Question 26453: Solve for a, b, and c using addition-subtraction method
a - 3b - c = 10
5a - 2b + 2c = 6
3a + 2b + c = 13
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! Solve for a, b, and c using addition-subtraction method
a - 3b - c = 10
5a - 2b + 2c = 6
3a + 2b + c = 13
To solve a system of 3 equations in 3 unknowns using the
addition-subtraction.
Case 1. When one of the equations contains only one unknown.
Solve for the unknown in that equation. Substitute its value
in the other two equations and you have two equations and two
unknowns. Solve that system of two equations in two unknowns.
Thios will give you the other two unknowns.
Case 2. When one of the equations contains only two unknowns.
Pick one of the equations with only two unknowns.
Observe which unknown is missing in it.
Eliminate that unknown from the other two equations
This gives you a second equation with the same two unknowns as the one
you picked.
Solve this system for those two unknowns.
Substitute these values in one of the original three equations which
contains the remaining unknown and solve to find its value.
Case 3. When all three equations contain three unknowns
Pick any pair of equations and eliminate an unmnown.
Pick a different pair of equations and eliminate THAT VERY SAME UNKNOWN.
This gives you two equations in two unknowns.
Solve for those two unknowns.
Substitute these values in any one of the original equations and solve
for the third unknown.
a - 3b - c = 10
5a - 2b + 2c = 6
3a + 2b + c = 13
This is case 3 because all three equations contain all three unknowns.
Pick any pair of equations and eliminate an unmnown.
Pick the 1st and 3rd and eliminate c
a - 3b - c = 10
3a + 2b + c = 13
Add them and the c's will cancel:
4a - b = 23
Pick a different pair of eqwuations and eliminate THAT VERY SAME UNKNOWN.
Pick the 1st and 2nd and eliminate c
a - 3b - c = 10
5a - 2b + 2c = 6
Multiply ther first equationm thru by 2 so the c's will cancel:
2a - 6b - 2c = 20
5a - 2b + 2c = 6
Add those two equations:
7a - 8b = 26
This gives you two equations in two unknowns.
4a - b = 23
7a - 8b = 26
Solve for those two unknowns.
Multiply the first equation thru by -8 so the b's will cancel:
-32a + 8b = -184
7a - 8b = 26
Add those two equations
-25a = -158
a = 158/25 = 6.32
Substitute 6.32 for a in 4a - b = 23
4a - b = 23
4(6.32) - b = 23
25.28 - b = 23
-b = -2.28
b = 2.28
Substitute these values in any one of the original equations and solve
for the third unknown.
Substitute 6.32 for a and 2.28 for b in the first original equation
and solve for c:
a - 3b - c = 10
6.32 - 3(2.28) - c = 10
6.32 - 6.84 - c = 10
-.52 - c = 10
-c = 10.52
c = 10.52
Solution: (a, b, c) = (6.32, 2.28, -10.52)
Edwin McCravy
AnlytcPhil@aol.com
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