Linear Inequalities

Help Questions

SAT Math › Linear Inequalities

Questions 1 - 3
1

Which number line shows the solution to $2x - 5 > 1$?

Open circle at 3, arrow to the right

Open circle at 2, arrow to the right

Closed circle at 2, arrow to the left

Closed circle at 3, arrow to the right

Explanation

The question asks which number line represents the solution to 2x - 5 > 1. To solve, first add 5 to both sides: 2x > 6. Then divide both sides by 2: x > 3. Since we're dividing by a positive number, the inequality symbol remains unchanged. The solution x > 3 requires an open circle at 3 (since 3 is not included) with an arrow pointing to the right (representing all numbers greater than 3). A common error is using a closed circle when the inequality is strict (> or <) rather than inclusive (≥ or ≤). Remember that strict inequalities always use open circles to show the boundary point is excluded.

2

What is the solution to the inequality $4x - 7 \ge 2x + 5$?

$x \ge 6$

$x \le 6$

$x \ge -1$

$x \le -1$

Explanation

1. Subtract $2x$ from both sides: $2x - 7 \ge 5$.

2. Add 7 to both sides: $2x \ge 12$.

3. Divide by 2: $x \ge 6$.

SAT Strategy

On inequality problems, often you can pick a number that satisfies the potential answer. If you test $x=0$, the inequality becomes $-7 \ge 5$, which is False. Therefore, the solution cannot include 0. This eliminates B, C, and D (since 0 is less than 6, greater than -1, etc). Only A excludes 0.

3

Which of the following is not a possible value for $x$ given the inequality above?

12

15

24

25

Explanation

In order to simplify this inequality, we'll want to start by subtracting 12 from both sides to arrive at 

$\frac{1}{4}x >3$

If we then multiply both sides of the inequality by 4 to cancel the coefficient in front of $x$, we get to 

$x>12$

Thus, 12 is not a possible value of x given the simplified inequality. 

Note - we could also solve this question by plugging each option in for $x$, but this route is likely somewhat more time consuming, so stay flexible in your approach on a question-to-question basis!