Solving Linear Inequalities on the SAT
Inequalities work almost exactly like equations, with one rule that the SAT tests relentlessly, because it's so easy to forget.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
You isolate the variable the same way you would in an equation. The catch: whenever you multiply or divide both sides by a negative number, the inequality sign flips direction. Compound inequalities (two bounds at once) get handled together, not as two separate problems.
How to solve one, step by step
Example: solve $-2x + 3 > 11$.
- Subtract 3 from both sides: $-2x > 8$.
- Divide by $-2$ and flip the sign: $x < -4$.
The mistakes that cost points
- Forgetting to flip the sign. Dividing or multiplying by a negative reverses the inequality. Skip it and your whole answer is backwards.
- Splitting a compound inequality. Treating $-3 < 2x \leq 8$ as two unrelated problems loses the connection between the bounds.
- Flipping for a negative term that isn't a multiplier. Only flip when you multiply or divide by a negative, not when you simply add or subtract one.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The inequality $-2x + k > 4$ has the solution $x < 3$. What is the value of $k$?
- A$k = 10$
- B$k = 2$
- C$k = 6$
- D$k = -2$
Show solution
Answer: A, $k = 10$. Substitute the boundary condition: when $x = 3$, the inequality becomes an equality. So $-2(3) + k = 4 \Rightarrow -6 + k = 4 \Rightarrow k = 10$. Verify the direction: $-2x + 10 > 4 \Rightarrow -2x > -6 \Rightarrow x < 3$ ✓. The flip occurs when dividing by $-2$, confirming $k = 10$.
Medium
If $px + 9 > 1$ and $p > 0$, which of the following describes all possible values of $x$?
- A$x < -\dfrac{8}{p}$
- B$x < \dfrac{8}{p}$
- C$x > -\dfrac{8}{p}$
- D$x > \dfrac{8}{p}$
Show solution
Answer: C, $x > -\dfrac{8}{p}$. Subtract $9$ from both sides: $px > 1 - 9 = -8$. Divide both sides by $p$. Since $p > 0$, the inequality stays the same: $x > -\dfrac{8}{p}$. The numerator stays as $-8$. Do not drop the negative sign.
Hard
If $px + 18 > 6$ and $p < 0$, which of the following describes all possible values of $x$?
- A$x < \dfrac{12}{p}$
- B$x > \dfrac{12}{p}$
- C$x < -\dfrac{12}{p}$
- D$x > -\dfrac{12}{p}$
Show solution
Answer: C, $x < -\dfrac{12}{p}$. Subtract $18$ from both sides: $px > 6 - 18 = -12$. Divide both sides by $p$. Since $p < 0$, flip the inequality: $x < -\dfrac{12}{p}$. The numerator stays as $-12$. Do not drop the negative sign.
Find your exact gaps
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