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Algebra

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Find the product CD of the integers C and D for which
$\frac{C}{x-3}+\frac{D}{x+8}=\frac{4x-25}{x^2+5x-24}$
for all real values of x except -8 and 3.

Jun 19, 2022

1+0 Answers

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Multiplying the entire equation by $$x^2 + 5x - 24$$ gives us: $$(x + 8)C + (x-3)D = 4x - 25$$. Note that $$x^2 + 5x - 24 = (x - 3) ( x+8)$$

Expanding the left-hand side gives us: $$Cx + 8C + Dx - 3D = 4x - 25$$

Note that the only 2 terms with x in them must sum to 4x, so we have the equation: $$Cx + Dx = 4x$$, meaning $$C + D = 4$$

Also note that the only 2 constants must sum to -25, giving us the equation: $$8C - 3D = -25$$.

Now, we have the following system:

$$8C - 3D = -25$$

$$C + D = 4$$

However, solving this system yields $$C = -{13 \over 11}$$ and $$D = {57 \over 11}$$

Thus... there are no integer solutions, and no product as well....

Here is the WA link: https://www.wolframalpha.com/input?i2d=true&i=Divide%5BC%2Cx+-+3%5D+%2B+Divide%5BD%2Cx+%2B+8%5D+%3D+Divide%5B4x+-+25%2CPower%5Bx%2C2%5D%2B5x+-+24%5D

Jun 19, 2022