SOLUTION: Hi, was wondering if anyone could help me with the following system of equations problem. Steps on how to solve fully would be ideal also. A-B+C=1 2A-2B=0 3A+B-3C=2 Any hel

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Question 190977: Hi, was wondering if anyone could help me with the following system of equations problem. Steps on how to solve fully would be ideal also.
A-B+C=1
2A-2B=0
3A+B-3C=2
Any help would be most appreciated,
-Thanks, -Nick.
Found 3 solutions by ankor@dixie-net.com, solver91311, stanbon:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website!
A - B + C = 1
2A -2B = 0
3A + B -3C = 2
:
multiply the 1st equation by 2, subtract the 2nd equation
2A - 2B + 2C = 2
2A - 2B + 0C = 0
------------------ subtraction eliminates A & B, find C
2C = 2
C = 1
:
Substitute 1 for C in the 1st and 3rd equations
A - B + 1 = 1
3A +B -3(1) = 2
:
A - B = 1 - 1
3A +B = 2 + 3
:
A - B = 0
3A +B = 5
------------Adding eliminates B
4A = 5
A =
:
Find B using 2A - 2B = 0; therefore: B = A
B =
:
Solution: A = , B = , C = 1
;
:
Check solution in the 3rd equation:
3A + B -3C = 2
3() + - 3(1) = 2
+ - 3 = 2
- 3 = 2
5 - 3 = 2

Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

Look at the middle equation:

Now use that fact in the first equation:

Use this information in the last equation:

Check:
, yep

, yep

, yep

John

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
A -B + C = 1
2A -2B + 0 = 0
3A +B -3C = 2
---------------
Subtract 2 times the 1st from the 2nd; Subtract 3 times the 1st from the 3rd:
A -B + C = 1
0 +0 -2C = -2
0 +4B -6C = -1
------------------------
The 2nd equation tells us that C = 1
--------------------------
Substitute that into the 3rd equation to solve for "B":
4B - 6 = -1
4B = 5
B = 5/4
--------------------------
Substitute those values into the 1st equation to solve for "A":
A - (5/4) + 1 = 1
A = 5/4
============================
Cheers,
Stan H.