System of equation..help!?

Question

i understand how to do systems of equations, but this one's a little complicated.
9a+7b=--30 (negative 30)
8b+5c=11
--3a+10c=73 (negative 3a)
when i solve this by elimination, it'll give me no solution, and when i solve it by substitution it'll give me infinite solution. and in the back of the book the answers are (--1,--3,7) a=--1, b=--3, c=7. i can't find an example for this in the book! please, somebody help me.

well, duh the book is right, you used a scientific calculator?

Answer 1

9a + 7b = -30 --- 1
8b+5c = 11 ------ 2
-3a+10c=73 ----- 3

(3) - 2x(2)
-3a - 16b = 51 --- 4

3x(4)+(1)
-41b = 123
b = -3

sub in b in (4)
-3a +48 = 51
-3a = 3
a = -1

sub b in 2
-24 + 5c = 11
5c = 35
c = 7

Answer 2

9a+7b=-30
7b=-30-9a
b=(-30-9a)/7

Use this in the next equation
8((-30-9a)/7)+5c
(-240-72a)/7+5c=11
(-72/7)a+5c=317/7

Then use elimination after taking thrid equation divided by -2

(-72/7)a+5c=317/7
+((3/2)a-5c= -73/2)
________________
(-123/14)a=123/14
-123a=123
a=-1

then solve first using a

9(-1)+7b=-30
-9+7b=-30
7b=-21
b=-3

then solve second equation with this

8(-3)+5c=11
-24+5c=11
5c=35
c=7

Answer 3

The book is right that is all I can say I used my calculator to solve three unkowns and it worked.