# 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 \(m=2,\)

\[\begin{align} -3 &\leq -5m + 7\\ -3 &\leq -5(2)+7\\ -3 &\leq -3\\ \end{align}\]

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

\[\begin{align} -3 &\leq -5m + 7\\ -3 &\leq -5(0)+7\\ -3 &\leq 7\\ \end{align}\]

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 \(m=3,\)

\[\begin{align} -3 &\leq -5m + 7\\ -3 &\leq -5(3)+7\\ -3 &\leq -8\\ \end{align}\]

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{align} 2x&<11\\

x&<5.5\\ \end{align}\]Similarly, divide both sides by \(7\) in the second inequality to solve for \(x\):

\[\begin{align} 7x&>29\\

x&>4.14, \end{align}\]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/