Add.
-$\frac { 2 x + 8 } { x } + \frac { 6 x - 9 } { 7 x }$
A) $\frac { 8 x + 47 } { 7 x }$
B) $\frac { 20 x - 65 } { 7 x }$
C) $\frac { 20 x + 47 } { 7 x }$
D) $\frac { 20 x + 47 } { 7 x ^ { 2 } }$

Question

Quizplus · Accepted Answer

To add these fractions, we need a common denominator. The least common multiple of the denominators is 7x. So, we convert the first fraction using the rule that $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$.
\begin{align*}
\frac{2x+8}{x}+\frac{6x-9}{7x} &= \frac{(2x+8)\cdot7x}{x\cdot7x}+\frac{(6x-9)\cdot x}{7x\cdot x}\
&= \frac{14x^2+56x}{7x^2}+\frac{6x^2-9x}{7x^2}\
&= \frac{20x^2+47x}{7x^2}\
&= \frac{20x+47}{7x} \quad\text{(after simplifying)}
\end{align*}
Therefore, the answer is C.

Add - 2x+8x+6x−97x\frac { 2 x + 8 } { x } + \frac { 6 x - 9 } { 7 x }x2x+8​+7x6x−9​

Add
- $\frac { 2 x + 8 } { x } + \frac { 6 x - 9 } { 7 x }$