# algebra

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how do you solve this?

x+4/x^2-x-6 + 3x-2/x^2+3x+2

• algebra -

We have 2 fractions here. The first thing you want to do is equal the entire expression to 0 (in order to solve for x).

So we get that:

((x+4)/(x^2-x-6))+ ((3x-2)/(x^2+3x+2))=0

Now, we can transfer one of the fractions to the other side of the equation. We get that:

((x+4)/(x^2-x-6)) = ((2-3x)/(x^2+3x+2))

Now, if we mulitply each of the denominators of the fractions with the numerators at the other side of the equation, we get that:

(x+4)*(x^2+3x+2) = (2-3x)*(x^2-x-6)
When we use the distributive rule, we find that:

x^3+3x²+2x+4x²+12x+8=2x²-2x-12-3x^3+3x²+18x
=> x^3+7x²+14x+8=-3x^3+5x²+16x-12

Now, if we transfer one side to the other, we get that:

4x^3+2x²-2x+20=0
if we divide both sides by 2, we get that:

2x^3+x²-x+10=0

Now, I don't know if you're familiar with factoring a polynomial, but that is in fact the best way to go. I won't go into depth into this subject, but when you factor this polynomial, you get the equation:

(x+2)*(2x²-3x+5)=0

When we calculate the discriminant for the second factor in the equation, we find that it equals -31, so its negative. This means that the second factor (2x²-3x+5) has no further solutions. So, the only time this equation becomes zero is when x=-2 (because the first factor of the equation (x+2) becomes zero when x=-2).

So -2 is a solution for this equation. Our work doesn't stop here. Since we had some polynomials in denominators in thr original equation, we have to check that the denominators don't become zero for x=-2. Otherwise, -2 isn't a solution for the equation. When we fill in -2 in both denominators, we notice they both become zero. So, since -2 isn't a solution for the equation after all, we find that the equation has no solutions.

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