solve by elimination

Question

x+y-z=4

3x+2y+4z=17

-x+5y+z=8

Answer 1

x+y-z=4

3x+2y+4z=17

-x+5y+z=8

x=4-y+z

x=5y+z-8

4-y+z=5y+z-8

4-y=5y-8

4=6y-8

12=6y

y=2

x+y-z=4

x+2-z=4

x-z=2

x=z+2

3x+2y+4z=17

3(z+2)+2(2)+4z=17

6+3z+4+4z=17

7z+10=17

z=1

x=1+2

x=3

x=3

y=2

z=1

Answer 2

Hi Jimena - to solve this series of equations, we need to first solve for one of the variables and plug it into the other equations. There are a few ways to do this, but you can see here that the first and third equations have x and -x (as well as -z and z) respectively. So let's start by adding those two equations together:

Step 1:

x + y - z = 4

+

-x + 5y + z = 8

...

6y = 12 --> y = 2

Step 2: let's plug this back into the first equation.

x + y - z = 4

x + 2 - z = 4

x - z = 2

x = z + 2

Step 3: let's plug this value of x into the equations we haven't used yet - equation two.

3x + 2y + 4z = 17

3(z+2) + 2(2) + 4z = 17

3z + 6 + 4 + 4z = 17

7z = 7

z = 1

Step 4: put it all together.

We know x = z+2, y = 2, z = 1... so x = 3

And there's our answer:

x = 3

y = 2

z = 1