Mathematics: HELP! If the area of a rectangle is 10x^2+3x-4 and the width is 2x-1, what is the side length of the rectangle?

HELP! If the area of a rectangle is 10x^2+3x-4 - id1023411037276, levicorey846

comeonmanim

27.03.2020

The solution of the equation is d=1.

Step-by-step explanation:

Given : Equation $-\frac{3d}{d^2-2d-8}+\frac{3}{d-4}=-\frac{2}{d+2}$

To find : What is the solution to the equation?

Solution :

Step 1 - Write the equation

$-\frac{3d}{d^2-2d-8}+\frac{3}{d-4}=-\frac{2}{d+2}$

Step 2 - Factor the quadratic equation and take RHS to LHS

$-\frac{3d}{(d+2)(d-4)}+\frac{3}{d-4}+\frac{2}{d+2}=0$

Step 3 - Take the Least common denominator (LCD)

$\frac{-3d+3(d+2)+2(d-4)}{(d+2)(d-4)}=0$

Step 4 - Solve the numerator and take denominator to other side

-3d+3d+6+2d-8=0

2d=2

d=1

Therefore, The solution of the equation is d=1.

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