Consider the linear equation in x shown below, where k is a real constant: \frac{x - k}{2}… — GRE General Test

Consider the linear equation in

x

shown below, where

k

is a real constant:

\frac{x - k}{2} - \frac{2x + 1}{3} = \frac{k(x + 1)}{6} - 1

Which of the following statements must be true? Select all such statements.

If $k = 1$ , then $x = 0$ is the unique solution to the equation.Answer
B
If $k = -1$ , the equation has infinitely many solutions.
If $k = 0$ , the solution to the equation is $x = 4$ .Answer
D
If $k = 3$ , the solution to the equation is $x = 2$ .
If $k = 2$ , the solution to the equation is $x = -\frac{4}{3}$ .Answer

Answer

The correct statements are: if

k = 1

, then

x = 0

is the unique solution; if

k = 0

, the solution is

x = 4

; and if

k = 2

, the solution is

x = -\frac{4}{3}

Multiplying the equation by

6

yields

3(x - k) - 2(2x + 1) = k(x + 1) - 6

. Expanding and collecting terms in

x

leads to

(k + 1)x = 4(1 - k)

. Testing the proposed values of

k

shows that when

k = 1

2x = 0

gives

x = 0

; when

k = 0

x = 4

; and when

k = 2

3x = -4

gives

x = -\frac{4}{3}

. Therefore, the statements corresponding to

k = 1

k = 0

, and

k = 2

are all correct.

Step-by-Step Solution

Clear fractions by multiplying the entire equation by the least common denominator,

6

3(x - k) - 2(2x + 1) = k(x + 1) - 6

Eliminating fractions simplifies algebraic expansion and grouping of like terms.

Expand all expressions and combine like terms on both sides.

3x - 3k - 4x - 2 = kx + k - 6 \implies -x - 3k - 2 = kx + k - 6

Distribute terms carefully to prevent sign errors.

Isolate terms containing

x

on one side and parameter/constant terms on the other side.

-x - kx = k + 3k - 6 + 2 \implies -x(k + 1) = 4k - 4 \implies (k + 1)x = 4(1 - k)

Factoring out

x

provides the general solution form

x = \frac{4(1 - k)}{k + 1}

for

k \neq -1

Evaluate each specified value of

k

against

(k + 1)x = 4(1 - k)

For

k = 1

2x = 0 \implies x = 0

.
For

k = -1

0x = 8 \implies

no solution.
For

k = 0

1x = 4 \implies x = 4

.
For

k = 3

4x = -8 \implies x = -2

.
For

k = 2

3x = -4 \implies x = -\frac{4}{3}

Direct substitution verifies which given statements are true.

Key Concept

Solving linear equations with parameters and analyzing existence and uniqueness of solutions.

Estimated Time:2m 30s

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