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6x - 4y + 5z = 31
5x + 2y + 2z = 13
x + y + z = 2
1. Pick a letter to eliminate.
2. Pick a pair of equations and
eliminate that letter
3. Pick a different pair of equations
and eliminate the SAME letter you
eliminated in step 1.
4. Now you have a system of two
equations in two unknowns. Solve
for the two unknowns.
5. Substitute these values in one of
the original equations to find
the value of the first unknown
eliminated in step 1.
6x - 4y + 5z = 31
5x + 2y + 2z = 13
x + y + z = 2
1. Pick a letter to eliminate.
I arbitrarily pick y.
2. Pick a pair of equations and
eliminate that letter
I arbitrarily pick the first
and third equations to eliminate
y.
To eliminate y I multiply the 3rd
equation through by 4 to make the
y-term become +4y so it will cancel
out with the -4y term in the 1st
equation:
6x - 4y + 5z = 31
4[ x + y + z = 2]
6x - 4y + 3z = 31
4x + 4y + 4z = 8
-------------------
10x + 9z = 39
3. Pick a different pair of equations
and eliminate the SAME letter you
eliminated in step 1.
I arbitrarily pick the second
and third equations to eliminate
the same letter y.
To eliminate y I multiply the 3rd
equation through by -2 to make the
y-term become -2y so it will cancel
out with the +2y term in the 2nd
equation:
5x + 2y + 2z = 13
-2[ x + y + z = 2]
5x + 2y + 2z = 13
-2x - 2y - 2z = -4
--------------------
3x = 9
x = 3
4. Now you have a system of two
equations in two unknowns. Solve
for the two unknowns.
The system we have is
10x + 9z = 39
x = 3
So we substitute x = 3 in
10x + 9z = 39
10(3) + 9z = 39
30 + 9z = 39
9z = 9
z = 1
5. Substitute these values in one of
the original equations to find
the value of the first unknown
eliminated in step 1.
I arbitrarily pick the 3rd original
equation and substitute x = 3 and
z = 1 into:
x + y + z = 2
3 + y + 1 = 2
4 + y = 2
y = -2
So the solution is:
(x, y, z) = (3, -2, 1)
Edwin
