Question

Solve the following system equations:

2x + 4y – 3z = –7
3x + y + 4z = –12
x + 3y + 4z = 4

POSIBBLE ANSWERS:

(–6, 2, 1)

(6, 2, 1)

(6, –2, –1)

(–6, –2, –1)
...?

Answer 1

The solution is (–6, 2, 1). (Option A)

Explanation:

Given three equations

2x + 4y - 3z = -7 → (1)

3x + y + 4z = -12 → (2)

x + 3y + 4z = 4 → (3)

we have to find the solution of above equations.

By elimination method

Multiply equation (2) by 4 and then subtracting from (1), we get

(2x + 4y - 3z+7)-4(3x + y + 4z + 12)=0

⇒ -10x-19z=41 → (4)

Multiply equation (2) by 3 and then subtracting from (3), we get

(x + 3y + 4z - 4)-3(3x + y + 4z + 12)=0

⇒ -8x-8z=40 ⇒ x+z=-5 → (5)

Solving (4) and (5), we get

-10x-19z-41+10(x+z+5)=0

⇒ -9z=-9 ⇒ z=1

⇒ x+1=-5 ⇒ x=-6

and (3) implies -6 + 3y + 4 = 4 ⇒ y=2

Hence, the solution of above 3 equations will be (x,y,z)=(-6,2,1)

Hence, option (1) is correct.

Answer 2

(1).. 2x + 4y - 3z = -7
(2).. 3x + 1y + 4z = -12
(3).. 1x + 3y + 4z = 4

******************
ok.. we're going to put this into matrix form.. where we drop x, y, and z
2.. 4.. -3.. | ... -7
3.. 1.. .4.. | ... -12
1.. 3.. .4 . | ... 4
ok?

now.. I'll designate the rows (1), (2), (3).. ok?
new (1) = -2x(3) + (1)
new (2) = -3x(3) + (2)
ie...
0..-2..-11. | .. -15
0..-8.. -8.. | ... -24
1.. 3.. .4 . | ... .+4

new row (2) = old (2) / -8
0..-2..-11. | .. -15
0.. 1.. 1.. | ... +3
1.. 3.. .4 . | ... +4

new (1) = 2xold(2) + 1
new (3) = -3xold(2) + 3
0.. 0..-9 .. | ... -9
0.. 1.. 1.. | ... +3
1.. 0.. .1 . | ... -5

new row(1) = old row 1 /-9
0.. 0.. .1 .. | ... +1
0.. 1.. 1.. | ... +3
1.. 0.. .1 . | ... -5

new (2) = old(2) - old(1)
new (3) = old(3) - old(1)
0.. 0.. .1 .. | ... +1
0.. 1.. 0.. | ... +2
1.. 0.. .0 . | ... -6

so.. (x,y,z) = (-6, +2, +1)..

the last answer is correct