SOLUTION: Systems of equations in three variables HELP!! a+b=3 -b+c=3 a+2c=10

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Question 1052089: Systems of equations in three variables HELP!!
a+b=3
-b+c=3
a+2c=10
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
How do you WANT to solve it?

Simpler system:

Elimination passage to keep c.

---------Try this in the second equation of your list.

Answer by ikleyn(52786)   (Show Source):
You can put this solution on YOUR website!
.
Systems of equations in three variables HELP!!
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a + b = 3 (1) - b + c = 3 (2) a + 2c = 10 (3) Add equations (1) and (2) (both sides). You will get a + c = 6. (4) Distract equation (4) from the equation (3). You will get c = 10-6 = 4. So, you just found the solution for c: c=4. Substitute c=4 into equation (2). You will get -b + 4 = 3. Hence, b = 4-3 = 1. Substitute b=1 into equation (1). You will get a + 1 = 3. Hence, a = 3-1 = 2. Answer. a=2, b=1, c=4.

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!
Systems of equations in three variables HELP!!
a+b=3
-b+c=3
a+2c=10

a + b = 3 ------- eq (i)
- b + c = 3 ------- eq (ii)
a + 2c = 10 ------- eq (iii)
a + c = 6 ------- Adding eqs (ii) & (i) ------- eq (iv)
------ Subtracting eq (iv) from eq (iii)
- b + 4 = 3 ------- Substituting 4 for c in eq (ii)
- b = 3 - 4
- b = - 1
b = , or
a + 1 = 3 ------- Substituting 1 fro b in eq (i)
a = 3 - 1, or
Again, you need to IGNORE JOGS/JOSG's solution, if it can be called that.