Question

writing an inequality in two variables given a table of values
x | y
-4 | -1
-2 | 4
3 | -3
3 | -4
which linear inequality could represent the given table of values?
○ $y < -2x + 3$
○ $y \leq -2x + 3$
○ $y > -\frac{1}{2}x - 3$
○ $y \leq -\frac{1}{2}x - 3$

Explanation:

Step1: Test the first option \( y < -2x + 3 \)

For \( x = -4 \), \( -2x + 3 = -2(-4)+3 = 8 + 3 = 11 \). The \( y \)-value is \( -1 \), and \( -1 < 11 \), which works. For \( x = -2 \), \( -2x + 3 = -2(-2)+3 = 4 + 3 = 7 \). The \( y \)-value is \( 4 \), and \( 4 < 7 \), which works. For \( x = 3 \), \( -2x + 3 = -2(3)+3 = -6 + 3 = -3 \). The \( y \)-values are \( -3 \) and \( -4 \). But \( -3 \) is not less than \( -3 \) (it's equal), so this option fails for \( (3, -3) \).

Step2: Test the second option \( y \leq -2x + 3 \)

For \( x = -4 \), \( -2x + 3 = 11 \), \( -1 \leq 11 \) (works). For \( x = -2 \), \( -2x + 3 = 7 \), \( 4 \leq 7 \) (works). For \( x = 3 \), \( -2x + 3 = -3 \). The \( y \)-values are \( -3 \) (which is equal, so \( -3 \leq -3 \) works) and \( -4 \leq -3 \) (works). But let's check the third and fourth options to be sure.

Step3: Test the third option \( y > -\frac{1}{2}x - 3 \)

For \( x = -4 \), \( -\frac{1}{2}x - 3 = -\frac{1}{2}(-4)-3 = 2 - 3 = -1 \). The \( y \)-value is \( -1 \), and \( -1 \) is not greater than \( -1 \) (it's equal), so this option fails for \( (-4, -1) \).

Step4: Test the fourth option \( y \leq -\frac{1}{2}x - 3 \)

For \( x = -4 \), \( -\frac{1}{2}x - 3 = -1 \), \( -1 \leq -1 \) (works). For \( x = -2 \), \( -\frac{1}{2}x - 3 = -\frac{1}{2}(-2)-3 = 1 - 3 = -2 \). The \( y \)-value is \( 4 \), and \( 4 \leq -2 \) is false, so this option fails for \( (-2, 4) \).

Answer:

\( y \leq -2x + 3 \) (Option: \( y \leq -2x + 3 \))