Question 847742
For this right triangle,
1.{{{A^2+B^2=13^2}}}
and 
{{{A+B+13=30}}}
2.{{{A+B=17}}}
From eq. 2,
{{{A=17-B}}}
Substitute into eq. 1,
{{{(17-B)^2+B^2=169}}}
{{{289-34B+B^2+B^2=169}}}
{{{2B^2-34B+120=0}}}
{{{B^2-17B+60=0}}}
{{{(B-5)(B-12)=0}}}
Two solutions:
{{{B-5=0}}}
{{{B=5}}}
Then,
{{{A=17-5=12}}}
and
{{{B-12=0}}}
{{{B=12}}}
then
{{{A=17-12=5}}}