which ordered pair makes both inequalities true?\ny > -2x + 3\ny ≤ x - 2\n(0,0)\n(0,-1)\n(1,1)

Explanation:

Step1: Test (0,0)

For $y > - 2x+3$, substitute $x = 0$ and $y = 0$: $0>-2\times0 + 3$, i.e., $0>3$ (False). For $y\leq x - 2$, substitute $x = 0$ and $y = 0$: $0\leq0 - 2$, i.e., $0\leq - 2$ (False).

Step2: Test (0,-1)

For $y > - 2x+3$, substitute $x = 0$ and $y=-1$: $-1>-2\times0 + 3$, i.e., $-1>3$ (False). For $y\leq x - 2$, substitute $x = 0$ and $y=-1$: $-1\leq0 - 2$, i.e., $-1\leq - 2$ (False).

Step3: Test (1,1)

For $y > - 2x+3$, substitute $x = 1$ and $y = 1$: $1>-2\times1+3$, i.e., $1>1$ (False). For $y\leq x - 2$, substitute $x = 1$ and $y = 1$: $1\leq1 - 2$, i.e., $1\leq - 1$ (False).

Since no given ordered - pair makes both inequalities true, there is an error in the problem - setup or the options provided. If we assume there are other options not shown, we would continue testing ordered pairs by substituting the $x$ and $y$ values into the two inequalities $y > - 2x+3$ and $y\leq x - 2$ one by one.

If we assume the problem has a solution among the options after re - checking the inequalities and options, we note that we can also solve the system of inequalities graphically. The solution of the system $y > - 2x+3$ is the region above the line $y=-2x + 3$ (dashed line as the inequality is strict) and the solution of $y\leq x - 2$ is the region below or on the line $y=x - 2$ (solid line as the inequality is non - strict). The intersection of these two regions gives the set of ordered pairs that satisfy both inequalities.

However, based on the given options:

Answer:

None of the given options (0,0), (0, - 1), (1,1) make both inequalities true.