Question 1063646
Start with the general equation of a line,
{{{ax+by=c}}}
So you know (3,2) is on the line,
{{{a(3)+b(2)=c}}}
{{{c=3a+2b}}}
and the x intercept is,
{{{ax+b(0)=c}}}
{{{ax=c}}}
{{{x=c/a}}}
and the y intercept is,
{{{a(0)+by=c}}}
{{{by=c}}}
{{{y=c/b}}}
and their sum,
{{{c/a+c/b=12}}}
{{{c(1/a+1/b)=12}}}
Combining with the previous equation,
{{{(3a+2b)(1/a+1/b)=12}}}
{{{(3a+2b)(b+a)=12ab}}}
{{{3ab+3a^2+2b^2+2ba=12ab}}}
{{{3a^2-7ab+2b^2=0}}}
{{{(a-2b)(3a-b)=0}}}
Two solutions:
{{{a-2b=0}}}
{{{a=2b}}}
So then,
{{{c=3a+2b}}}
{{{c=3(2b)+2b}}}
{{{c=6b+2b}}}
{{{c=8b}}}
So the solution line looks like,
{{{2bx+by=8b}}}
{{{highlight(2x+y=8)}}}
and
{{{3a-b=0}}}
{{{3a=b}}}
So,
{{{c=3a+2(3a)}}}
{{{c=3a+6a}}}
{{{c=9a}}}
So this solution line looks like,
{{{ax+3ay=9a}}}
{{{highlight(x+3y=9)}}}