Question 1103438
{{{system((n+2)/(d+3)=3/4,(n-3)/(d-1)=1/4)}}}



USING ALGEBRA.
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{{{4(n+2)=3(d+3)}}}
{{{4n+8=3d+9}}}
{{{4n-3d=1}}}
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{{{4(n-3)=d-1}}}
{{{4n-12=d-1}}}
{{{4n-d=11}}}
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{{{system(4n-3d=1,4n-d=11)}}}
{{{(4n-d)-4n+3d=11-1}}}
{{{2d=10}}}
{{{highlight(d=5)}}}
 

{{{4n-15=1}}}
{{{4n=16}}}
{{{n=4}}}
OR
{{{4n-5=11}}}
{{{4n=11+5}}}
{{{4n=16}}}
{{{highlight(n=4)}}}


The fraction {{{n/d=highlight(4/5)}}}.




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What happens if you take each of those in reverse?

{{{n/d=(3-2)/(4-3)=1/1=1}}}
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{{{n/d=(1+3)/(4+1)=4/5}}}
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NO GOOD!