In the xy-plane, a point (x, y) is a solution to the system of inequalities below. y \leq … — SAT

In the $xy$ -plane, a point $(x, y)$ is a solution to the system of inequalities below.

y \leq -3x + 14

y \leq 4x - 7

What is the maximum possible value of $y$ ?

Answer: 5

Answer

The maximum possible value of

y

is 5.

The maximum value of

y

is found at the intersection of the two boundary lines,

y = -3x + 14

and

y = 4x - 7

. Solving the system of equations gives

x = 3

and

y = 5

. Because the inequalities are both of the 'less than or equal to' type, any point in the solution set must lie on or below both lines. For

x < 3

, the value of

y

is limited by

y \leq 4x - 7

, which is strictly less than 5. For

x > 3

, the value of

y

is limited by

y \leq -3x + 14

, which is also strictly less than 5. Thus, the maximum value of

y

is 5.

Step-by-Step Solution

Set the two boundary equations equal to find their intersection point.

-3x + 14 = 4x - 7

The maximum value of

y

under these constraints occurs at the intersection of the two boundary lines.

Solve the equation for

x

x = 3

Isolating

x

determines the x-coordinate of the intersection point.

Substitute

x = 3

back into one of the boundary equations to solve for

y

y = 5

This yields the y-coordinate of the intersection point, which is the upper bound for

y

in the solution set.

Key Concept

Determining the maximum value of a variable in the solution set of a system of linear inequalities by finding the intersection of their boundary lines.

Estimated Time:1m 30s

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