Question

determine whether each ordered pair is a solution or not a solution to this system of inequalities.
$y \geq -x$
$2x + y > 2$
keyboard help
ordered pairs: (-5,6), (2,3), (0,0), (2,-2), (4,-5), (5,-6)
columns: yes, no

Explanation:

To determine if an ordered pair \((x, y)\) is a solution to the system \(

$$\begin{cases}y \geq -x \\ 2x + y > 2\end{cases}$$

\), we substitute \(x\) and \(y\) into both inequalities and check if both are satisfied.

For \((-5, 6)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = -5\), \(y = 6\):
\(6 \geq -(-5) \implies 6 \geq 5\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = -5\), \(y = 6\):
\(2(-5) + 6 = -10 + 6 = -4\).
\(-4 > 2\) (False).

Since the second inequality fails, \((-5, 6)\) is not a solution.

For \((2, 3)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 2\), \(y = 3\):
\(3 \geq -2\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 2\), \(y = 3\):
\(2(2) + 3 = 4 + 3 = 7\).
\(7 > 2\) (True).

Both inequalities hold, so \((2, 3)\) is a solution.

For \((0, 0)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 0\), \(y = 0\):
\(0 \geq 0\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 0\), \(y = 0\):
\(2(0) + 0 = 0\).
\(0 > 2\) (False).

Since the second inequality fails, \((0, 0)\) is not a solution.

For \((2, -2)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 2\), \(y = -2\):
\(-2 \geq -2\) (True, since \(-2 = -2\)).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 2\), \(y = -2\):
\(2(2) + (-2) = 4 - 2 = 2\).
\(2 > 2\) (False, since \(2 = 2\) and we need \(>\)).

Since the second inequality fails, \((2, -2)\) is not a solution.

For \((4, -5)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 4\), \(y = -5\):
\(-5 \geq -4\) (False, since \(-5 < -4\)).

The first inequality fails, so \((4, -5)\) is not a solution.

For \((5, 6)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 5\), \(y = 6\):
\(6 \geq -5\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 5\), \(y = 6\):
\(2(5) + 6 = 10 + 6 = 16\).
\(16 > 2\) (True).

Both inequalities hold, so \((5, 6)\) is a solution.

Final Classification:

• Yes (solutions): \((2, 3)\), \((5, 6)\)
• No (not solutions): \((-5, 6)\), \((0, 0)\), \((2, -2)\), \((4, -5)\)

To match the table format:

Ordered Pair	Yes/No
\((2, 3)\)	Yes
\((0, 0)\)	No
\((2, -2)\)	No
\((4, -5)\)	No
\((5, 6)\)	Yes

(If the task is to drag each pair to "Yes" or "No", then:

• Drag \((2, 3)\) and \((5, 6)\) to "Yes".
• Drag \((-5, 6)\), \((0, 0)\), \((2, -2)\), \((4, -5)\) to "No".)

Answer:

To determine if an ordered pair \((x, y)\) is a solution to the system \(

$$\begin{cases}y \geq -x \\ 2x + y > 2\end{cases}$$

\), we substitute \(x\) and \(y\) into both inequalities and check if both are satisfied.

For \((-5, 6)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = -5\), \(y = 6\):
\(6 \geq -(-5) \implies 6 \geq 5\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = -5\), \(y = 6\):
\(2(-5) + 6 = -10 + 6 = -4\).
\(-4 > 2\) (False).

Since the second inequality fails, \((-5, 6)\) is not a solution.

For \((2, 3)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 2\), \(y = 3\):
\(3 \geq -2\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 2\), \(y = 3\):
\(2(2) + 3 = 4 + 3 = 7\).
\(7 > 2\) (True).

Both inequalities hold, so \((2, 3)\) is a solution.

For \((0, 0)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 0\), \(y = 0\):
\(0 \geq 0\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 0\), \(y = 0\):
\(2(0) + 0 = 0\).
\(0 > 2\) (False).

Since the second inequality fails, \((0, 0)\) is not a solution.

For \((2, -2)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 2\), \(y = -2\):
\(-2 \geq -2\) (True, since \(-2 = -2\)).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 2\), \(y = -2\):
\(2(2) + (-2) = 4 - 2 = 2\).
\(2 > 2\) (False, since \(2 = 2\) and we need \(>\)).

Since the second inequality fails, \((2, -2)\) is not a solution.

For \((4, -5)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 4\), \(y = -5\):
\(-5 \geq -4\) (False, since \(-5 < -4\)).

The first inequality fails, so \((4, -5)\) is not a solution.

For \((5, 6)\):

Step 1: Check \(y \geq -x\)

Substitute \(x = 5\), \(y = 6\):
\(6 \geq -5\) (True).

Step 2: Check \(2x + y > 2\)

Substitute \(x = 5\), \(y = 6\):
\(2(5) + 6 = 10 + 6 = 16\).
\(16 > 2\) (True).

Both inequalities hold, so \((5, 6)\) is a solution.

Final Classification:

• Yes (solutions): \((2, 3)\), \((5, 6)\)
• No (not solutions): \((-5, 6)\), \((0, 0)\), \((2, -2)\), \((4, -5)\)

To match the table format:

Ordered Pair	Yes/No
\((2, 3)\)	Yes
\((0, 0)\)	No
\((2, -2)\)	No
\((4, -5)\)	No
\((5, 6)\)	Yes

(If the task is to drag each pair to "Yes" or "No", then:

• Drag \((2, 3)\) and \((5, 6)\) to "Yes".
• Drag \((-5, 6)\), \((0, 0)\), \((2, -2)\), \((4, -5)\) to "No".)