Question 225110
The following relatiosn can be set up from the given information:
{{{A=5B+2}}}
{{{AB=3}}}
We can plug the first equation into the second:
{{{(5B+2)B=3}}}
Distribute the B:
{{{5B^2+2B=3}}}
Move the 3 to the other side
{{{5B^2+2B-3=0}}}
This factors down to:
{{{(5B-3)(B+1)=0}}}
Set these each equal to zero and solve:
{{{5B-3=0}}}, {{{B+1=0}}}
{{{5B=3}}}, {{{B=-1}}}
{{{B=3/5}}}
So, we have two possible answers, {{{B=3/5}}}, {{{B=-1}}}.
Plug this back into the first equation:
{{{A=5(3/5)+2}}}
{{{A=5}}}
and
{{{A=5(-1)+2}}}
{{{A=-3}}}

Thus, the sets of equations are: {{{A=5}}}, {{{B=3/5}}} AND {{{A=-3}}}, {{{B=-1}}}.