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Solve the following linear equation. 2x − y + z = 8, x + y − z = 10, x + y + 2z = 12
x=3−1,y=6,z=314
x=−6,y=3−14,z=31
x=32,y=−6,z=314
x=6,y=314,z=32
x=6,y=314,z=32
Add equation (1) and (2) directly to eliminate y and z, getting 3x = 18, which gives x = 6. Then use x = 6 to simplify (2) and (3) to find y and z.
System of equations: 1) 2x - y + z = 8, 2) x + y - z = 10, 3) x + y + 2z = 12
2x−y+z=8, x+y−z=10, x+y+2z=12
Add equation (1) and (2) directly to eliminate y and z, getting 3x = 18, which gives x = 6. Then use x = 6 to simplify (2) and (3) to find y and z.
Students often make sign errors when adding or subtracting the equations to eliminate variables.
Solve for x
Add equation (1) and equation (2) to eliminate y and z: (2x - y + z) + (x + y - z) = 8 + 10.
3x=18⟹x=6
Solve for z
Subtract equation (2) from equation (3) to eliminate x and y: (x + y + 2z) - (x + y - z) = 12 - 10.
3z=2⟹z=32
Solve for y
Substitute x = 6 and z = 2/3 into equation (2): 6 + y - 2/3 = 10.
y=4+32=314
D is correct because the values are x = 6, y = 14/3, and z = 2/3.
This type of system is similar to solving Kirchhoff's Laws in circuit analysis or calculating net forces in mechanics.
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