For what value of k does the equation \frac{1}{5}(kx - 3) - \frac{1}{3}(2x - 5) = 2 have a… — ACT

For what value of $k$ does the equation $\frac{1}{5}(kx - 3) - \frac{1}{3}(2x - 5) = 2$ have a solution of $x = 7$ ?

Answer: 4

The value of

k

4

Substituting

x = 7

reduces the equation to

\frac{1}{5}(7k - 3) - 3 = 2

. Adding 3 to both sides yields

\frac{1}{5}(7k - 3) = 5

. Multiplying by 5 gives

7k - 3 = 25

. Adding 3 gives

7k = 28

, which results in

k = 4

Substitute

x = 7

into the equation.

\frac{1}{5}(7k - 3) - \frac{1}{3}(2(7) - 5) = 2

Since

x = 7

is given as a solution, substituting it into the equation must make the equality true.

Evaluate and simplify the expression in the second term.

\frac{1}{3}(14 - 5) = \frac{1}{3}(9) = 3

Follow the order of operations by simplifying the expression inside the parentheses first.

Isolate the fractional term containing the variable

k

\frac{1}{5}(7k - 3) - 3 = 2 \implies \frac{1}{5}(7k - 3) = 5

Add 3 to both sides of the equation to eliminate the subtraction of 3.

Clear the fraction and solve the remaining linear equation for

k

7k - 3 = 25 \implies 7k = 28 \implies k = 4

Multiply both sides by 5 to eliminate the denominator, add 3 to isolate the term with

k

, and divide by 7.

Solving multi-step linear equations containing parameters and fractions.

Estimated Time:1m 30s

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