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!!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

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. 

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