Chapter 1.3, Problem 35AYU

In Problems 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$\frac{2}{x+2}+\frac{3}{x-1}=\frac{-8}{5}$

To determine

To find: The solution for the equation $\frac{2}{x+2} + \frac{3}{x-1} = \frac{-8}{5}$ algebraically and verify using a graphing utility.

Answer to Problem 35AYU

Solution:

$x\text{ = − 4, }\frac{\text{−1}}{\text{8}}$

Explanation of Solution

Given:

The equation $\frac{2}{x+2} + \frac{3}{x-1} = \frac{-8}{5}$

Algebraic solution:

$\frac{2}{x+2} + \frac{3}{x-1} = \frac{-8}{5}$

$\frac{2\left(x-1\right)+3\left(x+2\right)}{\left(x+2\right)\left(x-1\right)} = \frac{-8}{5}$Take LCM on left hand side

$\frac{2x-2+3x+6}{x^2-x+2x-2} = \frac{-8}{5}$

$\frac{5x+4}{x^2+x-2} = \frac{-8}{5}$

$5\left(5x+4\right) = -8(x^2+\text{<mi>x</mi>−2})$Cross multiply both sides

$25x + 20 = -8x^2 - 8x + 16$

$25x + 20 + 8x^2 = -8x^2 - 8x + 16 + 8x^2$Add $8x^2$ to both sides

$8x^2 + 25x + 20 = \text{− <mn>8</mn><mi>x</mi>} + 16$

$8x^2 + 25x + 20 + 8x = \text{−<mn>8</mn><mi>x</mi>} + 16 + 8x$Add $8x$ to both sides

$8x^2 + 33x + 20 = 16$

$8x^2 + 33x + 20-16 = 16-16$Subtract 16 from both sides

$8x^2 + 33x + 4 = 0$

$8x^2 + 32x + x + 4 = 0$Splitting the middle term

$\text{ <mn>8</mn><mi>x</mi>}\left(x\text{ + 4}\right)\text{ + }\left(x\text{ + 4}\right)\text{ = 0}$Take $8x$ common from first two terms and group $x\text{ + 4}$

$\left(x\text{ + 4}\right)\left(8x\text{ + 1}\right)\text{ = 0}$Take $\text{(<mi>x</mi> + 4)}$ common factor

$\left(x\text{ + 4}\right)\text{ = 0 }$ or $\left(8x\text{ + 1}\right)\text{ = 0}$

$x\text{ + 4 − 4 = 0 − 4}$ or $8x\text{ + 1 − 1 = 0 − 1}$

$x\text{ = −4}$ or $8x\text{ = −1}$

$x\text{ = −4}$ or $x\text{ = }\frac{-1}{8}$

Check:

Let $x\text{ = − 4}$ in the expression in $x$ on the left side of the equation

$\frac{2}{x+2} + \frac{3}{x-1} = \frac{2}{-4+2} + \frac{3}{-4-1}$

$= \frac{2}{-2} + \frac{3}{-5}$

$= \frac{5+3}{-5}$Take LCM

$= \frac{-8}{5}$

Since the expression on the left side give $\frac{-8}{5}$ which is on the right side, the solution $x\text{ = − 4}$ checks

Let $x = \frac{-1}{8}$ on the left side of the equation

$\frac{2}{x+2} + \frac{3}{x-1} = \frac{2}{\frac{-1}{8}+2} + \frac{3}{\frac{-1}{8}-1}$

$= \frac{2}{\frac{-1+16}{8}} + \frac{3}{\frac{-1-8}{8}}$Take LCM in the denominator

$= \frac{2.8}{15} + \frac{3.8}{-9}$

$= \frac{16}{15} + \frac{24}{-9}$

$= \frac{\left(16\right)\left(-9\right)+\left(24\right)\left(15\right)}{\left(15\right)\left(-9\right)}$Take LCM

$= \frac{-144+360}{-135}$

$= \frac{216}{-135}$Divide the numerator and denominator by 3

$= \frac{72}{-45}$Divide the numerator and denominator by 9

$=\frac{-8}{5}$

Since the expression on the left side gives $\frac{-8}{5}$ which is on the right side, the solution $x = \frac{-1}{8}$checks.