SOLUTION: 2x+y+z = 5 x+3z = 14 -2x-3y+2z = 2

Question 120975: 2x+y+z = 5
x+3z = 14
-2x-3y+2z = 2
Answer by madhan_math(34) (Show Source): You can put this solution on YOUR website!
Given equations are
2x+y+z = 5 ------(1)
x+3z = 14--------(2)
-2x-3y+2z = 2----(3)
First we solve equations (1) and (3),
eqn.(1)*3+eqn.(3)we get,(since cancelling of y)
4x+5x=17--------(4)
Now we solve equations (4) and (2)
eqn.(4)-eqn.(2)*4 (Since cancelling of x), we get,
z=39/7 and substitute this value in equation (2) then we get x=-19/7.
Finally we substitute the x and z values in equation (1) we get
y=34/7.
Therefore, the values are
x=-19/7,
y= 34/7
z= 39/7 (If you like check these values are correct substitute these values in any equation and verify)
RELATED QUESTIONS
Solve each system 2x+y+z=14 -x-3y+2z=-2... (answered by MathLover1)

Extrememly confused, help please! Solve each system in three variables... (answered by stanbon)

Solve the system using any algebraic method. 1. x-y+z=5 2y+3z=14 -3y+2z=5... (answered by aswathytony)

x-3y+2z = 6 2x-y+3z=5 z + y -z =... (answered by ikleyn)

solve each system 4x+8y+z=2 x+7y-3z=-14... (answered by venugopalramana)

2x-3y+z=2 x+3y+2z=2... (answered by Alan3354)

6x+3y-2z=1 2x+y-3z=4... (answered by fners)

solve each system x+y-z= -1 4x-3y+2z= 16... (answered by venugopalramana)

2x + y - 2z = -1 3x - 3y - z = 5 x - 2y + 3z =... (answered by josgarithmetic,Alan3354)