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# What did I do wrong?

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5/(1+x) = 3 – 4/(x-1)

5/(1+x) = 3(x-1)/x-1 – 4/(x-1)

5/(1+x) = (3x– 7)/(x-1)

5(x-1)/(1+x)(x-1) = (3x– 7)(1+x)/(x-1)(1+x)

(5x-5)/(x-1+x^2-x) = (3x+3x^2-7-7x)/(x-1+x^2-x)

5x-5 = 3x^2 - 7 - 4x

(3x^2 - 9x – 12)/3 = 0

x^2 – 3x – 4 = 0

x = - (3/2) ± √ (3/2)^2 + 4

x = -1,5 ± √ 6,25

x = -1,5 ± 2.5

x1 = 1

x2 = -4

x1 = -0,21

x2 = 3,21

Where did I go wrong? :D

Guest May 6, 2017
#1
+2

Solve for x:
5/(x + 1) = 3 - 4/(x - 1)

Bring 3 - 4/(x - 1) together using the common denominator x - 1:
5/(x + 1) = (3 x - 7)/(x - 1)

Cross multiply:
5 (x - 1) = (x + 1) (3 x - 7)

Expand out terms of the left hand side:
5 x - 5 = (x + 1) (3 x - 7)

Expand out terms of the right hand side:
5 x - 5 = 3 x^2 - 4 x - 7

Subtract 3 x^2 - 4 x - 7 from both sides:
-3 x^2 + 9 x + 2 = 0

Divide both sides by -3:
x^2 - 3 x - 2/3 = 0

x^2 - 3 x = 2/3

x^2 - 3 x + 9/4 = 35/12

Write the left hand side as a square:
(x - 3/2)^2 = 35/12

Take the square root of both sides:
x - 3/2 = sqrt(35/3)/2 or x - 3/2 = -sqrt(35/3)/2

x = 3/2 + sqrt(35/3)/2 or x - 3/2 = -sqrt(35/3)/2

Answer: | x = 3/2 + sqrt(35/3)/2         or          x = 3/2 - sqrt(35/3)/2

Guest May 6, 2017
#2
+7348
+2

5/(1+x) = 3 – 4/(x-1)

x1 = -0,21

x2 = 3,21

Where did I go wrong? :D

$$\frac{5}{1+x}=3-\frac{4}{x-1}$$

$$\frac{5}{1+x}+\frac{4}{x-1}=3$$

$$\frac{5x-5+4+4x}{x^2-1}=3$$

$${9x-1}=3x^2-3$$

$$3x^2-9x-2=0$$

$$x^2-3x-\frac{2}{3}=0$$

$$x_1=1.5+\sqrt{2.25+\frac{2}{3}}$$

$$x=1.5\pm\sqrt{2.91\overline{66}}$$

$$x_1=3.2078\\x_2=-0.2078$$

!

asinus  May 6, 2017
edited by asinus  May 6, 2017
#3
+2

This is it, thank you! =)