Question 400010: I am attempting to solve system of linear equations through elimination. I have an entire list of them due but if you can help me promptly with this one, I should be able to use it as a guide with the others. Thank you in advance.
2x-y+z=1
x+y-7=2
3x-y+z=0
Found 3 solutions by Edwin McCravy, ewatrrr, stanbon:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!
Rules:
1. Pick 2 of the equations, and pick a letter to eliminate, and multiply
the equations through as is necessary to make the letter cancel out.
2. Take the equation you did not use in step 1, together with one of the
equations you used in step 1, and eliminate the SAME letter you eliminated
in step 1.
3. The results of step 1 and step 2 give you two equations in two letters.
Solve them for the two letters they contain.
4. Substitute those two values in one of the original equations and solve
for the remaining letter (the one you eliminated in steps 1 and 2.
---------------------
2x - y + z = 1
x + y - 7z = 2
3x - y + z = 0
1. Pick 2 of the equations, and pick a letter to eliminate, and multiply
the equations through as is necessary to make the letter cancel out.
I will pick the 2nd and 3rd equations and pick y to eliminate because that's
easy since the y's cancel without multiplying either one through.
x + y - 7z = 2
3x - y + z = 0
---------------
4x - 6z = 2
[Notice that since all the numbers in that equation are even, we can
divide that equation through by 2.]
2x - 3z = 1
2. Take the equation you did not use in step 1, together with one of the
equations you used in step 1, and eliminate the SAME letter you eliminated
in step 1.
The equation I did not use in step 1 is
2x - y + z = 1
I will take it together with this one that I did use in step 1:
x + y - 7z = 2
I will eliminate the letter y because that's the same letter I chose to
eliminate in step 1.
I don't need to multiply either equation through by anything because
the y's cancel as they are:
2x - y + z = 1
x + y - 7z = 2
---------------
3x - 6z = 3
[Notice that since all the numbers in that equation are multiples of 3,
we can divide that equation through by 3.]
x - 2z = 1
3. The results of step 1 and step 2 give you two equations in two letters.
Solve them for the two letters they contain.
2x - 3z = 1
x - 2z = 1
To make the x's cancel multiply the second equation through by -2 and add
corresponding terms:
2x - 3z = 2
-2x + 2z = -1
--------------
-z = 1
z = -1
Substitute -1 for z in
x - 2z = 1
x - 2(-1) = 1
x + 2 = 1
x = -1
4. Substitute those two values in one of the original equations and solve
for the remaining letter (the one you eliminated in steps 1 and 2.
So we substitute x = -1 and z = 1 in one of the original equations, say,
x + y - 7z = 2
-1 + y - 7(-1) = 2
-1 + y + 7 = 2
y + 6 = 2
y = -4
So the solution is (x,y,z) = (-1,-4,-1)
Edwin
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi
solving system of linear equations through elimination
2x-y+z=1
x+y-7=2
3x-y+z=0
2x-y+z=1
3x-y+z=0
-2x +y -z=-1 |Mulitplying 1st equation by -1, prior to adding the equations
3x -y +z=0
x = -1
x+y-7=2 |Substituting x = -1 into an equation to solve for y
-1 + y = 9
y = 10
3x-y+z=0 |Substituting x = -1 and y = 10 into an equation to solve for z
-3 -10 + z = 0
z = 13
CHECKING our Answer***
2x - y + z =1
-2 -10 + 13 = 1
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! 2x-y+z=1
x+y-7z=2
3x-y+z=0
-----------------
Move the 2nd equation to the top of the list to get:
x+y-7z=2
2x-y+z=1
3x-y+z=0
--------------
Subtract 2R1 from R2
Subtract 3R1 from R3
-------
2x-y+z = 1
0-3y+15z = -3
0-4y+22z = -6
-------
Subtract R3 from R2 to get:
2x-y+z = 1
0+ y-7z = 3
0-4y+22z = -6
--------
Add 4R2 to R3 to get:
2x-y+z = 1
0+ y-7z = 3
0 +0-6z = +6
-----
Divide R3 by -6 to get:
2x-y +z = 1
0+ y -7z = 3
0 +0 + z = -1
===================
Substitute z = -1 into R2 and into R1
2x- y - 1 = 1
0 + y + 7 = 3
0 + 0 + z = -1
---
Solve for "y" in R2 to get:
2x- y -1 = 1
0+ y + 0 = -4
0 + 0+ z = -1
---
Substitute y = -4 into R1 to get:
2x + 4 -1 = 1
0+ y + 0 = -4
0 + 0+ z = -1
---
Solve R1 for "x":
x + 0 + 0 = -1
0 + y + 0 = -4
0 + 0 + z = -1
======================
You will find more examples using Google
on the web. Search for "solving linear systems".
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cheers,
Stan H.
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