Take each answer choice $(x, y)$ and plug $x$ and $y$ into $y \ge 2x - 3$. Decide if the statement is true or false for each point.

For any point that works in the first inequality, substitute into $y < x + 1$. Be careful with the direction of the inequality sign and remember that $<$ does not allow equality.

Cross out any point that fails one of the inequalities. The remaining point (if you checked carefully) will be the one in the solution set.

In Desmos, enter y >= 2x - 3. You should see the line $y = 2x - 3$ with the region above and on the line shaded.

Enter y < x + 1. This will show the line $y = x + 1$ with the region below but not on that line shaded.

Look for the part of the plane where the two shaded regions overlap. That overlapping region represents all points that satisfy both inequalities.

Type each answer choice as a point in Desmos, like (2, 0), (1, -1), (0, 2), and (3, 5). See which point lies inside the overlapping shaded region; that point is in the solution set.

A point $(x, y)$ is in the solution set of a system of inequalities only if it makes every inequality in the system true when you substitute $x$ and $y$ into each one. Here, each answer choice is a point; we just need to test them one by one.

The first inequality is $y \ge 2x - 3$.

For each point, plug in its $x$ and $y$ values:

• For $(2, 0)$: check if $0 \ge 2(2) - 3$.
• For $(1, -1)$: check if $-1 \ge 2(1) - 3$.
• For $(0, 2)$: check if $2 \ge 2(0) - 3$.
• For $(3, 5)$: check if $5 \ge 2(3) - 3$.

Decide for each point whether this inequality is true or false. Any point that makes this false cannot be in the solution set.

Now use the second inequality: $y < x + 1$.

For any point that passed the first inequality, plug into this one:

• Use its $x$ and $y$ values and check if $y < x + 1$ is true.

Remember: the symbol $<$ means strictly less than, so if $y$ equals $x + 1$, it does not satisfy this inequality.

Compare your results:

• Some points fail the first inequality $y \ge 2x - 3$ and are eliminated.
• Among the points that satisfy the first inequality, check which also satisfy $y < x + 1$.

The only point that makes both inequalities true is $(1, -1)$, so $(1, -1)$ is in the solution set.