In the system of equations below, a is a constant. \begin{aligned} ax + 3y &= 5 \\ 2x - y … — SAT

In the system of equations below, $a$ is a constant.

\begin{aligned} ax + 3y &= 5 \\ 2x - y &= 1 \end{aligned}

If the system has a solution $(x, y)$ that lies in Quadrant IV of the $xy$ -plane, which of the following could be the value of $a$ ?

12Answer
B
8
C
-4
D
-8

Answer

The correct answer is 12. For the solution

(x, y)

to lie in Quadrant IV, the

x

-coordinate must be positive and the

y

-coordinate must be negative. Isolating

y

in the second equation gives

y = 2x - 1

. Substituting this into the first equation yields

x = \frac{8}{a+6}

. For

x

to be positive, we must have

a > -6

. Substituting

x

back into the expression for

y

gives

y = \frac{10-a}{a+6}

. For

y

to be negative when

a > -6

, we must have

10 - a < 0

, which simplifies to

a > 10

. Among the choices, only 12 is greater than 10.

Step-by-Step Solution

Identify the signs of coordinates in Quadrant IV.

x > 0

and

y < 0

In Quadrant IV of the coordinate plane, the horizontal coordinate is positive and the vertical coordinate is negative.

Express

y

in terms of

x

from the second equation.

y = 2x - 1

This facilitates substituting

y

into the first equation to solve for

x

Substitute

y = 2x - 1

into the first equation and solve for

x

x = \frac{8}{a + 6}

To find the coordinates of the intersection point in terms of the parameter

a

Substitute

x = \frac{8}{a + 6}

back into the expression for

y

y = \frac{10 - a}{a + 6}

To express the

y

-coordinate of the intersection point in terms of the parameter

a

Apply the conditions

x > 0

and

y < 0

to solve for

a

a > 10

For

x > 0

, we need

a + 6 > 0 \implies a > -6

. For

y < 0

given

a + 6 > 0

, we need

10 - a < 0 \implies a > 10

Evaluate the options to find a value of

a

that satisfies the condition.

Among the options (12, 8, -4, -8), only 12 is greater than 10.

Key Concept

Solving systems of linear equations with parameters and applying quadrant constraints

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