What is the value of $2x+3y$ if $x+y=7$ and $x^2-y^2=21$?

Question

If $x+y=7$ and $x^2-y^2=21$, then what is $2x+3y$?

I solved it like this:

\begin{align} y& =7-x \\ x^2-(7-x)^2-21&=0 \\ x^2-49+14x-x^2-21&=0 \\ 14x&=70 \\ x&=5 \end{align}

Then I solved for $y$ and I got $2$. I plugged in the values of $x$ and $y$ to $2x+3y$, and I got $16$.

Is this correct?

Answer 1

Hint: A slightly easier way. $$x^2-y^2=(x+y)(x-y)=7(x-y)=21 \implies x-y=3$$ But your solution is correct.

Answer 2

Quicker way is to note that $$ 21=(x-y)(x+y)=7(x-y) $$ so $$ x-y=3 $$ Combined with $x+y=7$, we get that $2x=10$ i.e. $x=5$ and $y=2$.

So your answer is correct.