Answer:
[tex]\frac{4x^2+40x+81}{3x^2+19x+28}[/tex]
Step-by-step explanation:
We have been given an expression [tex]\frac{x+8}{3x+7}+\frac{x+7}{x+4}[/tex]. 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:
[tex]\frac{(x+8)(x+4)}{(3x+7)(x+4)}+\frac{(x+7)(3x+7)}{(x+4)(3x+7)}[/tex]
Using FOIL we will get,
[tex]\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}[/tex]
[tex]\frac{x^2+12x+32}{3x^2+12x+7x+28}+\frac{3x^2+7x+21x+49}{3x^2+7x+12x+28}[/tex]
[tex]\frac{x^2+12x+32}{3x^2+19x+28}+\frac{3x^2+28x+49}{3x^2+19x+28}[/tex]
Since he denominators are equal, so we can add numerators as:
[tex]\frac{x^2+12x+32+3x^2+28x+49}{3x^2+19x+28}[/tex]
[tex]\frac{4x^2+40x+81}{3x^2+19x+28}[/tex]
Therefore, the sum of our given expression would be [tex]\frac{4x^2+40x+81}{3x^2+19x+28}[/tex].
