Question

Simplify the sum. State any restrictions on the variables. help with work
(x+2)/(x+3) + (10x)/(x^2-9)

Answer 1

Look to factor out (x^2-9) first because it is a difference of two squares. What does it factor out to?

Answer 2

You want to form instead of a sum of terms on the left-hand side of the equation, a single ratio (or fraction) of polynomials.

Answer 3

Therefore looking for a common denominator polynomial is what you want to do.

Answer 4

the problem is that you need both denominators to be the same

Answer 5

Note that you can factor x^2 - 9 into (x + 3)(x - 3).

Answer 6

So we have
\[\frac{ x-2 }{ x+3 }+\frac{ 10x }{ \left( x-3 \right) \left( x+3 \right)}\]

Answer 7

Now we are missing a (x -3) on the left fraction so multiply top and bottom by (x -3) and get:

Answer 8

\[\frac{ \left( x-2 \right)\left( x-3 \right) }{ \left( x-3 \right) \left( x+3 \right)}+\frac{ 10x }{\left( x-3 \right) \left( x+3 \right)}\]

Answer 9

So, add the fraction together now that the denominators are the same:

Answer 10

\[\frac{ \left( x-2 \right)\left( x-3 \right)+10x }{ \left( x-3 \right)\left( x+3 \right) }\]

Answer 11

FOIL out the parentheses and simplfy the top and you should be done with that part.

Answer 12

For the restrictions, you need to figure out when the denominator might equal 0, because that is not allowed. You can't divide by 0.

Answer 13

So, (x+3) cannot be zero, which means x cannot be -3.
But, x^2 -9 cannot be zero either, which means x^2 cannot be 9 so x cannot be +3 or -3.