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4.4 — Greater Than?
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Compare expressions, solve inequalities, and recognize when the information given does not determine which expression is greater.
1. In class: A Develop Understanding Task
What does an inequality tell you?
An inequality compares two quantities. \(x<2\) means “x is less than 2”; \(x>2\) means “x is greater than 2.” The symbols \(\le\) and \(\ge\) include equality: \(x\ge2\) means “x is greater than or equal to 2.”
Read the quantities in the order written. \(2<x\) and \(x>2\) say the same thing. When you exchange the two sides, reverse the symbol so it continues to point toward the smaller quantity.
A solution is a value that makes the inequality true. The solution set contains all such values. Solving an inequality means describing that whole set, rather than finding just one value that works.
Keep the comparison true
| Operation on both sides | What happens to the inequality? |
|---|---|
| Add or subtract the same number | Keep its direction. |
| Multiply or divide by the same positive number | Keep its direction. |
| Multiply or divide by the same negative number | Reverse its direction. |
For example, \(2<5\), but \(-2>-5\). Multiplying both numbers by −1 reverses their order on the number line. The reversal also applies to \(\le\) and \(\ge\).
Worked example 1 — Compare x with 2
Original problem — Main 3: Given \(10-2x<6\), compare \(x\) and 2.
Subtract 10 from both sides, then divide both sides by −2 and reverse the symbol. The answer is x is greater than 2.
Every number greater than 2 works; 2 itself does not. For example, \(x=3\) gives \(10-2(3)=4<6\), while \(x=2\) gives \(6<6\), which is false.
Sometimes more information is needed
In Main 8, knowing \(5>4\) is not enough to compare \(5x\) with \(4x\):
- If \(x>0\), then \(5x>4x\).
- If \(x<0\), then \(5x<4x\).
- If \(x=0\), both expressions are 0.
So the comparison depends on the sign of \(x\).
Worked example 2 — Compare fractions with an unknown denominator
Original problem — Main 9: Given \(5>4\), compare \(\frac5x\) and \(\frac4x\).
Dividing both sides by \(x\) keeps or reverses the comparison depending on its sign:
- If \(x>0\), then \(\frac5x>\frac4x\). For example, with \(x=2\), \(\frac52>2\).
- If \(x<0\), then \(\frac5x<\frac4x\). For example, with \(x=-2\), \(-\frac52<-2\).
When \(x=0\), neither fraction is defined. The relationship cannot be determined without knowing whether \(x\) is positive or negative.
Worked example 3 — Overlapping ranges do not establish an order
Original problem — Main 10: Given \(0<x<10\) and \(0<y<12\), compare \(x\) and \(y\).
The bounds allow both orders: \(x=8,y=3\) gives \(x>y\), while \(x=3,y=8\) gives \(x<y\). They also allow equality, such as \(x=y=5\). The relationship cannot be determined without more information about the values of \(x\) and \(y\).
Try the other comparisons yourself. Use these answers to check both your result and your reasoning:
| Main problem | Comparison and reason |
|---|---|
| 1 | \(x-y>y-x\). Since \(y<x\), \(x-y\) is positive and \(y-x\) is negative. |
| 2 | \(x>5\). Add 3 to both sides of \(2x-3>7\), then divide by 2. |
| 4 | \(1>x\). Dividing \(4x\le0\) by 4 gives \(x\le0\). |
| 5 | Cannot be determined. If \(n>0\), \(n>-n\); if \(n<0\), \(n<-n\); if \(n=0\), they are equal. Knowing that \(n\) is an integer does not specify its sign. |
| 6 | \(x+a>y+a\). Adding the same number preserves the order, even when \(a\) is negative. |
| 7 | \(x-a>y-a\). Subtracting the same number preserves the order. |
| 11 | \(n\ge1\). Write \(27=3^3\). Since powers of 3 increase as their exponents increase, \(3^{n+2}\ge3^3\) implies \(n+2\ge3\), hence \(n\ge1\). Equality is possible when \(n=1\). |
2. Ready — Story equations and inequality claims
For the two stories, write “fixed charge + charge per unit × number of units = total charge,” then solve as in Section 4.1. For the true/false statements, use the inequality rules above. Try the problems before checking the key.
Check your work — Ready 1–7 answer key
Equation: \(10+0.20s=50\), where \(s\) is the area painted in square feet
Answer: \(s=200\) square feet
Equation: \(200+4p=324\), where \(p\) is the number of people
Answer: \(p=31\) people
Statement: “The notation \(12<x\) means the same thing as \(x<12\). It works just like \(12=x\) and \(x=12\).”
False. \(12<x\) means \(x>12\). Exchanging the sides of an inequality requires reversing its direction.
Statement: “The inequality \(-2(x+10)\ge75\) says the same thing as \(-2x-20\ge75\). I can multiply by −2 on the left side without reversing the inequality symbol.”
True. This distributes the −2 already present in the expression. It rewrites the left side without changing its value; it does not multiply both sides of the inequality by −2.
Statement: “When solving the inequality \(10x+22<2\), the second step should say \(10x>-20\) because I added −22 to both sides and I got a negative number on the right.”
False. Subtracting 22 preserves the symbol, giving \(10x<-20\). A negative result is not a reason to reverse it.
Statement: “When solving the inequality \(-5x\ge45\), the answer is \(x\le-9\) because I divided both sides of the inequality by a negative number.”
True. Dividing both sides by −5 reverses \(\ge\) to \(\le\).
Statement: “The words that describe the inequality \(x\ge100\) are ‘x is greater than or equal to 100.’”
True. The symbol includes both values greater than 100 and 100 itself.
3. Set — Solve and check inequalities
Use the same simplifying steps as for equations, remembering to reverse the symbol when multiplying or dividing by a negative number. For a proposed value, substitute it into the original inequality to check whether it belongs to the solution set.
Worked example 4 — Clear the fractions first
Original problem — Set 19: Solve \(\frac{3(x-4)}{12}\le\frac{2x}{3}\).
Multiply both sides by positive 12, keeping the inequality’s direction:
Distribute, subtract \(3x\) from both sides, then divide both sides by 5. Write the result with \(x\) first as \(x\ge-\frac{12}{5}\). Exchanging the sides changes \(\le\) to \(\ge\) while preserving the meaning.
For the number-line graphs, use an open circle for \(<\) or \(>\), and a filled circle for \(\le\) or \(\ge\). Shade all values that satisfy the inequality. Study the examples, try Set independently, then check your work.
Check your work — Set 8–20 answer key
Solutions and proposed values — 8–11
\(2x<12\), so \(x<6\)
Given \(x=6\): no; \(2(6)-9=3\), and \(3<3\) is false.
\(4x>-12\), so \(x>-3\)
Given \(x=-5\): no; \(4(-5)+25=5\), and \(5>13\) is false.
\(6x\le-24\), so \(x\le-4\)
Given \(x=-10\): yes; \(6(-10)-4=-64\le-28\).
\(3x\ge0\), so \(x\ge0\)
Given \(x=1\): yes; \(3(1)-5=-2\ge-5\).
Solutions and number-line graphs — 12–17
Plots not shown. Endpoint and shading descriptions are provided to check your graphs.
\(x\le-2\)
Filled circle at −2; shade left.
\(-3x>6\), so \(x<-2\)
Open circle at −2; shade left.
\(x<-2\)
Open circle at −2; shade left.
\(x>-\frac32\)
Open circle at \(-\frac32\); shade right.
\(x<-15\)
Open circle at −15; shade left.
\(x\le35\)
Filled circle at 35; shade left.
Multi-step inequalities — 18–20
\(-5>x+3\)
\(-8>x\), or \(x<-8\)
\(3(x-4)\le8x\)
\(-12\le5x\)
\(x\ge-\frac{12}{5}\)
\(2x-6\le3x-2\)
\(-6\le x-2\)
\(-4\le x\), or \(x\ge-4\)
4. Go — Solve systems by substitution
A solution of a system makes both equations true. For a refresher, see Section 3.2. Substitution replaces a variable in one equation with an equal value or expression from the other equation. This leaves one equation in one variable.
Worked example 5 — Substitute an expression for y
Original problem — Go 25:
Replace \(y\) in the second equation with \(2x-5\) from the first:
Then use either original equation to find \(y\): \(y=13+8=21\). The solution is \((13,21)\). Checking the first equation gives \(21=2(13)-5\).
Try the remaining systems on your own before opening the key. Give each solution as an ordered pair \((x,y)\).
Check your work — Go 21–26 answer key
Solution: \((4,5)\)
Substitute \(y=5\): \(-x+5=1\), so \(x=4\).
Solution: \((8,-20)\)
Substitute \(x=8\): \(40+2y=0\), so \(y=-20\).
Solution: \((15,5)\)
First \(y=5\). Substitute into the second equation: \(4x-10=50\), so \(x=15\).
Solution: \((4,11)\)
First \(x=4\). Then \(16-y=5\), so \(y=11\).
Solution: \((13,21)\)
Substitute \(y=2x-5\) into \(y=x+8\): \(2x-5=x+8\), so \(x=13\) and \(y=21\).
Solution: \((3,-20)\)
First \(x=3\). Then \(15+y=-5\), so \(y=-20\).
Additional learning resources
| Resource | What it supports |
|---|---|
| Khan Academy — Two-step inequalities | Solving inequalities with addition, subtraction, multiplication, and division; supports the main task and Set. |
| The Organic Chemistry Tutor — How To Solve Linear Inequalities, Basic Introduction, Algebra | Worked inequalities, number-line graphs, and interval notation. Focus on the linear inequalities and graphs; compound inequalities go beyond this section. |
| Khan Academy — Systems of equations with substitution | Replacing a variable with an expression from the other equation; supports Go. |
Teacher’s review of the source packet
No substantial mathematical error was found. Main 5, 8–10 deliberately allow different comparisons depending on the values; these are reasoning tasks, not missing-data mistakes. Ready 3–7 deliberately include false statements for students to evaluate.
Coverage — Section 4.4
| Packet location | Where supported |
|---|---|
| Main 1–11 · pp. 14–15 | Inequality meanings and rules, worked comparisons, sign-dependent cases, and visible checking answers |
| Ready 1–7 · p. 16 | Equation-modeling reminder and true/false key preserving statement form |
| Set 8–20 · pp. 16–17 | Solving and checking, clearing fractions, number-line conventions, and a worked key; required plots omitted from the key |
| Go 21–26 · p. 18 | Substitution explanation, worked system, and ordered-pair answers with substitution steps |
Source: Mathematics Vision Project, Secondary Mathematics One, Module 4: Equations and Inequalities (May 2016), Section 4.4, printed pp. 14–18; teacher notes for 4.4. Packet equations and short excerpts are attributed under CC BY 4.0. Equations use MathJax.
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