SOLUTION: I need help in solving these two math problems. Here is the first one. Add. simplify if possible: 9u/u2-49 + u/u-7 This is adding here no division is included. That's just wher

Question 68970: I need help in solving these two math problems. Here is the first one.
Add. simplify if possible: 9u/u2-49 + u/u-7 This is adding here no division is included. That's just where it is underline top and bottom portions.
Next problem is:
Subtract. simplify if possible: a/a^2+5a+6 - 2/a^2+3a+2 There is no dividing with this one either. Just the underline top and bottom portions.
I will greatly appreciate if you can help with these two problems.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a/a^2+5a+6 - 2/a^2+3a+2
Fractions can only be subtracted if they have the same denominator.
Factor each denominator to get;
=a/[(a+3)(a+2)] - 2/[(a+1)(a+2)]
The least common denominator (lcd) is (a+1)(a+2)(a+3)
Rewrite each fraction with the lcd as its denominator:
a(a+1)/lcd - 2(a+3)/lcd
Since they have the same denominator subtract the numerators
and write over the lcd, as follows:
=[a^2+a-2a-6]/lcd
=(a^2-a-6)/lcd
Factor to get:
=[(a-3)(a+2)]/[(a+1)(a+2)(a+3)]
Cancel the common factor of (a+2) to get:
=(a-3)/[(a+1)(a+3)]
-------------------------------
9u/u2-49 + u/u-7
=9u/[(u-7)(u+7)] + u/(u-7)
The lcd is (u-7)(u+7)
Rewrite the fractions with the lcd as their denominator:
=[9u + u(u+7)]/lcd
=(u^2+16u]/lcd
=(u(u+16))/[(u-7)(u+7)]
Cheers,
Stan H.

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