## Algebra 1

$a=-3$ and $b=2$
The right hand side (RHS) of the given equation can be equivalently written as : $y^{2} = 1$ X $y^{2} =x^{0}y^{2}$ (since $x^{0} = 1$) Now, $x^{3+a}y^{b} = x^{0}y^{2}$ Comparing the left hand side(LHS) and the right hand side(RHS) of the equation with respect to the exponents of $x$ and $y$,we get : $x^{3+a} = x^{0}$ and $y^{b} = y^{2}$ This implies : 1] $3+a=0$ , hence $3 + a - 3 = 0 - 3$ , hence $a = -3$ 2] $b = 2$ Final result : $a=-3$ and $b=2$