systems of linear Equations: 2x-y-z=7,x-y+z=2,x+y-3z=5

I

Ironhawk

New member
Joined
Nov 20, 2008
Messages
10
This is my problem: 3 equations in 3 variables.

2x-y-z=7
x-y+z=2
x+y-3z=5[/u
I know we have to add two equations together and eliminate Z. I have done it in 3 different combinations and could only eliminate Z in one Equation.

1)2x-y-z+7
2)x-y+z=2
3x-2y=9

2)x-y+z=2
3)x+y-3z=5
2x-2z=7

1)2x-y-z=7
3)x+y-3z=5
3x-4z=12

Where do I go from here? I know I only need two of the but can't get rid of Z. Thank you for you help and Patience.
 
Re: systems of linear Equations

Ironhawk said:
This is my problem: 3 equations in 3 variables.

2x-y-z=7
x-y+z=2
x+y-3z=5[/u
I know we have to add two equations together and eliminate Z. I have done it in 3 different combinations and could only eliminate Z in one Equation.

1)2x-y-z+7
2)x-y+z=2
3x-2y=9

2)x-y+z=2
3)x+y-3z=5
2x-2z=7

1)2x-y-z=7
3)x+y-3z=5
3x-4z=12

Where do I go from here? I know I only need two of the but can't get rid of Z. Thank you for you help and Patience.
Click to expand...


To solve a system of three equations in three variables, you try to eliminate ONE of the variables (which one does not matter....it does not have to be z!

You have successfully eliminated y from two of the equations in the second and third steps of your work, and have come up with

2x - 2z = 7
3x - 4z = 12

Since this is a system of two equations in two variables, you should be able to solve it and find the values of x and z. (Hint...you might want to start by multiplying both sides of the first equation by -2, then add the result to the second equation.) Once you know the values of x and z, you can substitute those values into one of the original equations which contains all three variables, and solve for y.

If you REALLY need to eliminate z from the original set of equations, you could multiply the original second equation by 3, and add the result to the original third equation:

3x - 3y + 3z = 6
x + y - 3z = 5
------------------
4x - 2y = 11

That, together with the work you've done, would give you two equations involving x and y. But, if the goal is to solve the system, it does not matter which variable you eliminate...you just need to come up with two equations using the same two variables.
 
Re: systems of linear Equations

ok, so then I continue like this?:

-2(2x-2z)=-2(7)
3x-4z=12

=-4x+4z=-14
3x-4z=12
-x=-2
x=2

Then I plug x into what portion of the equation?
 
Re: systems of linear Equations

Ironhawk said:
ok, so then I continue like this?:

-2(2x-2z)=-2(7)
3x-4z=12

=-4x+4z=-14
3x-4z=12
-x=-2
x=2

Then I plug x into what portion of the equation?
Click to expand...

Ok...you have x = 2.

3x - 4z = 12

SUBSTITUTE 2 for x:

3(2) - 4z = 12

Now continue to find the value of z.

When you've got values for both x and z, choose one of the original equations that has all three variables in it. Substitute the values you have found for x and z; the only variable remaining in the equation will be y, and you should be able to solve for y.

It sounds like you could use a good review of solving systems of TWO equations in TWO variables before you proceed with systems of more equations with more variables!!
 
Re: systems of linear Equations

After inputing all the numbers I came up with

Z=-6/4
y=-3/2
x=2
Is this correct?

3(2)-4z=12
6-4z=12
-4z=6
z=-6/4

Then plug in : 2(2)-y-(-6/4)=7
 
Ironhawk said:
... I came up with

z = -6/4
y = -3/2
x = 2

Is this correct? ...
Click to expand...


Yes, but why do you report z as -6/4, instead of reducing this fraction to lowest terms?

 
You must log in or register to reply here.
Top