These four are all denominators of fractions. I multiply each numerator by the denominator that it doesn't have and my answer is wrong....
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5 & x are factors of 5x, so it is (5x)(x + 2)
Sorry, a guy at work was talking to me so I had (2 + 2), should have been (x + 2)
I don't know your numerators, so I will use fake ones.
First term:
A/x. You need this to be over (5x)(x + 2). You already have the x, so multiply top & bottom by 5(x + 2) [note: you multiply top & bottom by same thing, that is same as multiplying by 1 so doesn't change your number]
A/x * [5(x + 2)]/[5(x + 2)] = [5A(x + 2)]/[(5x)(x + 2)]
Notice you ended up with (5x)(x + 2) on bottom. Now continue with rest.
Second term:
B/(x + 2), This one your denom is missing the 5x, so
B/(x + 2) * 5x/5x = 5Bx/[(5x)(x + 2)]
Third term (I won't explain in as much detail from here, but hopefully you see the point):
C/5 * [x(x + 2)]/[x(x + 2)] = [Cx(x + 2)]/[(5x)(x + 2)]
Forth term:
D/(5x) * (x + 2)/(x + 2) = [D(x + 2)]/[(5x)(x + 2)]
Now all are over (5x)(x + 2) so you can just get rid of that (by multiplying all tops by same thing) to get
5A(x + 2) - 5Bx = Cx(x + 2) + D(x + 2)
Sorry, a guy at work was talking to me so I had (2 + 2), should have been (x + 2)
I don't know your numerators, so I will use fake ones.
First term:
A/x. You need this to be over (5x)(x + 2). You already have the x, so multiply top & bottom by 5(x + 2) [note: you multiply top & bottom by same thing, that is same as multiplying by 1 so doesn't change your number]
A/x * [5(x + 2)]/[5(x + 2)] = [5A(x + 2)]/[(5x)(x + 2)]
Notice you ended up with (5x)(x + 2) on bottom. Now continue with rest.
Second term:
B/(x + 2), This one your denom is missing the 5x, so
B/(x + 2) * 5x/5x = 5Bx/[(5x)(x + 2)]
Third term (I won't explain in as much detail from here, but hopefully you see the point):
C/5 * [x(x + 2)]/[x(x + 2)] = [Cx(x + 2)]/[(5x)(x + 2)]
Forth term:
D/(5x) * (x + 2)/(x + 2) = [D(x + 2)]/[(5x)(x + 2)]
Now all are over (5x)(x + 2) so you can just get rid of that (by multiplying all tops by same thing) to get
5A(x + 2) - 5Bx = Cx(x + 2) + D(x + 2)