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How would I go by simplifying this: $$\frac{a-b}{2}+\frac{a+b}{3}-\frac{b-a}{4}$$ Also, this: $$\frac{a^2-16^2}{2a+8b}$$

Tried looking around, but letters in equations just fuzzles me.

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Should that $16^2$ be $16b^2$? –  Gerry Myerson Nov 11 '13 at 9:10
    
For the first, try putting everything over a common denominator and then adding the numerators. For the second, after Gerry's correction, try factoring the numerator. –  Jaycob Coleman Nov 11 '13 at 9:11
    
Gerry Myerson, it's 16². & Jaycob, could you please explain what that means? –  user1622951 Nov 11 '13 at 9:13
    
You'd have to type @JaycobColeman to give me a notification. Both have been demonstrated now, but here and here do a good job explaining common denominators and factoring I think. –  Jaycob Coleman Nov 11 '13 at 19:10

4 Answers 4

$$\frac{a-b}{2}+\frac{a+b}{3}-\frac{b-a}{4}=\frac{6(a-b)+4(a+b)-3(b-a)}{12}=$$ $$=\frac{6a-6b+4a+4b-3b+3a}{12}=\frac{13a-5b}{12}$$ and $$\frac{a^2-16b^2}{2a+8b}=\frac{a^2-(4b)^2}{2(a+4b)}=\frac{(a-4b)(a+4b)}{2(a+4b)}=\frac{a-4b}{2}$$ and as you ask $$\frac{a^2-16^2}{2a+8b}=\frac{(a-16)(a+16)}{2(a+4b)}$$

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The second one looks like a complication, not a simplification, to me. –  Gerry Myerson Nov 11 '13 at 12:20
    
I don't see where is complication –  Adi Dani Nov 11 '13 at 19:14
    
It could be said that $(x-y)(x+y)$ is a complication of $x^2-y^2$ in that it requires several more syllables to write. –  Gerry Myerson Nov 11 '13 at 22:49
    
I think the question is defective, since OP insist that question I try to answer it but simplification in that case is not possible. As you commented there instead of $16$ must be $16b^2$ (in that case question make sense). –  Adi Dani Nov 11 '13 at 23:09

2, 3 and 4 all go into 12. This means that all of the fractions can be written with a denominator of 12.

(a-b)/2=6(a-b)/12,

(a+b)/3=4(a+b)/12 and

(b-a)/4=3(b-a)/12.

After giving them all an equal denominator, they can be added:

6(a-b)/12 + 4(a+b)/12 - 3(b-a)/12

=6(a-b)+4(a+b)-3(b-a)/12

From there, just simplify:

=6(a-b)+4(a+b)+3(a-b)/12

=9(a-b)+4(a+b)/12

=9a-9b+4a+4b/12

=13a-5b/12

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You missed several parentheses. Was that on purpose ? –  Claude Leibovici Nov 11 '13 at 9:59

For the second one, the difference of two squares is used:

a²-b²=(a+b)(a-b)

The question then becomes

(a+4)(a-4)/2a+8b

=(a+4)(a-4)/2(a+4b)

Which is why Gerry asked if it was 16b², because the a+4b would cancel out nicely

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(1) The first step is to find out the common factor of the denominator. It is obvious that the common factor of $2,3,4$ is $12$. Then given all the fraction below an equal denominator and adding the numerator:

\begin{align*} \\ \frac{a-b}{2}+\frac{a+b}{3}-\frac{b-a}{4} &= \frac{6(a-b)}{12}+\frac{4(a+b)}{12}-\frac{3(b-a)}{12} \\ &= \frac{6(a-b)+4(a+b)-3(b-a)}{12} \\ &= \frac{13a-5b}{12} \end{align*}

(2) For the second one, if we rewrite it as follow. It is obviously that there is not common factors between the nominator and denominator.Hence, $\frac{a^2-16}{2a+8b}$ is already a simplest form.

$$\frac{a^2-16}{2a+8b}=\frac{(a-4)(a+4)}{2(a+4b)}$$


NOTE why $16b^2$ is a reasonable guess? Note that $a^2-16b^2 = a^2-(4b)^2=(a-4b)(a+4b)$. Since $(a+4b)$ is a common factor between nominator and denominator, so we can just cancel it out as follow:

$$\frac{a^2-16b^2}{2a+8b}=\frac{(a-4b)(a+4b)}{2(a+4b)}=\frac{a-4b}{2}$$

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But OP insists it's $16^2$, not $16b^2$. –  Gerry Myerson Nov 11 '13 at 12:19
    
Thanks a ton! But as Gerry says, it's 16² not 16b². –  user1622951 Nov 11 '13 at 17:25
    
@user1622951 I rewrite my answer. I think the second fraction is already in the simplest form if it is $16^2$, not $16b^2$ –  sundaycat Nov 11 '13 at 18:37

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