+0

# solution

0
173
2

x+5/x-1 +x+8/x+1=8x+9/x^2-1

i tried a few times keep getting the wrong answer

Guest Sep 17, 2017
Sort:

#1
0

Solve for x :
(x + 5)/(x - 1) + (x + 8)/(x + 1) = (8 x + 9)/(x^2 - 1)

Hint: | Look for a polynomial to multiply both sides by in order to clear fractions.
Multiply both sides by x^2 - 1:
(x + 1) (x + 5) + (x - 1) (x + 8) = 8 x + 9

Hint: | Write the quadratic polynomial on the left hand side in standard form.
Expand out terms of the left-hand side:
2 x^2 + 13 x - 3 = 8 x + 9

Hint: | Move everything to the left-hand side.
Subtract 8 x + 9 from both sides:
2 x^2 + 5 x - 12 = 0

Hint: | Factor the left hand side.
The left-hand side factors into a product with two terms:
(x + 4) (2 x - 3) = 0

Hint: | Find the roots of each term in the product separately.
Split into two equations:
x + 4 = 0 or 2 x - 3 = 0

Hint: | Look at the first equation: Solve for x.
Subtract 4 from both sides:
x = -4 or 2 x - 3 = 0

Hint: | Look at the second equation: Isolate terms with x to the left hand side.
x = -4 or 2 x = 3

Hint: | Solve for x.
Divide both sides by 2:
x = -4                   or                     x = 3/2

Guest Sep 17, 2017
#2
+1794
0

$$x+\frac{5}{x}-1+x+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1$$

$$2x+\frac{5}{x}-1+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1$$

$$2x+\frac{5}{x}-0+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1$$

$$2x+\frac{5}{x}+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1$$

$$2x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1$$

$$2x+\frac{13}{x}-8x=8x+\frac{9}{{x}^{2}}-1-8x$$

$$-6x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1-8x$$

$$-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0x$$

$$-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0$$

$$-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1$$

$$-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{9}{{x}^{2}}-1-\frac{9}{{x}^{2}}$$

$$-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{0}{{x}^{2}}-1$$

$$-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=0-1$$

$$-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=-1$$

$$-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=-1+1$$

$$-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=0$$

$${x}^{2}(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1)=0\times{x}^{2}$$

$$-6{x}^{3}+\frac{13{x}^{2}}{x}-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}$$

$$-6{x}^{3}+13x-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}$$

$$-6{x}^{3}+13x-9+1{x}^{2}=0\times{x}^{2}$$

$$-6{x}^{3}+13x-9+{x}^{2}=0\times{x}^{2}$$

$$-6{x}^{3}+13x-9+{x}^{2}=0$$

$$-6{x}^{3}+{x}^{2}+13x-9=0$$

Since you cannot factor any further, the only way I know of to finish solving this equation is by graphing.

Click on the following link to view the graph: https://www.desmos.com/calculator/srwqsdcfr8

When looking at the graph, you find that x ≈ -1.669416

gibsonj338  Sep 17, 2017
edited by gibsonj338  Sep 17, 2017

### 9 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details