calculus please answer

Question

Solve the following system of linear equations: 

3x1 + 2x2 + 4x3 = −1 

2x1 − x2 + 2x3 = −2 

−x1 + x2 + 2x3 = 2

the numbers after the variables are subscripts they just werent computing

Answer 1

change the variables to x,y, and z

3x + 2y + 4z = -1

2x - y + 2z = -2

-x + y + 2z = 2

eliminates x as follows....

2*eq3 + eq2

3*eq3 + eq1

------------------

5y + 10z = 5

y + 6z = 2

-5 * eq2 + eq1

-----------------

-20z = -5

z = 1/4

5y + 10/4 = 5 and 5y + 10(1/4) = 5

5y + 5/2 = 5 5y + 10/4 = 5

5y = 5/2 which is the same equation

y = 1/2

3x + 2y + 4z = -1

3x + 2(1/2) + 4(1/4) = -1

3x + 1 + 1 = -1

3x + 2 = -1

3x = -3

x = -1

the solution is (x=-1, y = 1/2, z = 1/4)

Answer 2

x1 = x

x2 = y

x3 = z

(to avoid confusion)

[1] 3x + 2y + 4z = -1

[2] 2x - 1y + 2z = -2

[3] -1x + 1y + 2z = 2

add equation [2] and [3] results in:

[4] 1x + 4z = 0

add equation [1] and 2*(equation [2]) results in

[5] 7x + 8z = -5

you now have a set of 2 equations and 2 variables

solve by subbing the solution to equation [4] 1x = -4z

into equation [5] 7(-4z) + 8z = -5 to get a value for z

sub the value for z into equation [4] to get a value for x

sub both values into either [1], [2] or [3] to fine a value for y

https://www.khanacademy.org/math/algebra-home/alg-system-of-equations/alg-systems-with-three-variables/v/systems-of-three-variables