Question

Solve the following system of equations.
2a – 3b + c = 10
2a – 2b – 2c = 2
a + 3b + 2c = –1


(2, 3, 4)

(1, –2, 2)

(3, –1, 0)

(–1, 2, 3)

Answer 1

Can someone please help me?

Answer 2

is it 1, -2, 2

Answer 3

ok - add equn 1 to equn. 3 to eliminate b
3a + 3c = 9...............(1)
multiply second equn by 1.5:
3a - 3b - 3c = 3
add this to equn. 3:
4a - c = 2..........(2)
multiply (2) by 3:
12a - 3c = 6........(3)
add (1) and (3):
15a = 15

a = 1 and 4 - c = 2 so c = 2

sub a=1 c=2 in first equn.:
2*1- 3b + 2= 10
3b = 4 - 10 = -6
b = -2

correct option is (1,-2, 2)