Question

Solve the following system of equations.

Answer 1

2a – 3b + c = 10
2a – 2b – 2c = 2
a + 3b + 2c = –1

Answer 2

A) (2, 3, 4)

B) (–1, 2, 3)

C) (1, –2, 2)

D) (3, –1, 0)

Answer 3

Call the equations (i), (ii) and (iii).

(i)-(ii) is: -b + 3c = 8 (iv)
(ii)- 2(iii) is: -8b - 6c = 4 (v)

Now you have equations (iv) and (v), with only b and c as variables.

Answer 4

(iv): -b + 3c = 8
(v): -8b - 6c = 4

2(iv)+(v) is: -10b = 20 (vi)

Solution of equation (vi) is found by dividing by -10, it gives: b = -2.

Put b = -2 in (iv) or (v).
Now calculate c.

Put b and c in (i), (ii) or (iii).
Now calculate a.

It is a lot of work, but it is rewarding :D

Answer 5

What?

Answer 6

We don't even have to calculate a and c, because we found b=-2. The only answer that fits is C.