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
$4$
$-4$ Cevap
D
Both $-4$ and $4$
E
$16$

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

System Solvability and Linear Consistency

Tahmini Süre:2m 0s

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