If $\frac{m^{(x^2)}}{m^{(y^2)}}=m^{10}$ and $x+y=5$, then what is the value of $x-y$?

Question

Given that $$\frac {m^{(x^2)}}{m^{(y^2)}}=m^{10}\quad \text{and}\quad x+y=5$$

what is the value of $x-y$?

Since both of the bases are the same, I subtracted m2 from m2 and made that equal to m^10, Since all of the bases are the same, I removed them and factored everything down to (x+y)(x-y)=10. Since we know (x+y)=5 I substituted and subtracted. My final answer was 5. I just wanted clarification if this was the correct answer

Answer 1

Well, look here \begin{align*} &~\frac{m^{x^{2}}}{m^{y^{2}}} = m^{10}\\ \implies &~m^{x^{2}-y^{2}} = m^{10} \\ \implies &~ x^2 - y^2 = 10\\ \implies &~5\cdot(x - y) = 10\quad \text{as } (x + y) = 5\\ \implies &~ (x - y) = 2 \end{align*}

Answer 2

This is my first time using this site, so please forgive me if I am doing something wrong.

By the rules of exponents $m^{x^2-y^2}=m^{10}$, so $x^2-y^2=(x+y)(x-y)=10$. Since $x+y=5$, $x-y=2$.