Posted by **Laura** on Thursday, March 7, 2013 at 3:48pm.

solve the system of equations in three variables

-3x+y-z=-10

- math -
**Dr. Jane**, Thursday, March 7, 2013 at 3:49pm
We need two more equations.

- math -
**Laura**, Thursday, March 7, 2013 at 3:51pm
-3x+y-z=-10

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

2x+3y-2z=-5

sorry.

- math -
**Dr. Jane**, Thursday, March 7, 2013 at 3:59pm
Your goal is to take these 3 equations in 3 variables down to 2 equations in 2 variables then the problem becomes much easier.

you have to decide which variable to eliminate first and you have to do this twice.

Looking at the problem. I might try to eliminate the y's

I will take -2 times the first equation and add it to the second equation.

6x -2y + 2z = 20

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

This reduces to 2x + 5z = 19 Save this equation..

Now, I have to use the 3rd equation

2x + 3y -2z = 5

It is easier for me to eliminate the 3y if I use the first equation again.

I multiply the first equation by -3

2x + 3y -2z = -5

9x -3y+3z =30

add these two: 11x + z = 25

Now, go to the problem you saved

2x + 5z = 19

11x + z = 25

Multiply the second equation by -5 and that will help you eliminat the z's. solve for x.

Take your answer for x and subsitute into one of these two equations to find z.

Take your answers for x and z and substitute these values into one of the original equations to find y.

Once you have x, y, z... check all 3 in each of the three equations to make sure you haven't made a mistake.

It is really, really easy to make a mistake with these equations. If you get lost. Just start over try eliminating a different variable. Sometimes that helps.

- math -
**Steve**, Thursday, March 7, 2013 at 4:04pm
y = 3x+z-10

substituting values from the first into the others,

-4x+2(3x+z-10)+3z = -1

2x+3(3x+z-10)-2z = -5

or

2x+5z = 19

11x+z = 25

Now, z=25-11x, so

2x + 5(25-11x) = 19

-53x = -106

x = 2

z = 25-11x = 3

y = 3x+z-10 = -1

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