SOLUTION: Hello I have a problem, but cant seem to find out what i am doing wrong. The equations are solutions to systems of linear equations in three variables.
2x+ y-3z=-12
3x-2y-z= 3
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Coordinate Systems and Linear Equations
-> SOLUTION: Hello I have a problem, but cant seem to find out what i am doing wrong. The equations are solutions to systems of linear equations in three variables.
2x+ y-3z=-12
3x-2y-z= 3
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Question 313259: Hello I have a problem, but cant seem to find out what i am doing wrong. The equations are solutions to systems of linear equations in three variables.
2x+ y-3z=-12
3x-2y-z= 3
-x+5y+2z=-3
I know you have to work only two equations at a time. My teacher took the class through it step by step but i'm thinking im making a mistake on my adding or subtracting because i keep coming out with the wrong answer.
THanks
Kayla Found 3 solutions by Alan3354, mananth, OmniMaestra:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! 2x+ y-3z=-12
3x-2y-z= 3
-x+5y+2z=-3
-------------
To use elimination, pick one of the variables to eliminate.
I'll eliminate z
Multiply eqn 2 by 2 and add it to eqn 3.
Then multiply eqn 2 by 3 and subtract it from eqn 1
-------------
3x-2y-z= 3 x2 --> 6x-4y-2z = 6
-x+5y+2z=-3
6x-4y-2z = 6
------------ Add
5x + y = 3 Eqn A
----------
2x+ y-3z=-12
3x-2y-z= 3 x3 --> 9x-6y-3z = 9
2x+ y-3z=-12
9x-6y-3z = 9
------------ Subtract
-7x+5y = -21 Eqn B
--------------------
Now eqns A and B are in x and y only
-7x+5y = -21 Eqn B
25x+5y = 15 Eqn A times 5
------------ Subtract
-32x = -36
x = 9/8
--------
Sub for x into Eqn A or B
5*(9/8) + y = 3
y = 3 - 45/8
y = -21/8
----------
Sub for x and y into eqn 2 (or any of the original 3)
3x-2y-z= 3
3*(9/8) - 2*(-21/8) - z = 3
z = -3 + 27/8 + 42/8 = -3 + 69/8
z = 45/8
You can put this solution on YOUR website! 2x+ y-3z=-12
3x-2y-z= 3
-x+5y+2z=-3
..
2x+ y-3z=-12 ( multiply by 2 to eliminate y)
3x-2y-z= 3.
..
4x+2y-6z+3x-2y-z= -24+3
7x-7z=-21.
x-z=-3
..
3x-2y-z= 3 ( multiply by 5 to eliminate y)
-x+5y+2z=-3 ( multiply by 2) and add
15x-10y-5z-2x+10y+4z=15-6
13x-z=9.
..
x-z=-3
13x-z=9
..
13x-13z-13x+z=-39-9
-12z=-48
z= 4.
..
plug the value of z in x-z = -3
x-4=-3
x=4-3
x=1.
..
plug the values of x & z in the equation 2x+ y-3z=-12
2+y-12=-12
y=-2
You can put this solution on YOUR website! 2x+ y-3z=-12
3x-2y-z= 3
-x+5y+2z=-3
Yes, you do have to take it 2 equations at a time. THere are some options to begin so we look for the option that will make our work easy.
I'm going to start with the first two equations and eliminate the y.
I'll multiply the first equation by 2:
Now I add down:
....Since all the terms are divisible by 7, I will simplify by dividing by 7
Next, I must eliminate the y's with a different set of two equations.
I'm going to do the first and third equations.
2x+ y-3z=-12
-x+5y+2z=-3
Multiply the top equation by -5.
Add down:
Use the two new equations and eliminate the x's
Now lets eliminate the z's.
Multiply the second equation by 17.
add down:
divide by 6
I'm going to use the two equations I created when I eliminated the y's and substitute 1 in place of the x's. Then solve both for z.
--->--> divide by 17. z= 4 --->
Now you know that x=1 and z=4. Use one of the original equations, substitute, then solve for y.
-x+5y+2z=-3
-1+5y+2*4=-3
-1+5y+8=-3
5y+7=-3
5y=-10
y=-2
x=1
y=-2
z=4
