Question

1. which ordered pair is a solution of the system of linear inequalities shown?

\

$$\begin{cases} 2x + 3y > 12 \\\\ x - y \\leq 1 \\end{cases}$$

\bigcirc (0, 5)
\bigcirc (0, 4)
\bigcirc (5, 0)
\bigcirc (5, 2)

Explanation:

Step1: Test (0,5)

Substitute \(x = 0\), \(y = 5\) into \(2x + 3y>12\): \(2(0)+3(5)=15>12\), true.
Substitute into \(x - y\leq1\): \(0 - 5=-5\leq1\), true. Wait, but let's check others too.

Step2: Test (0,4)

Substitute into \(2x + 3y>12\): \(2(0)+3(4)=12\), but \(12
ot>12\), false.

Step3: Test (6,0)

Substitute into \(2x + 3y>12\): \(2(6)+3(0)=12\), \(12
ot>12\), false.

Step4: Test (5,2)

Substitute into \(x - y\leq1\): \(5 - 2 = 3\), \(3
ot\leq1\), false. Wait, earlier (0,5) seemed to work? Wait, no, wait the problem's options—wait maybe I made a mistake. Wait, no, let's recheck (0,5): \(x - y = 0 - 5 = -5\leq1\), yes. \(2x + 3y = 15>12\), yes. But wait, let's check again. Wait, maybe the original problem's inequalities—wait, maybe I miscalculated (5,2). Wait, (5,2): \(x - y = 5 - 2 = 3\), which is not ≤1, so false. (0,5): both inequalities hold. Wait, but let's confirm again.

Wait, the system is \(

$$\begin{cases}2x + 3y>12\\x - y\leq1\end{cases}$$

\)

For (0,5):
First inequality: \(2(0)+3(5)=15>12\) ✔️
Second inequality: \(0 - 5=-5\leq1\) ✔️

For (0,4):
First inequality: \(2(0)+3(4)=12\), not >12 ❌

For (6,0):
First inequality: \(2(6)+3(0)=12\), not >12 ❌

For (5,2):
Second inequality: \(5 - 2 = 3\), not ≤1 ❌

So (0,5) is a solution. Wait, but let me check again. Wait, maybe the user made a typo, but according to the calculations, (0,5) works.

Answer:

(0, 5)