Let $f(x) = ax^7 + bx^3 + cx - 5.$  If $f(-7) = 7,$ then find $f(7).$
So if we plug in $x=-7$, we get $f(-7)=a(-7)^7+b(-7)^3+c(-7)-5=7$. So if we plug in $x=7$, we get $f(7)=a(7)^7+b(7)^3+c(7)-5$. And if we plug in $x=7$, we get $f(7)=a(7)^7+b(7)^3+c(7)-5$. Which can be simplified to $7^7a+7^3b+7c-5$. Right, so the question is asking us to find $7^7a+7^3b+7c-5$ if $-7^7a-7^3b-7c-5=7$. That's correct. So, let's add these two equations to get $-7^7a-7^3b-7c-5+7^7a+7^3b+7c-5=7+7^7a+7^3b+7c-5$. So, $-10=7+7^7a+7^3b+7c-5$. If we add $10$ to both sides, we get $0=17+7^7a+7^3b+7c$. So we found $f(7)$? We found $f(7)$!

