Question

add. state the sum in simplest terms. \\(\frac{4x}{12y^2} + \frac{5x}{8y}\\) \\(\bigcirc\\ \frac{x(4 + 5y)}{20y^2}, y \
eq 0\\) \\(\bigcirc\\ \frac{23x}{24y^3}, y \
eq 0\\) \\(\bigcirc\\ \frac{9x}{20y^3}, y \
eq 0\\) \\(\bigcirc\\ \frac{x(8 + 15y)}{24y^2}, y \
eq 0\\)

Explanation:

Step1: Find the LCD of denominators

The denominators are \(12y^2\) and \(8y\). Prime factorize: \(12y^2 = 2^2\times3\times y^2\), \(8y = 2^3\times y\). LCD is \(2^3\times3\times y^2=24y^2\).

Step2: Rewrite fractions with LCD

\(\frac{4x}{12y^2}=\frac{4x\times2}{12y^2\times2}=\frac{8x}{24y^2}\), \(\frac{5x}{8y}=\frac{5x\times3y}{8y\times3y}=\frac{15xy}{24y^2}\).

Step3: Add the fractions

\(\frac{8x}{24y^2}+\frac{15xy}{24y^2}=\frac{8x + 15xy}{24y^2}=\frac{x(8 + 15y)}{24y^2}\) (factor out \(x\)).

Answer:

\(\frac{x(8 + 15y)}{24y^2}, y
eq0\) (the fourth option: \(\boldsymbol{\frac{x(8 + 15y)}{24y^2}, y
eq0}\))