Posted by Laura on Thursday, March 7, 2013 at 3:48pm.
solve the system of equations in three variables
3x+yz=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+yz=10
4x+2y+3z=1
2x+3y2z=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+z10
substituting values from the first into the others,
4x+2(3x+z10)+3z = 1
2x+3(3x+z10)2z = 5
or
2x+5z = 19
11x+z = 25
Now, z=2511x, so
2x + 5(2511x) = 19
53x = 106
x = 2
z = 2511x = 3
y = 3x+z10 = 1