Question

which ordered pair makes both inequalities true?

$y \leq -x + 1$
$y > x$

graph of the inequalities with shaded regions

Explanation:

To solve this, we need to test ordered pairs in both inequalities \( y \leq -x + 1 \) and \( y > x \). Let's assume common test points (though the graph shows the overlapping region is where \( y > x \) and \( y \leq -x + 1 \) intersect, typically in the purple region). For example, take \((-2, 2)\):

Step 1: Test \( y \leq -x + 1 \)

Substitute \( x = -2 \), \( y = 2 \):
\( 2 \leq -(-2) + 1 \)
\( 2 \leq 2 + 1 \)
\( 2 \leq 3 \) (True)

Step 2: Test \( y > x \)

Substitute \( x = -2 \), \( y = 2 \):
\( 2 > -2 \) (True)

Another example: \((-1, 1)\):

• \( y \leq -x + 1 \): \( 1 \leq -(-1) + 1 = 2 \) (True)
• \( y > x \): \( 1 > -1 \) (True)

The overlapping region (purple) contains points where \( x \) is negative (since \( y > x \) and \( y \leq -x + 1 \) intersect left of the y - axis). A valid ordered pair is \(\boldsymbol{(-2, 2)}\) (or other points in the purple region).

Answer:

\((-2, 2)\) (or any ordered pair in the overlapping region of \( y > x \) and \( y \leq -x + 1 \), e.g., \((-1, 1)\), \((-3, 3)\) etc.)