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.

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

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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).

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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) \(\ \ 6\)

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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).

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

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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.

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Incorrect Choices:

(A), (C), (D), and (E)
Solution 2 offers a counter-example for each of these wrong choices.

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

• 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