# SAT Inequalities

To solve problems with inequalities on the SAT, you need to know:

- how to manipulate algebraic expressions
- the properties of inequality:

If $a<b$, then $a+c<b+c$ and $a-c<b-c.$

If $a<b$ and $c<d$, then $a+c<b+d.$

If $a<b$ and $c>0$, then $ac<bc$ and $\frac{a}{c}<\frac{b}{c}.$

If $a<b$ and $c<0$, then $ac>bc$ and $\frac{a}{c} > \frac{b}{c}.$Reversing the inequality:

If $a<b$, then $-a>-b.$

If $a>b$, then $-a<-b.$

If $a$ and $b$ are both positive or both negative and $a<b$, then $\frac{1}{a}>\frac{1}{b}.$

- the properties of numbers between $0$ and $1$

For $0<a<1$:

- If $0 < b < 1$, then $ba<a$.
- If $m$ and $n$ are integers, such that $1<n<m$, then $x<x^{n}<x^{m}$.
- $a<\sqrt{a}$.
- $a<1<\frac{1}{a}$.

- the number line notation

If a point is included in the interval, it is represented by a circle that is bubbled in. If it is excluded, it is represented by an open circle. The following examples illustrate this notation.

#### Contents

## Examples

Which of the following shows all the values of $m$ that satisfy the inequality $-3 \leq -5m + 7$?

Correct Answer: B

Solution 1:

Tip: Multiplying (or dividing) both sides of an inequality by a negative number reverses its sign.

We solve for $m$ and choose the option that corresponds to the answer.$\begin{array}{r c l l l} -3 &\leq& -5m +7 &\quad \text{given} &(1)\\ -10 &\leq& -5m &\quad \text{subtract}\ 7\ \text{from both sides} &(2)\\ 2 &\geq& m &\quad \text{divide both sides by}\ -2 &(3)\\ \end{array}$

Be careful in the last step. The sign reverses because we are dividing both sides of the inequality by a negative number. The only choice that represents all the numbers smaller than or equal to $2$ is choice (B).

Solution 2:

Tip: Look for short-cuts.

Tip: Plug and check.

For each answer choice, we could select a point from inside, outside, and at the end-points of the presented interval. We would plug them into the given inequality and look for a contradiction. If we reach one, we would eliminate that choice. Let's be selective about the points we test.Notice that $2$ is an end-point in four answer choices. In (A) and (B), it is included in the interval, whereas in (C) and (E), it is excluded. Although it isn't an end-point in choice (D), there it is also excluded. If, when plugged into the inequality, $m=2$ yields a true statement, then it should be included in the answer; if it yields a contradiction, then it should be excluded from it.

If $m=2,$

$\begin{aligned} -3 &\leq -5m + 7\\ -3 &\leq -5(2)+7\\ -3 &\leq -3\\ \end{aligned}$

This is true. So, $m=2$ should be included in the interval. Choices (C), (D), and (E) exclude it, and therefore should be eliminated. Now we need to test a point to the left of $2$ and a point to the right of $2$ to determine whether (A) or (B) is the correct answer.

An easy-to-work-with point to the left of $2$ is $m=0,$ for which the inequality gives us

$\begin{aligned} -3 &\leq -5m + 7\\ -3 &\leq -5(0)+7\\ -3 &\leq 7\\ \end{aligned}$

This is true, and therefore, $m=0$ should be included in the answer. Although choice (B) seems to be the correct one, we verify that a point to the right of $2$ yields a contradiction.

If $m=3,$

$\begin{aligned} -3 &\leq -5m + 7\\ -3 &\leq -5(3)+7\\ -3 &\leq -8\\ \end{aligned}$

This is indeed false, and therefore $m=3$ should be excluded from the answer. We eliminate choice (A).

Incorrect Choices:

(A)

Tip: Multiplying (or dividing) both sides of an inequality by a negative number reverses its sign.

If you forget to reverse the inequality sign in step $(3),$ you will get this wrong answer. See Solution 2 for a counter-example.

(C),(D), and(E)

Solution 2 offers a counter-example for each of these choices.

If $x$ is an integer such that $2x<11$ and $7x>29$, what is the value of $x$?

(A) $\ \ 2$

(B) $\ \ 3$

(C) $\ \ 4$

(D) $\ \ 5$

(E) $\ \ 6$

Correct Answer: D

Solution 1:

Tip: use a calculator.

Divide both sides by $2$ in the first inequality to solve for $x$:$\begin{aligned} 2x&<11\\ x&<5.5\\ \end{aligned}$

Similarly, divide both sides by $7$ in the second inequality to solve for $x$:

$\begin{aligned} 7x&>29\\ x&>4.14, \end{aligned}$

where we have rounded the result to the nearest hundredth.

The only integer greater than $4.14$ and smaller than $5.5$ is $5$.

Solution 2:

Tip: Plug and check.

(A) If $x=2$:$\begin{array}{c c c c r} 2x &<& 11 &\quad & \text{and} &\quad & 7x>29\\ 2\cdot 2&<&11&\quad & &\quad & 7\cdot 2>29\\ 4&<&11&\quad &&\quad &14>29\\ \end{array}$

The second inequality is not satisfied. Eliminate (A).

(B) If $x=3$:

$\begin{array}{c c c c r} 2x &<& 11&\quad & \text{and} &\quad & 7x>29\\ 2\cdot 3&<& 11 &\quad & &\quad & 7\cdot 3>29\\ 6&<&11 &\quad &&\quad &21>29 \end{array}$

The second inequality is not satisfied. Eliminate (B).

(C) If $x=4$:

$\begin{array}{c c c c r} 2x &<& 11&\quad & \text{and} &\quad & 7x>29\\ 2\cdot 4&<& 11 &\quad & &\quad & 7 \cdot 4>29\\ 8&<&11 &\quad &&\quad &28>29 \end{array}$

The second inequality is not satisfied. Eliminate (C).

(D) If $x=5$:

$\begin{array}{c c c c r} 2x &<& 11&\quad & \text{and} &\quad & 7x>29\\ 2 \cdot 5&<& 11 &\quad & &\quad & 7\cdot 5>29\\ 30&<&11 &\quad &&\quad &35>29 \end{array}$

Both inequalities are satisfied. This is the correct answer.

(E) If $x=6$:

$2x<11$ means that $2(6)<12,$ or that $12<12,$ which is false. Eliminate (E).

Incorrect Choices:

(A),(B),(C), and(E)

See Solution 2 for how to eliminate these choices by plugging and checking.

Let $m$ be a number such that $20<m<40$, and let $n$ be a number such that $50<n<80$. Which of the following represents all possible values of $m-n$?

(A) $\ \ -60 < m-n < -40$

(B) $\ \ -60 < m-n < -10$

(C) $\ \ -30< m-n < -40$

(D) $\ \ -30< m-n < 40$

(E) $\ \ -10 < m-n < 60$

Correct Answer: B

Solution 1:$m-n$ will be largest if $m$ is largest and $n$ is smallest. $m-n$ cannot be more than $40-50=-10$.

$m-n$ will be smallest if $m$ is smallest and $n$ is largest. $m-n$ cannot be less than $20-80=-60$.

Therefore, $-60<m<-10$.

Solution 2:

Tip: Plug and check.

Tip: Look for a counter-example.

This can be time-consuming because we are just guessing. Points near the end-points of intervals tend to yield contradictions, so we try those first. Here, we present a counter-example for each wrong choice.(A) If $m=39$ and $n=51$, then $m-n=39-51=-12,$ which is not in the range $-60 < m-n < -40$. Eliminate (A).

(C) If $m=39$ and $n=51$, then $m-n=39-51=-12,$ which is not in the range $-30< m-n < -40$. Eliminate (C).

(D) If $m=21$ and $n=79$, then $m-n=21-79=-58,$ which is not in the range $-30 < m-n < 40$. Eliminate (D).

(E) If $m=19$ and $n=49$, then $m-n=19-49=-30,$ which is not in the range $-10 < m-n < 60$. Eliminate (E).We couldn't find a counter-example for choice (B). Therefore, it is the correct answer.

Incorrect Choices:

(A),(C),(D), and(E)

Solution 2 offers a counter-example for each of these wrong choices.

## Review

If you thought these examples difficult and you need to review the material, these links will help:

- Linear Inequalities
- Numerical Inequalities
- Finding Solutions to Linear Inequalities
- One-Step Linear Inequalities
- Inequalities Solution Set
- Two-Sided Linear Inequalities
- Absolute Value Inequalities
- Rules of Exponents

$a^m \times a^n = a^{ m + n }$, $a^n / a^m = a^ { n - m }\ldots$ - SAT Reasoning Skills
- SAT Number Line
- SAT Absolute Value
- SAT Exponents

## SAT Tips for Inequalities

- Multiplying (or dividing) both sides of an inequality by a negative number reverses its sign.
- Know the properties of numbers between $0$ and $1$.
- Plug and check.
- Look for a counter-example.
- SAT General Tips

**Cite as:**SAT Inequalities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sat-inequalities/