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Question 117012: Trying my best to get a better understanding of solving linear systems but I have a long way to go. I need step by step examples to follow. -
solve the linear system without using matrices or determinants
x+y+z=2
x-y+3z=12
2x+5y+2z=-2
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Here's one way of doing it without using matrices or determinants.
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You are given 3 equations with three unknowns. For reference purposes call the equations
A, B, and C. Note that you can pair these equations as follows: A & B, B & C, and A & C.
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Pick one pair of these equations and by variable elimination, eliminate one of the variables.
This will leave you an equation with two unknowns. For reference call this new equation "D".
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Next pick one of the other pairs and, by variable elimination, eliminate THE SAME VARIABLE as
you did in the first pair. This will leave you an equation with two unknowns. For reference
call this new equation "E".
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Take the new equations D and E and solve those two equations for the two variables
that are left in them.
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Once you have two of the variables solved, return to one of the original three equations
(A, B, or C), substitute those two variables, and solve for the third variable.
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Sounds complex, but not really once you work a couple of examples and become familiar with
the process. Let's use this process to solve the example you posted.
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Your three equations are (labeled A, B, and C for reference):
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(A) x + y + z = 2
(B) x - y +3z = 12
(C)2x +5y +2z = -2
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Select on pair to eliminate a variable. Suppose you select pair A and B. An easy variable
to eliminate in this pair is "y" because you can eliminate that variable just by adding the
two equations vertically in columns. So you have:
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(A) x + y + z = 2
(B) x - y +3z = 12
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Add vertically in columns to drop the "y" terms and the result is:
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(D) 2x + 4z = 14
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Notice that this new equation is referenced as "D".
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Next pick another pair ... either pair B & C or pair A & C. Let's take pair B & C as follows:
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(B) x - y +3z = 12
(C)2x +5y +2z = -2
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Using variable elimination, eliminate variable "y" again because we need to get equation E
that has only variables x and z as does equation D.
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To eliminate variable y let's first multiply equation B by 5 (all terms multiplied by 5).
That changes equation B and the pair becomes:
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(B) 5x -5y +15z = 60
(C) 2x +5y + 2z = -2
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Add these two equations vertically and note that the -5y and +5y cancel. The equation
resulting from vertical addition (equation E) is:
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(E) 7x + 17z = 58
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This means that we now have equations D and E as follows:
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(D) 2x + 4z = 14
(E) 7x +17z = 58
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Now we have two equations with only two unknowns in each. We can solve this set of equations
by using substitution or variable elimination ... and just for sake of consistency,
let's use variable elimination. Multiply all the terms in equation D by -7 and multiply
all the terms in equation E by 2. These two multiplications change the pair of equations to:
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(D) -14x - 28z = -98
(E) +14x + 34z = 116
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Adding these two equations vertically in columns eliminates the terms containing x and
the resulting equation is:
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6z = 18
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Solve for z by dividing both sides of this resulting equation by z and you get:
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z = 18/6 = 3
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Now that we know that z = 3, we can return to any of the D or E equations, substitute
3 for z, and solve for x. Let's return to the D equation:
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(D) 2x + 4z = 14
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Substituting 3 for z results in:
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(D) 2x + 4(3) = 14
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Multiply out the 4 times 3 to get:
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(D) 2x + 12 = 14
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Subtract 12 from both sides to get rid of the 12 on the left side and you have:
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(D) 2x = 2
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Divide both sides by 2 to solve for x and you have:
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x = 1
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So now we know that z = 3 and x = 1. With that knowledge, we can return to any of the original
equations, replace z with 3 and x with 1, and solve for y. Let's return to the original
equation A:
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(A) x + y + z = 2
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Replace z with 3 and x with 1 to get:
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(A) 1 + y + 3 = 2
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Combine the numbers on the left side:
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(A) y + 4 = 2
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Subtract 4 from both sides and you finally arrive at:
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y = -2
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So the simultaneous solution of all three of the original equations is x = 1, y = -2, z = 3
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Make sure that you understand the process of pairing the original equations, eliminating
one of the variables from one pair, eliminating the same variable from another pair, solving
the resulting pair of equations that have only 2 of the variables, then working your way back
up the chain to get all the unknowns.
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Hope this gives you an organized and understandable way of working problems such as this
without using matrices or determinants.
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