Simplification of rational expression gone wrong(high school math)












1












$begingroup$


Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?



Original expression:
$$left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}$$



When I simplify the equations in parentheses I get:
$$frac{b^2-ab+a^2}{ab(a+b)}timesfrac{1}{a^{2}}$$



Which should result in:
$$frac{b^2+a(a-b)}{a^3b(a+b)}=frac{b^2-a}{a^3b}$$



That is of course not right. The correct answer according to the book is $frac{b-a}{a}$






rational-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your last equality is false. it is the low school maths.
    $endgroup$
    – hamam_Abdallah
    Dec 15 '18 at 20:59










  • $begingroup$
    Your derivation of the second factor is incorrect. You should find $frac{a^3+b^3}{(a+b)(b^3-a^2b)}=frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:04










  • $begingroup$
    Thank you so so much @Mindlack this solved it easily!
    $endgroup$
    – Zae
    Dec 15 '18 at 21:34


















1












$begingroup$


Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?



Original expression:
$$left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}$$



When I simplify the equations in parentheses I get:
$$frac{b^2-ab+a^2}{ab(a+b)}timesfrac{1}{a^{2}}$$



Which should result in:
$$frac{b^2+a(a-b)}{a^3b(a+b)}=frac{b^2-a}{a^3b}$$



That is of course not right. The correct answer according to the book is $frac{b-a}{a}$






rational-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your last equality is false. it is the low school maths.
    $endgroup$
    – hamam_Abdallah
    Dec 15 '18 at 20:59










  • $begingroup$
    Your derivation of the second factor is incorrect. You should find $frac{a^3+b^3}{(a+b)(b^3-a^2b)}=frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:04










  • $begingroup$
    Thank you so so much @Mindlack this solved it easily!
    $endgroup$
    – Zae
    Dec 15 '18 at 21:34
















1












1








1





$begingroup$


Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?



Original expression:
$$left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}$$



When I simplify the equations in parentheses I get:
$$frac{b^2-ab+a^2}{ab(a+b)}timesfrac{1}{a^{2}}$$



Which should result in:
$$frac{b^2+a(a-b)}{a^3b(a+b)}=frac{b^2-a}{a^3b}$$



That is of course not right. The correct answer according to the book is $frac{b-a}{a}$






rational-functions







share|cite|improve this question









$endgroup$




Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?



Original expression:
$$left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}$$



When I simplify the equations in parentheses I get:
$$frac{b^2-ab+a^2}{ab(a+b)}timesfrac{1}{a^{2}}$$



Which should result in:
$$frac{b^2+a(a-b)}{a^3b(a+b)}=frac{b^2-a}{a^3b}$$



That is of course not right. The correct answer according to the book is $frac{b-a}{a}$






rational-functions




rational-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 20:56









ZaeZae

82




82












  • $begingroup$
    Your last equality is false. it is the low school maths.
    $endgroup$
    – hamam_Abdallah
    Dec 15 '18 at 20:59










  • $begingroup$
    Your derivation of the second factor is incorrect. You should find $frac{a^3+b^3}{(a+b)(b^3-a^2b)}=frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:04










  • $begingroup$
    Thank you so so much @Mindlack this solved it easily!
    $endgroup$
    – Zae
    Dec 15 '18 at 21:34




















  • $begingroup$
    Your last equality is false. it is the low school maths.
    $endgroup$
    – hamam_Abdallah
    Dec 15 '18 at 20:59










  • $begingroup$
    Your derivation of the second factor is incorrect. You should find $frac{a^3+b^3}{(a+b)(b^3-a^2b)}=frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:04










  • $begingroup$
    Thank you so so much @Mindlack this solved it easily!
    $endgroup$
    – Zae
    Dec 15 '18 at 21:34


















$begingroup$
Your last equality is false. it is the low school maths.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 20:59




$begingroup$
Your last equality is false. it is the low school maths.
$endgroup$
– hamam_Abdallah
Dec 15 '18 at 20:59












$begingroup$
Your derivation of the second factor is incorrect. You should find $frac{a^3+b^3}{(a+b)(b^3-a^2b)}=frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result.
$endgroup$
– Mindlack
Dec 15 '18 at 21:04




$begingroup$
Your derivation of the second factor is incorrect. You should find $frac{a^3+b^3}{(a+b)(b^3-a^2b)}=frac{a^2-ab+b^2}{b(b+a)(b-a)}$. So the $a^2-ab+b^2$ and $b(b+a)$ will cancel out when you make the division and you will get the final result.
$endgroup$
– Mindlack
Dec 15 '18 at 21:04












$begingroup$
Thank you so so much @Mindlack this solved it easily!
$endgroup$
– Zae
Dec 15 '18 at 21:34






$begingroup$
Thank you so so much @Mindlack this solved it easily!
$endgroup$
– Zae
Dec 15 '18 at 21:34












1 Answer
1






active

oldest

votes


















0












$begingroup$

Go slowly. Do it in blocks.



First block:
begin{align}
frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}
&=frac{b}{a(a+b)}-frac{b-a}{b(a+b)} \[4px]
&=frac{b^2-a(b-a)}{ab(a+b)} \[4px]
&=frac{a^2-ab+b^2}{ab(a+b)}
end{align}



Second block:
begin{align}
frac{a^2}{b^3-a^2b}+frac{1}{a+b}
&=frac{a^2}{b(b^2-a^2)}+frac{1}{a+b} \[4px]
&=frac{a^2}{b(b-a)(b+a)}+frac{1}{a+b} \[4px]
&=frac{a^2+b(b-a)}{b(b-a)(b+a)} \[4px]
&=frac{a^2-ab+b^2}{b(b-a)(b+a)}
end{align}



Putting the pieces together:
$$
left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}=
frac{a^2-ab+b^2}{ab(a+b)}frac{b(b-a)(b+a)}{a^2-ab+b^2}=
frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
    $endgroup$
    – Zae
    Dec 16 '18 at 9:58











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Go slowly. Do it in blocks.



First block:
begin{align}
frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}
&=frac{b}{a(a+b)}-frac{b-a}{b(a+b)} \[4px]
&=frac{b^2-a(b-a)}{ab(a+b)} \[4px]
&=frac{a^2-ab+b^2}{ab(a+b)}
end{align}



Second block:
begin{align}
frac{a^2}{b^3-a^2b}+frac{1}{a+b}
&=frac{a^2}{b(b^2-a^2)}+frac{1}{a+b} \[4px]
&=frac{a^2}{b(b-a)(b+a)}+frac{1}{a+b} \[4px]
&=frac{a^2+b(b-a)}{b(b-a)(b+a)} \[4px]
&=frac{a^2-ab+b^2}{b(b-a)(b+a)}
end{align}



Putting the pieces together:
$$
left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}=
frac{a^2-ab+b^2}{ab(a+b)}frac{b(b-a)(b+a)}{a^2-ab+b^2}=
frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
    $endgroup$
    – Zae
    Dec 16 '18 at 9:58
















0












$begingroup$

Go slowly. Do it in blocks.



First block:
begin{align}
frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}
&=frac{b}{a(a+b)}-frac{b-a}{b(a+b)} \[4px]
&=frac{b^2-a(b-a)}{ab(a+b)} \[4px]
&=frac{a^2-ab+b^2}{ab(a+b)}
end{align}



Second block:
begin{align}
frac{a^2}{b^3-a^2b}+frac{1}{a+b}
&=frac{a^2}{b(b^2-a^2)}+frac{1}{a+b} \[4px]
&=frac{a^2}{b(b-a)(b+a)}+frac{1}{a+b} \[4px]
&=frac{a^2+b(b-a)}{b(b-a)(b+a)} \[4px]
&=frac{a^2-ab+b^2}{b(b-a)(b+a)}
end{align}



Putting the pieces together:
$$
left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}=
frac{a^2-ab+b^2}{ab(a+b)}frac{b(b-a)(b+a)}{a^2-ab+b^2}=
frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
    $endgroup$
    – Zae
    Dec 16 '18 at 9:58














0












0








0





$begingroup$

Go slowly. Do it in blocks.



First block:
begin{align}
frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}
&=frac{b}{a(a+b)}-frac{b-a}{b(a+b)} \[4px]
&=frac{b^2-a(b-a)}{ab(a+b)} \[4px]
&=frac{a^2-ab+b^2}{ab(a+b)}
end{align}



Second block:
begin{align}
frac{a^2}{b^3-a^2b}+frac{1}{a+b}
&=frac{a^2}{b(b^2-a^2)}+frac{1}{a+b} \[4px]
&=frac{a^2}{b(b-a)(b+a)}+frac{1}{a+b} \[4px]
&=frac{a^2+b(b-a)}{b(b-a)(b+a)} \[4px]
&=frac{a^2-ab+b^2}{b(b-a)(b+a)}
end{align}



Putting the pieces together:
$$
left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}=
frac{a^2-ab+b^2}{ab(a+b)}frac{b(b-a)(b+a)}{a^2-ab+b^2}=
frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.






share|cite|improve this answer









$endgroup$



Go slowly. Do it in blocks.



First block:
begin{align}
frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}
&=frac{b}{a(a+b)}-frac{b-a}{b(a+b)} \[4px]
&=frac{b^2-a(b-a)}{ab(a+b)} \[4px]
&=frac{a^2-ab+b^2}{ab(a+b)}
end{align}



Second block:
begin{align}
frac{a^2}{b^3-a^2b}+frac{1}{a+b}
&=frac{a^2}{b(b^2-a^2)}+frac{1}{a+b} \[4px]
&=frac{a^2}{b(b-a)(b+a)}+frac{1}{a+b} \[4px]
&=frac{a^2+b(b-a)}{b(b-a)(b+a)} \[4px]
&=frac{a^2-ab+b^2}{b(b-a)(b+a)}
end{align}



Putting the pieces together:
$$
left(frac{b}{a^{2}+ab}-frac{b-a}{b^2+ab}right)times left(frac{a^2}{b^3-a^2b}+frac{1}{a+b}right)^{-1}=
frac{a^2-ab+b^2}{ab(a+b)}frac{b(b-a)(b+a)}{a^2-ab+b^2}=
frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 22:59









egregegreg

183k1486205




183k1486205












  • $begingroup$
    Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
    $endgroup$
    – Zae
    Dec 16 '18 at 9:58


















  • $begingroup$
    Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
    $endgroup$
    – Zae
    Dec 16 '18 at 9:58
















$begingroup$
Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
$endgroup$
– Zae
Dec 16 '18 at 9:58




$begingroup$
Thank you so much for taking the time to write it all out! A comment already pointed out my mistake, and this illustrates everything perfectly.
$endgroup$
– Zae
Dec 16 '18 at 9:58


















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