Anonymous

# How do I solve systems of three linear equations with three variables?

I know how to figure these out. But when I get decimals, big numbers, or something that looks wrong to me, I kinda freak out and just don’t do the work. I need someone to please walk me through the following problem so I can get an idea of how to properly figure out the rest. I know the steps I have to take like: write each equation in standard from Ax + By + Cz = D, choose a pair of equations and use the equations to eliminate a variable, choose any other pair of equations and eliminate the same variable as in step 2., two equations in two variables should be obtained from steps 2 and 3 and I must use the same method that I used before to solve the system. To solve for the third variable, substitute the values of the variables found in step 4 into any original equation containing the third variable. I know the steps, I just don’t understand how to get my answer.

1) x + y + z = -5

x – y + 5z = -29

3x +3y + 3z = -15

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I hate these problems they scare me to first start with the first two equations

x+y+z=-5

x-y+5z=-29 Eliminate the Y's first by subtracting the second from the first

you have x-4z= -24

now you have to work with the other two equations and also eliminate the y first

x-y+5z=-29

3x+3y+3z= -15 this one you have to get a common factor of 3 and 1 which is 3 and you have to use the -3 on the first equation so that when you subtract the bottom equation from the top the y's subtracted = 0

-3(x-y+5z= -29)

3x+3y+3z= -15 Now distribute the -3 on the first equation

-3x+3y+15z= 87

3x+3y+3z= -15 now subtract the the 2nd from the 1st

-6x+12z=72 Now you have to work with the equations that have resulted in elimnating the y's so you have to work with

x-4z=24

-6x+12z=72 Now with this one I'd start with eliminating the x's alls you have to do is find a common factor which is 6

6(x-4z=24)

-6x+12z=72 distribute the 6 on the first equation

6x-24z=144subtract 2nd equation from first

-6x+12z=72

-36z= 62 now divide 62 by -36

z= -2 now you can plug in the -2 into x-4z=24 or -6x+12z=72

so x-4(2)=24 x-8=24 add 8 to 24

x=32 now plug x and z into the first original equation

x+y+z= -5

32+y-2 = -5 subtract 2 from 32

30+y= -5 Now subtract 30 from -5

y= -25 so the final answer is :

x= 32 y= -25 and z= -2

Its been awhile so I hope this helps I tried I'm sorry if it doesn't good luck = )

Source(s): A in Algebra 2
• Well the process usually is:

1) either subract or add equation 1 from 2 to eliminate a term this will be equation 4.

2) subract or add equation 3 from either equation 1 or 2 to eliminate the same term this will be equation 5.

3) either subract or add equation 4 from 5 to eliminate a term, this will give you the value of one of the terms.

4) substitute that value into either equation 4 or 5 to get the other term.

5) finally substitute both of the terms you have determined into any of the original 3 equations and you should find your last variable.

In the case of your example you will not be able to find a solution, as it seems you have 2 equations with three unknowns. If you look at it equation 1 and 3 are essentially the same.

• There is no unique solution for this problem. The third equation is simply the first equation times 3 so you really only have 2 equations and 3 unknowns. You therefore have infinitely many solutions.

If you know linear algebra, you will actually solve these using a method called Gaussian elimination, which actually accounts for situations in which there are infinitely many solutions by looking at the pivots and free variables.

• This can not be solved. equations 1 and 3 are identical if you divide the third by 3 so in fact you only have two equations with three unknowns