Question

does the equation 5(x + 4) = \frac{1}{3}(24x + 15) have one solution, no solutions, or infinitely many solutions?
we can distribute the 5 to rewrite the left side of the equation without parentheses. you can think of this as (5 \cdot x) + (5 \cdot 4). rewrite the left side.
5(x + 4)
\square + \square
=
\frac{1}{3}(24x + 15)
=
\frac{1}{3}(24x + 15)

Explanation:

Step1: Distribute 5 on left side

$5(x+4) = 5x + 20$

Step2: Distribute $\frac{1}{3}$ on right side

$\frac{1}{3}(24x+15) = 8x + 5$

Step3: Set equal, solve for x

$5x + 20 = 8x + 5$
$20 - 5 = 8x - 5x$
$15 = 3x$
$x = \frac{15}{3} = 5$

Answer:

First, the expanded left side is $\boldsymbol{5x + 20}$.
The equation has one solution, which is $\boldsymbol{x=5}$.