Can someone me solve a-d? i have answers in the back of the book, but i can't figure out how they (they being the author of the textbook) got their answer. i'm really confused. ?

Alright. So for:
a) 4(x+7)+3(x+5) Here we need to distribute the number outside of the parenthesis to the terms inside. So after distributing, we get 4x+28+3x+15. Then simplify this equation to 7x+43.
b) 3(2a+4b)+9(3a-2b) Same here. We distribute the number outside of the parenthesis to the terms inside. After distributing, we get 6a+12b+27a-18b. Then simplify to get 33a-6b.
c) 2(3u-5)-3(2u+1) As expected, same here. We distribute the number outside of the parenthesis. But here, we have a negative. Remember to distribute that as well. So we get 6u-10-6u-3. Then we simplify and get -13.
d) b(-5v-4)-2b(v+4) -_- Getting a bit boring... We distribute the variable from outside the parenthesis into the terms in the parenthesis. We get -5bv-4b-2bv-8b. Then simplify to get -3bv-12b.

**If you don't understand distributing, just let me know!

(A) 4(x+7)+3(x+5)
4x+28+3x+15 **combine like terms**
7x+43

(B) 3(2a+4b)+9(3a-2b)
6a+12b+18a-18b **combine like terms**
24a-6b

(C) 2(3u-5)-3(2u+1)
6u-10-6u-3 **combine like terms**
-13 *6u cancels out

(D) b(-5v-4)-2b(v+4)
-5bv-4b-2bv-8b **combine like terms**
-7bv-12b