its says solve the equation allgebraically. Check for extraneous solutions. The equation is: 4x/x+4 + 3/x-1 = 15/x^2+3x-4. can anyone help?

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its says solve the equation allgebraically. Check for extraneous solutions. The equation is: 4x/x+4 + 3/x-1 = 15/x^2+3x-4. can anyone help?

Mathematics
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Can you put parenthesis under the correct denominators?
4x/(x+4) + 3/(x-1) = 15/(x^2+3x-4), is that what you mean?
Yes it is. So the first thing to realize is that on the right side, under the 15, is a factorable quadratic equation, so rewrite like: \[4x/(x + 4) + 3/(x -1) = 15/((x + 4)(x -1))\] We are going to multiply by (x + 4)(x - 1) on both sides, and note where they cancel appropriately: \[4x(x-1) + 3(x + 4) = 15\] multiply everything out, collect like terms, and move everything over to one side so we can then either factor of use the quadratic equation: \[4x^2 - 4x + 3x + 12 = 15 \rightarrow 4x^2 - x - 3 = 0\] no need for the quadratic equation, this factors into: (4x + 3)(x - 1) = 0 setting each term equals to 0 yields x = -3/4 and x = 1. Be sure to plug back into your original equaiton to see that this works

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thank you so much pyeh9, that was a really big help. :)

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