Question 942232
{{{A+B+C=165}}}
.
.
{{{A=2B-12}}}
.
..
{{{C=A+2B}}}
Substituting from above,
{{{C=(2B-12)+2B}}}
{{{C=4B-12}}}
Now substitute into the first equation and solve for {{{B}}},
{{{2B-12+B+4B-12=165}}}
{{{7B=189}}}
{{{B=27}}}
Then,
{{{A=2(27)-12}}}
{{{A=54-12}}}
{{{A=42}}}
and
{{{C=42+2(27)}}}
{{{C=42+54}}}
{{{C=96}}}