In the equation 12(6x - k) = 4x - (x + 5), k is a constant... | SAT Linear equations in one variable Question, Answer Explained

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Math · AlgebraHard

In the equation $\frac{1}{2}(6x - k) = 4x - (x + 5)$ , $k$ is a constant. For what value of $k$ does the equation have infinitely many solutions?

$-10$

$\frac{5}{2}$

$5$

$10$

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Answer: D

The left side simplifies to $3x - \frac{k}{2}$ . The right side simplifies to $4x - x - 5 = 3x - 5$ . The equation has infinitely many solutions when the two sides are identical, so $\frac{k}{2} = 5$ and $k = 10$ .

Choice C sets $k = 5$ , which forgets that the $\frac{1}{2}$ also multiplies $k$ .

Why the other choices are wrong

A. This drops the negative sign in front of the parentheses on the right side, which makes the constant +5 instead of -5.
B. This divides 5 by 2. Since k/2 must equal 5, k is 5 times 2.
C. This ignores the 1/2 in front of the parentheses. The constant on the left is -k/2, not -k.

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