Look at the graph below Which part of the graph best represents the solution set to the system of inequalities y ≥ x + 1 and y + x ≥ −1

ANSWER

Part A

EXPLANATION

The given inequalities are,

[tex]y \geqslant x + 1[/tex]

and

[tex]y + x \geqslant - 1[/tex]

To see which part of the graph best represent the solution set, choose a point from each part and substitute in to the inequalities.

If a point from a given part satisfies the inequalities simultaneously, then that part best represents the solution set.

Part A.

We choose
[tex](0,2)[/tex]
We plug in to the inequalities.

[tex]2 \geqslant 0 + 1[/tex]

[tex]\Rightarrow \: 2 \geqslant 1[/tex]
The above inequality is true.

We plug in to the second inequality.

[tex]2 + 0 \geqslant - 1[/tex]

.
[tex]\Rightarrow \: 2 \geqslant - 1[/tex]

This statement is also true.

Part B.

If we plug in
[tex](-2,0)[/tex]
in to the first statement, we get,

[tex] 0 \geqslant - 2 + 1[/tex]

This implies that,

[tex]0 \geqslant - 1[/tex]

This is true.

If substitute in to the second, we get,

[tex] 0 + - 2\geqslant - 1[/tex]

[tex]\Rightarrow \: - 2 \geqslant - 1[/tex]
This is false.

Part C

We plug
[tex](0,-2)[/tex]
in to the first inequality

[tex] - 2 \geqslant 0 + 1[/tex]
This means that,

[tex] - 2 \geqslant 1[/tex]
This is false.

We plug in to the second inequality,

[tex] - 2 + 0 \geqslant -1[/tex]

[tex] - 2 \geqslant -1[/tex]

False.

Part D also has the point

[tex](2,0)[/tex]

We put this point in to the first inequality to get,

[tex]0 \geqslant 2 + 1[/tex]
[tex]0 \geqslant 3[/tex]
This is false.

Then in to the second inequality.

[tex]0 + 2 \geqslant 1[/tex]
[tex]2 \geqslant -1[/tex]

This final statement is true.

Since the point from Part A satisfies both inequalities simultaneously, it represents the solution set.