Question

does the equation 5(x + 4) = \frac{1}{3}(24x + 15) have one solution, no solutions, or infinitely many solutions?
5(x + 4)
5x + 20
5x + 20
5x + 20
=
=
=
\frac{1}{3}(24x + 15)
\frac{1}{3}(24x + 15)
8x + 5
one solution
infinitely many solutions
no solutions

Explanation:

Step1: Expand left-hand side

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

Step2: Expand right-hand side

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

Step3: Rearrange to isolate x terms

$5x + 20 = 8x + 5$
$20 - 5 = 8x - 5x$

Step4: Solve for x

$15 = 3x$
$x = \frac{15}{3} = 5$

Answer:

One solution