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How do you solve $\frac{x - 2}{3 x + 5} = \frac{6}{x - 3}$ $(x-2)/(3x+5)=6/(x-3)$ and check for extraneous solutions?

1 Answer

Nov 9, 2016

$x = 24 or x = - 1$ $x = 24 " or " x = -1$

Explanation:

Identify the restrictions on $x$ $x$ before you even solve the equation.
The denominators may not be equal to 0.

$3 x + 5 \neq 0 and x - 3 \neq 0$ $3x +5 != 0" and "x-3!=0$
$3 x \neq - 5 \times \times \times \times \times x x \neq 3$ $3x !=-5color(white)(xxxxxxxxxxx)x!=3$
$x \neq - \frac{5}{3}$ $x != -5/3$

As there is one fraction on each side of the equal sign you can cross -multuiply.

$(x - 2) (x - 3) = 6 (3 x + 5)$ $(x-2)(x-3)= 6(3x+5)$

$x^{2} - 3 x - 2 x + 6 = 18 x + 30 \leftarrow$ $x^2 -3x-2x+6 = 18x+30" "larr$ a quadratic, so set =0

$x^{2} - 5 x - 18 x + 6 - 30 = 0$ $x^2 -5x-18x+6-30= 0$

$x^{2} - 23 x - 24 = 0 \leftarrow$ $x^2 -23x-24=0" "larr$ find factors of 24 which differ by 23

$(x - 24) (x + 1) = 0$ $(x-24)(x+1)=0$

$x = 24 or x = - 1$ $x = 24 " or " x = -1$

Both solutions are valid.

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