3’s worked examples. Try the problems first, then check your answers and justifications.
Check your work — Set 3–20 answer key
3–8 — Steps and justifications
3
| Inequality | Justification |
|---|---|
| \(-5x<35\) | Original inequality |
| \(x>-7\) | Divide both sides by −5 and reverse the symbol. |
4
| Inequality | Justification |
|---|---|
| \(x+68\ge75\) | Original inequality |
| \(x\ge7\) | Subtract 68 from both sides. |
5
| Inequality | Justification |
|---|---|
| \(2x-4\le10\) | Original inequality |
| \(2x\le14\) | Add 4 to both sides. |
| \(x\le7\) | Divide both sides by 2. |
6
| Inequality | Justification |
|---|---|
| \(5-4x\le17\) | Original inequality |
| \(-4x\le12\) | Subtract 5 from both sides. |
| \(x\ge-3\) | Divide both sides by −4 and reverse the symbol. |
7
| Inequality | Justification |
|---|---|
| \(\frac{x}{-3}>-\frac{10}{9}\) | Original inequality |
| \(x<\frac{10}{3}\) | Multiply both sides by −3 and reverse the symbol. |
8
| Inequality | Justification |
|---|---|
| \(2(x-3)\le3x-2\) | Original inequality |
| \(2x-6\le3x-2\) | Distribute 2. |
| \(-6\le x-2\) | Subtract 2x from both sides. |
| \(-4\le x\) | Add 2 to both sides. |
| \(x\ge-4\) | Exchange the sides and reverse the symbol. |
9–10 — Number-line solutions
Plots are not shown; use the endpoint and direction below to check yours.
| Problem | Solution | Graph |
|---|---|---|
| 9 | \(x>-12\) | Open dot at −12; shade right. |
| 10 | \(x>2\) | Open dot at 2; shade right. |
11–14 — Multi-step inequalities
\(x<-1\)
\(x\ge-\frac15\)
\(x\ge-\frac{11}{4}\)
\(x\le\frac32\)
15–20 — Isolate the requested variable
| Problem | Answer |
|---|---|
| 15 | \(C=\frac59(F-32)\) |
| 16 | \(r=\sqrt{\frac{3V}{\pi h}}\) |
| 17 | \(a=\frac{2A}{P}\) |
| 18 | \(b=y-mx\) |
| 19 | \(r=\frac{c}{2\pi}\) |
| 20 | \(d=2\sqrt{\frac{A}{\pi}}\) |
For the geometry formulas, use the nonnegative root for a radius or diameter; h and P are positive measurements.
4. Go — Graph linear systems
For a refresher, see Section 3.2: graphing and checking an intersection. Graph both lines, find their intersection, and substitute its coordinates into both equations. Try your graphs before checking the key.
Check your work — Go 21–23 answer key
Points of intersection are given below. Plots are not shown.
\((-4,-3)\)
\((2,-4)\)
\((1,8)\)
Additional learning resources
| Resource | What it helps with |
|---|---|
| Khan Academy — Multi-step inequalities | Solving inequalities and reversing the comparison when multiplying or dividing both sides by a negative number. |
| Khan Academy — Testing solutions to inequalities | Substituting a proposed value to see whether it makes an inequality true. |
Teacher’s review of the source packet
No substantial mathematical or instructional issues were identified in the student exercises.
Coverage
| Packet component | Where it is supported |
|---|---|
| Main 1–4 | Words, symbols, endpoint inclusion, and brief explained answers. |
| Main 5–8 | Worked solutions and explanations of solving and checking. |
| Main 9–10 | All-real/no-solution cases and an unknown coefficient’s sign. |
| Ready 1–2 | Review guidance and complete lettered answer key. |
| Set 3–20 | Linked review; justification tables for 3–8; all answers; graph descriptions for 9–10. |
| Go 21–23 | Linked graphing review and intersection answers. Required plots are not reproduced. |
Original prompts adapted from Mathematics Vision Project, Secondary Mathematics I, Module 4 student packet, Section 4.6, pp. 24–30, licensed under CC BY 4.0. Explanations, solutions, and the number-line illustration were prepared for this supplement.
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