Find the value of y
Now this could be solved if you identify the triple, the 3,4,5 tirangle.
or you can just do the algebra.
$\displaystyle (y+1)^2 + y^2 = 5^2$
$\displaystyle y^2+2y+1 + y^2 = 5^2$
$\displaystyle 2y^2+2y-24 = 0$
$\displaystyle y^2+y-12 = 0$
$\displaystyle (y+4)(y-3) = 0$
remember y is a length so it cannot be negative.
hope that help.
bobak