For what value of the constant k does the system of linear equations below have no solutio… — GMAT

For what value of the constant $k$ does the system of linear equations below have no solution?

\begin{aligned} (k - 2)x + 3y &= 6 \\ 4x + (k + 2)y &= 12 \end{aligned}

A
$-2$
B
$4$
$-4$ Answer
D
Both $-4$ and $4$
E
$16$

Answer

The constant value

k = -4

results in a system with no solution.

For a

2 \times 2

linear system

a_1 x + b_1 y = c_1

and

a_2 x + b_2 y = c_2

to have no solution, the equations must have proportional variable coefficients but non-proportional constant terms:

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

. Solving

\frac{k-2}{4} = \frac{3}{k+2}

gives

k^2 - 4 = 12

, so

k = \pm 4

. Substituting

k = -4

yields

-6x + 3y = 6

and

4x - 2y = 12

, which simplify to

-2x + y = 2

and

-2x + y = -6

. Since the constants

2

and

-6

differ, the lines are parallel and distinct, meaning the system has no solution.

Step-by-Step Solution

Set up the condition for parallel lines (equal slopes) by equating the ratio of the coefficients of

x

and

y

\frac{k - 2}{4} = \frac{3}{k + 2}

A system of two linear equations has either zero solutions (parallel, non-intersecting lines) or infinitely many solutions (coincident lines) when the slopes are equal.

Cross-multiply and solve the quadratic equation for

k

(k - 2)(k + 2) = 12 \implies k^2 - 4 = 12 \implies k^2 = 16 \implies k = 4 \text{ or } k = -4

Finding all values of

k

where the coefficient matrix determinant is zero.

Test

k = 4

in the original system.

(4 - 2)x + 3y = 6 \implies 2x + 3y = 6 \quad \text{and} \quad 4x + (4 + 2)y = 12 \implies 4x + 6y = 12

Dividing

4x + 6y = 12

2

yields

2x + 3y = 6

, which is identical to the first equation. Thus,

k = 4

yields infinitely many solutions.

Test

k = -4

in the original system.

(-4 - 2)x + 3y = 6 \implies -6x + 3y = 6 \quad \text{and} \quad 4x + (-4 + 2)y = 12 \implies 4x - 2y = 12

Simplifying both equations gives

-2x + y = 2

and

-2x + y = -6

. The lines have identical slopes but different constants, so they are parallel and distinct, producing no solution.

Key Concept

System Solvability and Linear Consistency

Estimated Time:2m 0s

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