SOLUTION: a+b=3 -b+c=3 a+2c=10

Algebra.Com
Question 14066: a+b=3
-b+c=3
a+2c=10

Answer by rahman(247)   (Show Source): You can put this solution on YOUR website!
a+b=3 ...............(1)
-b+c=3 ...............(2)
a+2c=10 ...............(3)
Use elimination method for (1) and (2):
a+b=3
-b+c=3
------ +
a+c=6 .............(4)
Use elimination method for (3) and (4):
a+2c=10
a+c=6
------ -
c=4
Substitute c=4 to (4):
a+c=6
a+4=6
a=6-4
a=2
Substitute a=2 to (1):
a+b=3
2+b=3
b=3-2=1
Thus, a=2, b=1, and c=4.
OK

RELATED QUESTIONS

Systems of equations in three variables HELP!! a+b=3 -b+c=3... (answered by josgarithmetic,ikleyn,MathTherapy)
2a + 3b + c + 2d = 20 4a + b + 2c + 5d = 50 a + 2b + 3c + d = 10 2a + 4b + 2c + 3d =... (answered by Fombitz,richwmiller,sonic1)
a + b + c + d = 7 2a - b + c - 2d= -13 3a + b - 2c - d = 5 4a - b + 3c - d = -10 (answered by math_helper)
solve by elimination a+b+c=-3 3b-c=4 2a-b-2c=-5 (answered by Fombitz)
Given log3 a = c and log3 b=2c, then solve for a SOLUTION take the base for common... (answered by owandera)
Evaluate the following expression for a = 6, b = 3 and c = 4. a ÷ b + 2c (answered by jim_thompson5910)
evaluate if a = 2, b = 5 and c = (-3): 3a to the second power - b + 2c... (answered by Deina)
solve each of the following equation. a-3=-5 3/b=9... (answered by mathtutor19)
Solve for a, b, and c using addition-subtraction method 

 a - 3b -  c = 10... (answered by AnlytcPhil)