Question

(X^2+3y) (3y+7) x=1 y=1

Answer 1

Okay for this I got 40.

You put forx=1 and for 1 you put also 1

like this
\[\left(1^2+3\left(1\right)\right)\left(3\left(1\right)+7\right)\]
\(\ 40\)