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does the equation $5(x + 4) = \\frac{1}{3}(24x + 15)$ have one solution…

Question

does the equation $5(x + 4) = \frac{1}{3}(24x + 15)$ have one solution, no solutions, or infinitely many solutions?\
first, lets try to isolate the variable, $x$. right now, $x$ is inside parentheses on the left and right sides. start by looking at the left side. how could we rewrite the left side of the equation without parentheses so that its easier to work with?\
$5(x + 4) = \frac{1}{3}(24x + 15)$\
options: distribute the 5 to the $x$ and to the 4; add $x$ to each side; add 4 to each side

Explanation:

Step1: Distribute on both sides

Left side: $5(x+4) = 5x + 20$
Right side: $\frac{1}{3}(24x+15) = 8x + 5$
Equation becomes: $5x + 20 = 8x + 5$

Step2: Isolate x terms

Subtract $5x$ from both sides:
$5x + 20 - 5x = 8x + 5 - 5x$
$20 = 3x + 5$

Step3: Isolate constant terms

Subtract 5 from both sides:
$20 - 5 = 3x + 5 - 5$
$15 = 3x$

Step4: Solve for x

Divide both sides by 3:
$\frac{15}{3} = \frac{3x}{3}$
$5 = x$

Answer:

The equation has one solution, $x=5$.
First step to rewrite the left side: Distribute the 5 to the x and to the 4