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

`(4x^2+40x+81)/(3x^2+19x+28)`

Explanation:

We have been given an expression
`(x+8)/(3x+7)+(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:

`((x+8)(x+4))/((3x+7)(x+4))+((x+7)(3x+7))/((x+4)(3x+7))`

Using FOIL we will get,

`(x*x+4x+8x+8*4)/((3x*x+3x*4+7x+7*4)+(x*3x+7x+7*3x+7*7)/(x*3x+7x+4*3x+7*4)`

`(x^2+12x+32)/(3x^2+12x+7x+28)+(3x^2+7x+21x+49)/(3x^2+7x+12x+28)`

`(x^2+12x+32)/(3x^2+19x+28)+(3x^2+28x+49)/(3x^2+19x+28)`

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

`(x^2+12x+32+3x^2+28x+49)/(3x^2+19x+28)`

`(4x^2+40x+81)/(3x^2+19x+28)`

Therefore, the sum of our given expression would be
`(4x^2+40x+81)/(3x^2+19x+28)`.

Answer 2

`(x+8)/(3x+7) + (x+7)/(x+4) \\ = ((x+8)(x+4)+(x+7)(3x+7))/((3x+7)(x+4)) \\ = ( x^(2) +12x+32+3 x^(2) +28x+49)/(3 x^(2) +19x+28) \\ = (4 x^(2) +40x+81)/(3 x^(2) +19x+28)`