How do you solve $\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{9}{x + 1}$ $\frac{3}{2x-4}-\frac{2}{x^{2}-x-2}=\frac{9}{x+1}$ ?

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How do you solve $\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{9}{x + 1}$ $\frac{3}{2x-4}-\frac{2}{x^{2}-x-2}=\frac{9}{x+1}$ ?

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Answer 1 · 2018-04-19T17:52:45

$x = \frac{7}{3}$ $x=7/3$

Explanation:

First we factor the denominators, then find a common denominator so we can subtract, then cross multiply and solve the resulting polynomial equation.

$\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{9}{x + 1}$ $frac{3}{2x-4} - frac{2}{x^2 - x -2 } = frac{9}{x+1}$

$\frac{3}{2 (x - 2)} - \frac{2}{(x + 1) (x - 2)} = \frac{9}{x + 1}$ $frac{3}{2(x-2)} - frac{2}{(x+1)(x-2)} = frac{9}{x+1}$

$\frac{3 (x + 1) - 2 (2)}{2 (x + 1) (x - 2)} = \frac{9}{x + 1}$ $frac{3(x+1) - 2(2)}{2(x+1)(x-2)} = frac{9}{x+1}$

$\frac{3 x - 1}{2 (x + 1) (x - 2)} = \frac{9}{x + 1}$ $frac{3x-1}{2(x+1)(x-2)} = frac{9}{x+1}$

$(3 x - 1) (x + 1) = 18 (x + 1) (x - 2)$ $(3x-1)(x+1) = 18(x+1)(x-2)$

$(x + 1) ((3 x - 1) - 18 (x - 2)) = 0$ $(x+1)( (3x - 1) - 18(x-2) ) = 0$

$(x + 1) (- 15 x + 35) = 0$ $(x+1)(-15 x + 35) = 0$

$x = - 1$ $x = -1$ or $x = \frac{35}{15} = \frac{7}{3}$ $x = 35/15 = 7/3$

Answer 2 · 2018-04-19T18:02:21

$x = \frac{7}{3}$ $x=7/3$

Explanation:

$\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{9}{x + 1}$ $3/(2x-4)-2/(x^2-x-2)=9/(x+1)$

Factorising

$2 x - 4 = 2 (x - 2)$ $2x-4=2(x-2)$

$x^{2} - x - 2 = x^{2} - 2 x + x - 2$ $x^2-x-2=x^2-2x+x-2$

$= x (x - 2) + (x - 2)$ $=x(x-2)+(x-2)$

$= (x + 1) (x - 2)$ $=(x+1)(x-2)$

$\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{3}{2 (x - 2)} - \frac{2}{(x + 1) (x - 2)}$ $3/(2x-4)-2/(x^2-x-2)=3/(2(x-2))-2/((x+1)(x-2))$

$\frac{3}{2 (x - 2)} - \frac{2}{(x + 1) (x - 2)} = \frac{9}{x + 1}$ $3/(2(x-2))-2/((x+1)(x-2))=9/(x+1)$

$\frac{3}{2 (x - 2)} - \frac{9}{x + 1} = \frac{2}{(x + 1) (x - 2)}$ $3/(2(x-2))-9/(x+1)=2/((x+1)(x-2))$

Multiplying throughout by $(x + 1) (x - 2)$ $(x+1)(x-2)$

$\frac{3}{2} (x + 1) - 9 (x - 2) = 2$ $3/2(x+1)-9(x-2)=2$

$\frac{3}{2} x + \frac{3}{2} - 9 x + 18 = 2$ $3/2x+3/2-9x+18=2$

$(\frac{3}{2} - 9) x + (\frac{3}{2} + 18 - 2) = 0$ $(3/2-9)x+(3/2+18-2)=0$

$- \frac{15}{2} x + \frac{35}{2} = 0$ $-15/2x+35/2=0$

$- 15 x + 35 = 0$ $-15x+35=0$

$15 x = 35$ $15x=35$

$x = \frac{35}{15}$ $x=35/15$

$x = \frac{7}{3}$ $x=7/3$

Check:

lhs=
$\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{3}{2 \times \frac{7}{3} - 4} - \frac{2}{{(\frac{7}{3})}^{2} - \frac{7}{3} - 2}$ $3/(2x-4)-2/(x^2-x-2)=3/(2xx7/3-4)-2/((7/3)^2-7/3-2)$

$= \frac{3}{\frac{14}{3} - 4} - \frac{2}{\frac{49}{9} - \frac{7}{3} - 2} = \frac{9}{14 - 12} - \frac{18}{49 - 21 - 18}$ $=3/(14/3-4)-2/(49/9-7/3-2)=9/(14-12)-18/(49-21-18)$

$= \frac{9}{2} - \frac{18}{10} = \frac{45}{10} - \frac{18}{10} = \frac{45 - 18}{10} = \frac{27}{10}$ $=9/2-18/10=45/10-18/10=(45-18)/10=27/10$

$r h s = \frac{9}{x + 1} = \frac{9}{\frac{7}{3} + 1} = \frac{27}{7 + 3} = \frac{27}{10}$ $rhs=9/(x+1)=9/(7/3+1)=27/(7+3)=27/10$

$l h s = r h s$ $lhs=rhs$

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How do you solve $\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{9}{x + 1}$ $\frac{3}{2x-4}-\frac{2}{x^{2}-x-2}=\frac{9}{x+1}$ ?

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How do you solve $\frac{3}{2 x - 4} - \frac{2}{x^{2} - x - 2} = \frac{9}{x + 1}$ $\frac{3}{2x-4}-\frac{2}{x^{2}-x-2}=\frac{9}{x+1}$ ?