Question 960221
1.{{{4a=3b}}}
.
.
{{{4a=a+b+c}}]
2.{{{3a=b+c}}}
From eq. 1,
{{{12a=9b}}}
From eq. 2,
{{{12a=4b+4c}}}
Subtract the two equations,
{{{12a-12b=9b-4b-4c}}}
{{{5b-4c=0}}}
{{{5b=4c}}}
{{{highlight(b=(4/5)c)}}}
Then,
{{{4a=3(4/5)c}}}
{{{highlight(a=(3/5)c)}}}