#(f@g)(x)# is equivalent to #f (g(x))#. We solve this problem just as we solve #f(x)#. But since it asks us to find out #f(g(x))#, in #f(x)#, each time we encounter x, we replace it with #g(x)#.

In the above problem, #f(x) = x+3#.

Therefore, #f(g(x))= g(x) +3#.

#=> (f@g)(x)= 2x - 7 + 3#

#=> (f@g)(x)= 2x - 4#

Basically, write the #g(x)# equation where you see the x in the #f(x)# equation.

#f@g(x)= (g(x))+3# Replace #g(x)# with the equation

#f@g(x)= (2x-7)+3#

#f@g(x)= 2x-7+3# we just took away the parentheses

#f@g(x)= 2x-4# Because the #-7+3=4#

This is it

#g@f(x)# would be the other way around

#g@f(x)= 2(x+3)-7#

now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them.

#g@f(x)= 2x+6-7#

Next, #+6-7 = -1#

#g@f(x) = 2x-1#

Its a lt easier than you think!