Question

What is the simplified form of the quantity of x plus 8, all over the quantity of 3x plus 7 + the quantity of x plus 7, all over the quantity of x plus 4?
x+8/3x+7 + x+7/x+4
PLEASE help!
I don't understand it at all!

Answer 1

$\frac{x+8}{3x+7} + \frac{x+7}{x+4} \\ = \frac{(x+8)(x+4)+(x+7)(3x+7)}{(3x+7)(x+4)} \\ = \frac{ x^{2} +12x+32+3 x^{2} +28x+49}{3 x^{2} +19x+28} \\ = \frac{4 x^{2} +40x+81}{3 x^{2} +19x+28}$

Answer 2

Answer:

$\frac{4x^2+40x+81}{3x^2+19x+28}$

Step-by-step explanation:

We have been given an expression $\frac{x+8}{3x+7}+\frac{x+7}{x+4}$. We are asked to find the simplified form of our given expression.

Since the denominators of our given expression is not equal, so we need to make a common denominator to add our given expression as:  

$\frac{(x+8)(x+4)}{(3x+7)(x+4)}+\frac{(x+7)(3x+7)}{(x+4)(3x+7)}$

Using FOIL we will get,  

$\frac{x*x+4x+8x+8*4}{(3x*x+3x*4+7x+7*4}+\frac{x*3x+7x+7*3x+7*7}{x*3x+7x+4*3x+7*4}$

$\frac{x^2+12x+32}{3x^2+12x+7x+28}+\frac{3x^2+7x+21x+49}{3x^2+7x+12x+28}$  

$\frac{x^2+12x+32}{3x^2+19x+28}+\frac{3x^2+28x+49}{3x^2+19x+28}$  

Since he denominators are equal, so we can add numerators as:

$\frac{x^2+12x+32+3x^2+28x+49}{3x^2+19x+28}$  

$\frac{4x^2+40x+81}{3x^2+19x+28}$  

Therefore, the sum of our given expression would be $\frac{4x^2+40x+81}{3x^2+19x+28}$.