Question

-2 (m+n-4)+5 (-2m+2n)+n (m+4n-5)

Answer 1

So if you want to add
we distribute
a(b+c)=ab+ac so
-2(m+n-4)=-2m-2n+8

5(-2m+2n)=-10m+10n

n(m+4n-5)=mn+4n^2-5n

so total we ahve
-2m-2n+8-10m+10n+mn+4n^2-5n
group like terms
4n^2+-2m-10m-2n+10n-5n+mn+8
add like temrs
4n^2-12m+3n+mn+8