Question

1. if \\(\frac{2x}{3} + \frac{y}{5} = 8\\), how would you solve for \\(y\\)?

\\(y = 8 + \frac{2x}{5}\\)
\\(y = \frac{40}{3x} + 8\\)
\\(y = 40 - \frac{10x}{3}\\)
\\(y = \frac{8 - \frac{2x}{3}}{5}\\)

Explanation:

Step1: Eliminate denominators

Multiply all terms by 15 (LCM of 3 and 5): $15 \cdot \frac{2x}{3} + 15 \cdot \frac{y}{5} = 15 \cdot 8$ → $10x + 3y = 120$

Step2: Isolate term with y

Subtract $10x$ from both sides: $3y = 120 - 10x$

Step3: Solve for y

Divide by 3: $y = \frac{120 - 10x}{3} = 40 - \frac{10x}{3}$

Answer:

$y = 40 - \frac{10x}{3}$