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$\frac{4 x}{x + 3} = \frac{37}{x^{2} - 9} - 3$ $(4x)/(x+3) = (37)/(x^2 -9) -3$

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Answer 1 · 2018-06-17T17:21:29

$x = - \frac{16}{7} and x = 4$ $color(blue)(x=-16/7 " and " x=4)$

Move all terms to the $L H S$ $LHS$ :

$\frac{4 x}{x + 3} - \frac{37}{x^{2} - 9} + 3 = 0$ $(4x)/(x+3)-37/(x^2-9)+3=0$

Factor denominator of second term:

$(x^{2} - 9) = (x^{2} - 3^{2}) = (x + 3) (x - 3)$ $(x^2-9)=(x^2-3^2)=(x+3)(x-3)$ ( difference of two squares )

$\frac{4 x}{x + 3} - \frac{37}{(x + 3) (x - 3)} + 3 = 0$ $(4x)/(x+3)-37/((x+3)(x-3))+3=0$

Add $L H S$ $LHS$ :

$\frac{4 x (x - 3) - 37 + 3 (x + 3) (x - 3)}{(x + 3) (x - 3)} = 0$ $(4x(x-3)-37+3(x+3)(x-3))/((x+3)(x-3))=0$

Since the denominator can't be zero, we concentrate on the numerator:

$4 x (x - 3) - 37 + 3 (x + 3) (x - 3) = 0$ $4x(x-3)-37+3(x+3)(x-3)=0$

Expand this:

$4 x^{2} - 12 x - 37 + 3 x^{2} - 27 = 0$ $4x^2-12x-37+3x^2-27=0$

Collecting like terms:

$7 x^{2} - 12 x - 64 = 0$ $7x^2-12x-64=0$

Factor:

$(7 x + 16) (x - 4) = 0 \Rightarrow x = - \frac{16}{7} and x = 4$ $(7x+16)(x-4)=0=>x=-16/7 and x=4$