Question

Find the value of z:

-x+y+2z=20
x-4y+4z=13
-2x-6y-3z= -10

Answer 1

add and subtract equations.

Answer 2

@wio teach me how

Answer 3

you use
3 var elimination

Answer 4

add the first and the second equation and what do you get ?

Answer 5

yeah .. @mth3v4 is right..

Answer 6

Make it a matrix\[
\begin{bmatrix}
-1&1&2 &20\\\
1&-4&4&13\\
-2&-6&-3&-10
\end{bmatrix}
\]

Answer 7

\[
R_1+R_2\to R_2
\]

Answer 8

Make it a matrix\[
\begin{bmatrix}
-1&1&2 &20\\\
0&-3&6&33\\
-2&-6&-3&-10
\end{bmatrix}
\]

Answer 9

\[
-2R_1+R_3\to R_3
\]

Answer 10

\[
\begin{bmatrix}
-1&1&2 &20\\\
0&-3&6&33\\
0&-8&-9&-50
\end{bmatrix}
\]

Answer 11

You keep going like this.

Answer 12

yay

Answer 13

-x + y + 2z = 20
x - 4y + 4z = 13
---------------add
0 - 3y + 6z = 33
-3y + 6z = 33

now add the 2nd and the third equation. But since we cancelled out the x in the first two equations, we want to make the x's cancel out these equations.
x - 4y + 4y = 13-->(2)
-2x - 6y - 3z = -10
-------------------
2x - 8y + 8y = 26 (result of multiplying by 2)
-2x - 6y - 3z = -10
-------------------add
0 - 14y + 5z = 16
-14y + 5z = 16

now take the answers you got and try to eliminate a variable
-3y + 6z = 33 -->(5)
-14y + 5z = 16 -->(-6)
--------------------
-15y + 30z = 165 (result of multiplying by 5)
84y - 30z = - 96 (result of multiplying by -6)
--------------------add
69y = 69
y = 1

now sub 1 in for y into either of the equations that does not have a x in it
-14y + 5z = 16
-14(1) + 5z = 16
-14 + 5z = 16
5z = 16 + 14
5z = 30
z = 6

now sub 1 in for y and 6 in for z into any of the original equations
-x + y + 2z = 20
-x + 1 + 2(6) = 20
-x + 1 + 12 = 20
-x + 13 = 20
-x = 20 - 13
-x = 7
x = -7

check...
x - 4y + 4z = 13
-7 - 4(1) + 4(6) = 13
-7 - 4 + 24 = 13
-11 + 24 = 13
13 = 13 (correct)

x = -7, y = 1, z = 6

Answer 14

wio is probably doing it the easier way by using the matrix.

Answer 15

however which way
you still are going to perform the same more likely operations ?