In the linear equation below, k is a constant: \frac{x - k}{2} + \frac{2x + 1}{5} = \frac{3x - 4}{10} If x must be a positive integer, which of the following values could be k? Select all such values.

In the linear equation below, k is a constant: \frac{x - k}{2} + \frac{2x + 1}{5} = \frac{… — GRE General Test

In the linear equation below,

k

is a constant:

\frac{x - k}{2} + \frac{2x + 1}{5} = \frac{3x - 4}{10}

x

must be a positive integer, which of the following values could be

k

? Select all such values.

A
$-6$
$6$ Answer
C
$10$
$12$ Answer
E
$15$

Answer

The values of

k

that satisfy the condition are

6

and

12

Clearing denominators gives

6x = 5k - 6

, or

x = \frac{5k - 6}{6}

. For

x

to be a positive integer,

5k - 6

must be positive and divisible by

6

. This requires

k

to be a positive multiple of

6

. The options

6

and

12

satisfy both criteria, yielding

x = 4

and

x = 9

respectively.

Step-by-Step Solution

Clear denominators by multiplying the entire equation by the least common denominator (LCD), which is 10.

5(x - k) + 2(2x + 1) = 3x - 4

Eliminating fractions simplifies further algebraic manipulation.

Expand terms and combine like terms on the left side of the equation.

5x - 5k + 4x + 2 = 3x - 4 \implies 9x - 5k + 2 = 3x - 4

Distribute constants across parentheses to isolate variable expressions.

Isolate

x

in terms of

k

6x = 5k - 6 \implies x = \frac{5k - 6}{6}

Express

x

as an explicit linear function of

k

to test divisibility and positivity conditions.

Determine constraints on

k

for

x

to be a positive integer.

5k

must be divisible by

6

(so

k

must be a multiple of

6

) and

5k - 6 > 0 \implies k > 1.2

Since

5

and

6

share no common factors,

k

must be a positive multiple of

6

Evaluate the choices against the constraints.

Among the options, only

6

and

12

are positive multiples of

6

, producing

x = 4

and

x = 9

respectively.

Both values produce integer solutions greater than zero.

Key Concept

Solving single-variable linear equations with rational coefficients and applying divisibility and inequality constraints.

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