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# solution

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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
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#1
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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
+1822
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

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