Question 1054130
{{{A+B+C=19}}}
.
.
{{{C=2A}}}
So then,
{{{A=C/2}}}
,
,
{{{C=(A+B)-3}}}
{{{B=C-A+3}}}
{{{B=C-C/2+3}}}
{{{B=C/2+3}}}
,
,
Substituting into the first equation,
{{{C/2+C/2+3+C=19}}}
{{{2C+3=19}}}
{{{2C=16}}}
{{{C=8}}}
So then,
{{{A=8/2}}}
{{{A=4}}}
and
{{{B=8/2+3}}}
{{{B=7}}}