How to simplify this?
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.
algebra-precalculus
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
add a comment |
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.
algebra-precalculus
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
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
add a comment |
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.
algebra-precalculus
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
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.
algebra-precalculus
algebra-precalculus
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
edited Nov 11 '13 at 10:01
Adi Dani
15.3k32246
15.3k32246
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
asked Nov 11 '13 at 9:06
user1622951user1622951
1022
1022
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
add a comment |
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
Should that $16^2$ be $16b^2$?
– Gerry Myerson
Nov 11 '13 at 9:10
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
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
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
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
add a comment |
5 Answers
5
active
oldest
votes
$$frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}=frac{6(a-b)+4(a+b)-3(b-a)}{12}=$$
$$=frac{6a-6b+4a+4b-3b+3a}{12}=frac{13a-5b}{12}$$
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
$$frac{a^2-16^2}{2a+8b}=frac{(a-16)(a+16)}{2(a+4b)}$$
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
1
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
add a comment |
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
answered Nov 11 '13 at 9:43
user102120user102120
555
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
add a comment |
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
answered Nov 11 '13 at 9:48
user102120user102120
555
add a comment |
(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:
$$frac{a^2-16b^2}{2a+8b}=frac{(a-4b)(a+4b)}{2(a+4b)}=frac{a-4b}{2}$$
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
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
add a comment |
Let us separate $a, ; b$
$$begin {aligned}frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}&=frac{a}{2}-frac b2 + frac{a}{3} +frac b3-frac{b}{4} + frac a4\
&= aleft(frac 12+frac 13 + frac 14right)+bleft(-frac 12 +frac 13 - frac 14right)\&=acdot frac{13}{12}-bcdot frac{5}{12}\&=frac{13a-5b}{12}
end{aligned}$$
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
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active
oldest
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$$frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}=frac{6(a-b)+4(a+b)-3(b-a)}{12}=$$
$$=frac{6a-6b+4a+4b-3b+3a}{12}=frac{13a-5b}{12}$$
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
$$frac{a^2-16^2}{2a+8b}=frac{(a-16)(a+16)}{2(a+4b)}$$
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
1
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
add a comment |
$$frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}=frac{6(a-b)+4(a+b)-3(b-a)}{12}=$$
$$=frac{6a-6b+4a+4b-3b+3a}{12}=frac{13a-5b}{12}$$
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
$$frac{a^2-16^2}{2a+8b}=frac{(a-16)(a+16)}{2(a+4b)}$$
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
1
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
add a comment |
$$frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}=frac{6(a-b)+4(a+b)-3(b-a)}{12}=$$
$$=frac{6a-6b+4a+4b-3b+3a}{12}=frac{13a-5b}{12}$$
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
$$frac{a^2-16^2}{2a+8b}=frac{(a-16)(a+16)}{2(a+4b)}$$
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
$$frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}=frac{6(a-b)+4(a+b)-3(b-a)}{12}=$$
$$=frac{6a-6b+4a+4b-3b+3a}{12}=frac{13a-5b}{12}$$
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
$$frac{a^2-16^2}{2a+8b}=frac{(a-16)(a+16)}{2(a+4b)}$$
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
edited Nov 11 '13 at 23:21
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
answered Nov 11 '13 at 10:15
Adi DaniAdi Dani
15.3k32246
15.3k32246
1
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
add a comment |
1
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
1
1
The second one looks like a complication, not a simplification, to me.
– Gerry Myerson
Nov 11 '13 at 12:20
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
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
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
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
add a comment |
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
answered Nov 11 '13 at 9:43
user102120user102120
555
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
add a comment |
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
answered Nov 11 '13 at 9:43
user102120user102120
555
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
add a comment |
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
answered Nov 11 '13 at 9:43
user102120user102120
555
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
answered Nov 11 '13 at 9:43
user102120user102120
555
answered Nov 11 '13 at 9:43
user102120user102120
555
answered Nov 11 '13 at 9:43
user102120user102120
555
answered Nov 11 '13 at 9:43
user102120user102120
555
555
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
add a comment |
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
You missed several parentheses. Was that on purpose ?
– Claude Leibovici
Nov 11 '13 at 9:59
add a comment |
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
answered Nov 11 '13 at 9:48
user102120user102120
555
add a comment |
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
answered Nov 11 '13 at 9:48
user102120user102120
555
add a comment |
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
answered Nov 11 '13 at 9:48
user102120user102120
555
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
answered Nov 11 '13 at 9:48
user102120user102120
555
answered Nov 11 '13 at 9:48
user102120user102120
555
answered Nov 11 '13 at 9:48
user102120user102120
555
answered Nov 11 '13 at 9:48
user102120user102120
555
555
add a comment |
add a comment |
(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:
$$frac{a^2-16b^2}{2a+8b}=frac{(a-4b)(a+4b)}{2(a+4b)}=frac{a-4b}{2}$$
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
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
add a comment |
(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:
$$frac{a^2-16b^2}{2a+8b}=frac{(a-4b)(a+4b)}{2(a+4b)}=frac{a-4b}{2}$$
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
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
add a comment |
(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:
$$frac{a^2-16b^2}{2a+8b}=frac{(a-4b)(a+4b)}{2(a+4b)}=frac{a-4b}{2}$$
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
(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:
$$frac{a^2-16b^2}{2a+8b}=frac{(a-4b)(a+4b)}{2(a+4b)}=frac{a-4b}{2}$$
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
edited Nov 11 '13 at 18:32
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
answered Nov 11 '13 at 10:26
SundayCatSundayCat
441418
441418
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
add a comment |
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
But OP insists it's $16^2$, not $16b^2$.
– Gerry Myerson
Nov 11 '13 at 12:19
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
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
@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
add a comment |
Let us separate $a, ; b$
$$begin {aligned}frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}&=frac{a}{2}-frac b2 + frac{a}{3} +frac b3-frac{b}{4} + frac a4\
&= aleft(frac 12+frac 13 + frac 14right)+bleft(-frac 12 +frac 13 - frac 14right)\&=acdot frac{13}{12}-bcdot frac{5}{12}\&=frac{13a-5b}{12}
end{aligned}$$
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
add a comment |
Let us separate $a, ; b$
$$begin {aligned}frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}&=frac{a}{2}-frac b2 + frac{a}{3} +frac b3-frac{b}{4} + frac a4\
&= aleft(frac 12+frac 13 + frac 14right)+bleft(-frac 12 +frac 13 - frac 14right)\&=acdot frac{13}{12}-bcdot frac{5}{12}\&=frac{13a-5b}{12}
end{aligned}$$
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
add a comment |
Let us separate $a, ; b$
$$begin {aligned}frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}&=frac{a}{2}-frac b2 + frac{a}{3} +frac b3-frac{b}{4} + frac a4\
&= aleft(frac 12+frac 13 + frac 14right)+bleft(-frac 12 +frac 13 - frac 14right)\&=acdot frac{13}{12}-bcdot frac{5}{12}\&=frac{13a-5b}{12}
end{aligned}$$
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
Let us separate $a, ; b$
$$begin {aligned}frac{a-b}{2}+frac{a+b}{3}-frac{b-a}{4}&=frac{a}{2}-frac b2 + frac{a}{3} +frac b3-frac{b}{4} + frac a4\
&= aleft(frac 12+frac 13 + frac 14right)+bleft(-frac 12 +frac 13 - frac 14right)\&=acdot frac{13}{12}-bcdot frac{5}{12}\&=frac{13a-5b}{12}
end{aligned}$$
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
answered Nov 22 '18 at 22:18
user376343user376343
2,9132823
2,9132823
add a comment |
add a comment |
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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