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# Help! I can't seem to remember how to solve these types of equations anymore

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Find all values of $$x$$ such that $$\dfrac{x}{x+4} = -\dfrac{9}{x+3}$$. If you find more than one value, then list your solutions in increasing order, separated by commas.

Oct 18, 2018

#1
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First of all we eill find domain of equation. x ≠ -4, x ≠  -3 because if x = -4,-3 we will have in denominator 0

After this we start

$$(x+3)(x) = -(x+4)(9) <=>$$

$$<=> x^2+3x = -9x -36 <=>$$

$$<=> x^2 + 12x + 36 = 0 <=>$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

x= -12/2 = -6 and

$$-6≠ -3,-6≠ -4$$

so $$x=-6$$ is the answer.

Hope it helps!

Oct 18, 2018

#1
+322
+2

First of all we eill find domain of equation. x ≠ -4, x ≠  -3 because if x = -4,-3 we will have in denominator 0

After this we start

$$(x+3)(x) = -(x+4)(9) <=>$$

$$<=> x^2+3x = -9x -36 <=>$$

$$<=> x^2 + 12x + 36 = 0 <=>$$

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

x= -12/2 = -6 and

$$-6≠ -3,-6≠ -4$$

so $$x=-6$$ is the answer.

Hope it helps!

Dimitristhym Oct 18, 2018