## Algebra 2 (1st Edition)

$45$ and $28$
Our equation is: $2\cdot35\cdot18=(35+x)(18+x)\\1260=630+x^2+53x\\x^2+53x-630=0\\(x+63)(x-10)=0$ Thus $x=10$ or $x=−63$ but $x$ has to be positive, thus $x=10$. Thus, the new dimensions are $45$ and $28$.