# algebra

posted by .

systems of linear equations in 3 variables

1/3x-2/3y+z=0
1/2x-3/4y+z=1/4
-2x-y+z=1

• algebra -

Mult 1st eq by 3; x-2y+3z = 0
Mult 2nd eq by 4: 2x - 3y + 4z = 1
Add 2nd and 3rd: -4y+5z = 2
Mut 1st by 2: 2x-4y + 6z = 0
5(-4y + 5z = 2) + -4(-5y + 7z = 1)
-20y + 25z = 10 + 20y - 28z = -4
-3z =6
z = -2
Sub into the 1st eq: x - 2y -6 =0
Sub into the 3rd eq: -2x-y-2 = 1
2(x-2y = 6) + (-2x-y=3)
2x - 4y = 12 + -2x -y = 3
-5y =15
y = -3
sub into the 3rd eq
-2x -(-3) +(-2) = 1
-2x +1 = 1
x = 0
(0,-3,-2)

• algebra -

It might be easier if you
a) multiplied the first equation by 3
b) the second equation by 4

## Similar Questions

1. ### College Algebra

Translate the problem into a pair of linear equations in two variables. Solve the equations using either elimination or substitution. State your answer for both variables. The perimeter of a rectangle is 72m. If the width were doubled …
2. ### algebra

Can anyone help me to understand algebra?
3. ### Algebra 2

Solving systems of linear equations in three variables. Ok so i don't really totally get these problems. Can you help me solve this one?
4. ### Algebra

Solving systems of linear equations in 3 variables... 1) 3x-2y+2z=1 2) 2x+5y-5z=7 3) 4x-3y+z=-3 I combined 1 and 3 to get -5x+4y=7 And then i combined 2 and 3 to get -13y+11z=-17. Can that be right?
5. ### Linear Algebra

Express the solutions of the following systems in terms of the free variables: x1 + 3*(x2) - 2*(x3) + 2*(x5) = 0 2*(x1) + 6*(x2) - 5*(x3) - 2*(x4) + 4*(x5) - 3*(x6) = -1 5*(x3) + 10*(x4) + 15*(x6) = 5 2*(x1) + 6*(x2) + 8*(x4) + 4*(x5) …
6. ### Algebra II

Right now we're doing systems of Linear Equations in Three Variables. It says to solve the following system: x - y + 3z = 4 x + 2y - z = -3 4x + 3y + 2z = -5 Whatever I try with this problem (and the others like it) it comes out wrong. …
7. ### algebra

I do not know how to solve the following systems of linear equations in 3 variables 1) x-2y+z=7 2) 2x+y-3z=-1 3) x-4y+3z=13
8. ### algebra

systems of linear equations in 3 variables 1/3x-2/3y+z=0 1/2x-3/4y+z=1/4 -2x-y+z=1 every time i do it is wrong and i cannot figure out what I'm doing wrong please help!!!
9. ### Algebra I DONT KNOW HOW TO DO 3 VARIABLES

Solving Linear Equations and Inequalities Solve by eliminating x (this is solving 3 linear equations) then substitute to the other two equation x + y + 5z =2 (1) 4x - 3y + 5z =3 (2) 3x - 2y + 5z=1 (3) my teacher gave me the answer …
10. ### Math

PLEASE CHACK MY ANSWER!!! Discussion for Applications of Linear Systems When solving a system of equations, how do you determine which method to use?

More Similar Questions