SAT Inequalities

Correct Answer: C

Solution 1:

Tip: Plug and check.
(A) If $a = -3$ , then $4a + 5 = 4 (-3) + 5 = -12 + 5 = -7$ . But this means that $-3 < -7 = 4a + 5,$ which can't be. Wrong choice.
(B) If $a = -2$ , then $4a + 5 = 4 (-2) + 5 = -8 + 5 = -3$ . We can't have $-3 < -3 = 4a + 5,$ so this is incorrect.
(C) If $a = -1$ , then $4a + 5 = 4 (-1) + 5 = -4 + 5 = 1$ . Since $-3 < 1 < 5,$ the given restriction is satisfied and this is the correct answer.
(D) If $a = 1$ , then $4a + 5 = 4 (1) + 5 = 4 + 5 = 9$ . But this means that $4a + 5 = 9 < 3,$ so this is incorrect.
(E) If $a = 2$ , then $4a + 5 = 4 (2) + 5 = 8 + 5 = 13$ . We can't have $4a+5 = 13 < 3,$ so this is incorrect.

Solution 2:

Let's solve the inequality by splitting it up.

First, we solve $-3 < 4a + 5$ .

$\begin{array} { l l l } - 3 & < 4a + 5 & \text{given} \\ -8 & < 4a & \text{subtract 5 from both sides} \\ -2 & < a & \text{divide by 4 on both sides} \\ \end{array}$

Next, we solve $4a + 5 < 3$ .

$\begin{array} { l l l } 4a + 5 & < 3 & \text{given} \\ 4a & < -2 & \text{subtract 5 from both sides} \\ a & < - \frac{2}{4} = - \frac{1}{2} & \text{ divide by 4 on both sides} \\ \end{array}$

Combining both of these inequalities, we get that $- 2 < a < \frac{1}{2}$ . Looking at the options, the only one that satisfies this is $a = - 1$ .

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Incorrect Choices:

(A) , (B) , (D) , and (E)
Look at Solution 1 for how to eliminate these choices by plugging and checking.