In the xy-plane, a point with coordinates (a, b) lies in the solution set of the system of… — SAT

In the $xy$ -plane, a point with coordinates $(a, b)$ lies in the solution set of the system of inequalities below:

\begin{aligned} y &\ge -2x + 8 \\ y &< \frac{1}{2}x + 3 \end{aligned}

If $a$ and $b$ are integers, what is the minimum possible value of $a$ ?

The correct value is 3. Substituting

x = 3

into the system gives

y \ge 2

and

y < 4.5

. Since the coordinate

b

must be an integer,

b

can be 2, 3, or 4, showing that a valid integer coordinate pair exists. For any integer

x < 3

, there are no integer solutions for

y

Find the intersection point of the boundary lines of the system of inequalities by setting their corresponding equations equal.

-2x + 8 = \frac{1}{2}x + 3 \implies 5 = 2.5x \implies x = 2

This determines the x-coordinate where the boundaries of the two inequality regions intersect.

Evaluate the system of inequalities at the intersection point

x = 2

y \ge -2(2) + 8 \implies y \ge 4

and

y < \frac{1}{2}(2) + 3 \implies y < 4

. This yields

4 \le y < 4

, which has no solution.

This checks if any valid solution exists at the boundary intersection point.

Test consecutive integer values of

x

greater than 2 to find the first value that allows at least one integer

y

Substituting

x = 3

gives

y \ge -2(3) + 8 \implies y \ge 2

and

y < \frac{1}{2}(3) + 3 \implies y < 4.5

. The integers that satisfy

2 \le y < 4.5

are

y = 2, 3,

and

4

Since

a = 3

is the smallest integer greater than 2 that yields integer solutions for

y

, it is the minimum possible value.

Identifying integer coordinate solutions in systems of linear inequalities

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