Question

Use the elimination method to solve the following system of equations.

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

(–1, –3, –2)

(1, –3, –2)

(1, 3, –2)

(1, 3, 2)

Answer 1

Combine the first and third equations and eliminate z variables:

2x – y + z = –3
3x – 3y – z = –4

5x - 4y = -7

Combine the first two equations and also eliminate z variables by multiplication:

-3(2x – y + z = –3)
2x + 2y + 3z = 2

-6x + 3y - 3z = 9
2x + 2y + 3z = 2

-4x + 5y = 11

Here, we have:

-4x + 5y = 11
5x - 4y = -7

Then,

4(-4x + 5y = 11)
5(5x - 4y = -7)

-16x + 20y = 44
25x - 20y = -35

9x = 9
x = 1

Substitute that value for either two-variable equation and solve for y:

5(1) - 4y = -7
5 - 4y = -7
-4y = -12
y = 3

Finally, substitute x and y values for either three-variable equation and solve for z!

2(1) – 3 + z = –3
2 - 3 + z = -3
-1 + z = -3
z = -2

Hence, the solution is (1,3,-2)

Answer 2

thank you!!

Answer 3

No problem :)