How do you subtract $(-4)/(9-x^2)-(2x+1)/(x^2-3x)$ ?

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Answer 1 · 2016-08-04T12:07:27

Here's what I got.

The first thing to do here is make sure that the two fractions have the same denominator.

Notice that the first denominator can be written as

$9 - x^2 = 3^2 - x^2 = (3-x) * (3+x)$

The second denominator can be written as

$x^2 - 3x = x * (x - 3)$

Now, you can rewrite this as

$x * (x-3) = - x * (3 - x)$

since

$-x * (3 -x ) = -3x - x * (-x) = x^2 - 3x$

This means that the second fraction can be rewritten s

$(2x+1)/(x^2 - 3x) = (2x + 1)/(-[x * (3-x)]) = - (2x+1)/(x (3-x))$

The original expression becomes

$-4/((3-x) * (3 + x)) - [ - (2x+1)/(x (3-x))]$

$-4/((3-x)(3 + x)) + (2x + 1)/(x(3-x))$

Now, in order to find the common denominator, you must multiply the first fraction by $1 = x/x$ and the second fraction by $1 = (3+x)/(3+x)$ .

This will get you

$-4/((3-x)(3+x)) * x/x + (2x+1)/(x(3-x)) * (3+x)/(3+x)$

$-(4x)/(x(3-x)(3+x)) + ((2x+1)(3+x))/(x(3-x)(3+x))$

Now you're ready to focus on the numerators. You have

$-4x + (2x+1)(3+x)$

$-4x + (6x + 2x^2 + 3 + x)$

$2x^2 + 3x + 3$

Put this back into the resulting expression to get

$(2x^2 + 3x + 3)/(x(3-x)(3+x))$

Keep in mind that you need to have $x != {0, +- 3}$ .

How do you subtract (-4)/(9-x^2)-(2x+1)/(x^2-3x)(-4)/(9-x^2)-(2x+1)/(x^2-3x)?