If given a system of equations which contain multiple variables:
(1) 2x - 3y + 4z = 11
(2) x + 2y - z = 4
(3) 5x + y + 2z = 21
The easiest way to solve for x, y, and z is to match the equations together. This is done by multiplying the entire equation by a given number, then adding the equations together successively.
STEP 1
First, match the first and second equations so that the first variable is removed. This is done by multiplying one or both of the equations so that the first respective variables ('x' in this case) nullify each other:
Multiply (2) by -2:
[x + 2y - z = 4] * -2 => [-2x -4y + 2z = -8]
and add (1) and (2) together:
+2x – 3y + 4z = 11
– 2x – 4y + 2z = -8
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0x - 7y + 6z = 3
This gives the equation (a). (a) will be used later on and should be remembered. This gives 4 equations to work with, (1), (2), (3), and (a).
(a) -7y + 6z = 3
STEP 2
Next, use the same method to match equations (2) and (3).
Multiply equation (2) by -5 so that 'x' cancels itself out:
[x + 2y -z = 4] * -5 => [-5x -10x +5z = -20]
add (2) and (3) together:
-5x - 10y + 5z = -20
+5x + y + 2z = 21
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0x – 9y + 7z = 1
this gives (b). (b) will be used later on and should be remembered. This gives 5 equations to work with, (1), (2), (3), (a), and (b).
(b) -9y + 7z = 1
STEP 3
Now match the new equations (a) and (b) to eliminate the next variable:
(a) [-7y + 6z = 3]
(b) [-9y + 7z = 1]
(a) * -9 => [63y – 54z = -27]
(b) * 7 => [-63y + 49z = 7]
add the newly-multiplied equations to rid them of 'y':
+63y – 54z = -27
-63y + 49z = 7
---------------------
-5z = -20 or 5z = 20
this gives the new equation (c):
(c) 5z = 20
Solve (c) to get 'z':
5z = 20
z=20/5
z = 4
Now that z is known to be equal to 4, (a) or (b) can be solved to get 'y':
-7y + 6z = 3
-7y + 6(4) = 3
-7y = 3 - 6(4)
-7y = -21
-y =-21/7
y = 21/7
y = 3
and finally with 'y', solve (2) to get 'x':
x + 2(3) – 4 = 4
x + 2(3) = 8
x = 8 – 6 = 2
x = 2
y = 3
z = 4
