In the system of linear equations below, c is a constant. \begin{aligned} 2x + 3y &= 12 \\… — SAT

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

\begin{aligned} 2x + 3y &= 12 \\ cx - y &= 6 \end{aligned}

If the system has a unique solution $(x, y)$ such that $x > 0$ and $y < 0$ , which of the following could be the value of $c$ ?

A
$-2$
B
$-1$
$\frac{1}{2}$ Answer
D
$2$

Answer

The option containing the value

\frac{1}{2}

is correct.

The correct answer is the option containing the value

\frac{1}{2}

. Solving the system of linear equations in terms of the constant

c

gives

x = \frac{30}{2+3c}

and

y = \frac{12c-12}{2+3c}

. For the solution

(x, y)

to lie in the fourth quadrant, we require

x > 0

and

y < 0

. The condition

x > 0

is satisfied when

2+3c > 0

, which simplifies to

c > -\frac{2}{3}

. Using this result, the condition

y < 0

requires the numerator of

y

to be negative, so

12c - 12 < 0

, which simplifies to

c < 1

. Combining these inequalities yields the interval

-\frac{2}{3} < c < 1

. The only value among the given options that falls within this interval is

\frac{1}{2}

Step-by-Step Solution

Express

y

in terms of

x

and the constant

c

using the second equation.

y = cx - 6

This allows for substitution into the first equation to solve for

x

Substitute the expression for

y

into the first equation and solve for

x

2x + 3(cx - 6) = 12 \implies (2 + 3c)x - 18 = 12 \implies (2 + 3c)x = 30 \implies x = \frac{30}{2 + 3c}

This isolates the variable

x

in terms of the constant

c

Substitute the expression for

x

back into the equation for

y

to express

y

in terms of

c

y = c\left(\frac{30}{2 + 3c}\right) - 6 = \frac{30c - 6(2 + 3c)}{2 + 3c} = \frac{12c - 12}{2 + 3c}

This isolates the variable

y

in terms of the constant

c

Apply the condition

x > 0

to find the constraint on

c

Since

x = \frac{30}{2 + 3c} > 0

and the numerator is positive, the denominator must also be positive:

2 + 3c > 0 \implies c > -\frac{2}{3}

This determines the lower bound for the constant

c

Apply the condition

y < 0

to find the constraint on

c

Since

y = \frac{12c - 12}{2 + 3c} < 0

and the denominator

2 + 3c

is positive, the numerator must be negative:

12c - 12 < 0 \implies c < 1

This determines the upper bound for the constant

c

Combine the inequalities to find the complete range for

c

and identify the matching option.

-\frac{2}{3} < c < 1

. The only value among the options that lies in this interval is

\frac{1}{2}

This identifies the correct option based on the mathematical constraints.

Key Concept

Solving systems of linear equations with parameters and applying quadrant boundary constraints.

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