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Re: If (3a+2b)/(3a+4b) [#permalink]
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Carcass wrote:
\(\frac{(3a+2b)}{(7a+4b)}=\frac{15}{32}\)

Cross multiply

\(96a+64b=105a+60b\)

\(4b=9a\)

\(a=\frac{4b}{a}\)

\(\frac{3a+b}{7b}\)=\(\frac{\frac{(3*4b}{9+b)}}{7b}\)=\(\frac{7b}{3*7b}\)=\(\frac{1}{3}\)



Carcass in the question the denominator is 3a and not 7a. I think it is a typo.

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Re: If (3a+2b)/(3a+4b) [#permalink]
Expert Reply
Ks1859 wrote:
Carcass wrote:
\(\frac{(3a+2b)}{(7a+4b)}=\frac{15}{32}\)

Cross multiply

\(96a+64b=105a+60b\)

\(4b=9a\)

\(a=\frac{4b}{a}\)

\(\frac{3a+b}{7b}\)=\(\frac{\frac{(3*4b}{9+b)}}{7b}\)=\(\frac{7b}{3*7b}\)=\(\frac{1}{3}\)



Carcass in the question the denominator is 3a and not 7a. I think it is a typo.

Regards


fixed
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Re: If (3a+2b)/(3a+4b) [#permalink]
Carcass wrote:
\(\frac{(3a+2b)}{(3a+4b)}=\frac{15}{32}\)

Cross multiply

\(96a+64b=105a+60b\)

\(4b=9a\)

\(a=\frac{4b}{9}\)

\(\frac{3a+b}{7b}\)=\(\frac{\frac{3*4b}{9+b}}{7b}\)=\(\frac{7b}{3*7b}\)=\(\frac{1}{3}\)

15*(3a+4b)=45a+60b
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Re: If (3a+2b)/(3a+4b) [#permalink]
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