Solve the following linear equations

x+y-2z=10,    3x+y-z=12,    x+2y-z=5

The correct answer is x = 3, y = −1, z = −4.

Step-by-step Explanation:

We are given the following system of linear equations:

$x+y-2z=10$      --- (Equation 1)

$3x+y-z=12$      --- (Equation 2)

$x+2y-z=5$      --- (Equation 3)

Step 1: Eliminate x using Equation 1 and Equation 3
Subtract Equation 1 from Equation 3:

$(x+2y-z)-(x+y-2z)=5-10$

$y+z=-5$      --- (Equation 4)

Step 2: Eliminate x using Equation 1 and Equation 2
Multiply Equation 1 by 3:

$3(x+y-2z)=3\left(10\right)$

$3x+3y-6z=30$      --- (Equation 5)

Now subtract Equation 5 from Equation 2:

$(3x+y-z)-(3x+3y-6z)=12-30$

$-2y+5z=-18$      --- (Equation 6)

Step 3: Solve the two-variable system (Equation 4 and Equation 6)
Multiply Equation 4 by 2:

$2(y+z)=2(-5)$

$2y+2z=-10$      --- (Equation 7)

Now add Equation 6 and Equation 7 to eliminate y:

$(-2y+5z)+(2y+2z)=-18+(-10)$

$7z=-28$

$z=-4$

Step 4: Find the values of y and x
Substitute $z=-4$ into Equation 4:

$y+(-4)=-5$

$y=-5+4$

$y=-1$

Now substitute $y=-1$ and $z=-4$ into Equation 1:

$x+(-1)-2(-4)=10$

$x-1+8=10$

$x+7=10$

$x=3$

Thus, the final solution is x = 3, y = −1, z = −4.