Question

Solve the following system equations:

Answer 1

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

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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 

Answer 2

Answer:

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

Step-by-step 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.