Question 55821: 2a-5b+c=5
3a+2b-c=17
4a+3b+2c=17
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 2a - 5b + c = 5
3a + 2b - c = 17
4a + 3b + 2c = 17
:
If we mult the 1st eq by -1 we can eliminate c from all three equation:
-2a + 5b - c = -5
3a + 2b - c = 17
4a + 3b + 2c = 17
---------------------add
5a + 10b = 29
:
Adding the 1st and 2nd equations eliminates c also:
2a - 5b + c = 5
3a + 2b - c = 17
-------------------add
5a - 3b = 22
:
WE have two equations with two unknowns a & b
5a + 10b = 29
5a - 3b = 22
--------------- subtract
13b = 7
b = 7/13
:
Find a using the equation 5a - 10b = 29, substitute 5/13 for b
5a + 10(7/13) = 29
5a + 70/13 = 29
5a = 377/13 - 70/13, (converted 29 to 13th)
5a = 307/13
a = (307/13) / 5
a = 307/65
:
Find c using the 1st equation, 2a - 5b + c = 5, substitute for a & b
2(307/65) - 5(7/13) + c = 5
:
614/65 - 5(35/65) + c = 5; convert 7/13 to 65th
:
614/65 - 175/65 + c = 5
:
439/65 + c = 5
:
c = 325/65 - 439/65; converted 5 to 65th
:
c = -114/65
:
So we end up with these nasty fractions (the guy who dreamed this up must be a sadist)
a = 307/65; b = 7/13; c = -114/65
:
Because of the tremendous opportunity for error here, I check these solutions using the matrix feature on the TI*83 and got the decimal equivalent for each one of these fractions.
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