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The solution of the system of linear equations is: x + y + z = 6, 2x − y + 3z = 14, −x + 2y −z = −2
x=3−4,y=34,z=6
x=34,y=3−4,z=5
x=35,y=3−5,z=6
x=3−5,y=35,z=−6
x=3−4,y=34,z=6
Add equation (1) and (3) to eliminate z: (x+y+z) + (-x+2y-z) = 6-2 => 3y = 4 => y = 4/3. Only options A and B have y = 4/3. Checking option A's x = -4/3 in equation (1): -4/3 + 4/3 + z = 6 => z = 6.
System of equations: x + y + z = 6 (1), 2x - y + 3z = 14 (2), -x + 2y - z = -2 (3)
x+y+z=6, 2x−y+3z=14, −x+2y−z=−2
Add equation (1) and (3) to eliminate z: (x+y+z) + (-x+2y-z) = 6-2 => 3y = 4 => y = 4/3. Only options A and B have y = 4/3. Checking option A's x = -4/3 in equation (1): -4/3 + 4/3 + z = 6 => z = 6.
Students often miscalculate the signs when adding or subtracting equations, especially with negative coefficients like -x or -z.
Step 1: Simplify by adding eq (1) and (3)
Adding equation (1) x+y+z=6 and equation (3) −x+2y−z=−2 eliminates x and z terms involving x: (x−x)+(y+2y)+(z−z)=6−2.
3y=4⟹y=34
Step 2: Solve for x and z using y = 4/3
Substitute y=4/3 into (1) x+4/3+z=6⟹x+z=6−4/3=14/3. Substitute into (3) −x+2(4/3)−z=−2⟹−x−z=−2−8/3=−14/3⟹x+z=14/3.
x+z=314
Step 3: Solve the system
Using (2) 2x−4/3+3z=14⟹2x+3z=14+4/3=46/3. We have the system x+z=14/3 and 2x+3z=46/3. From first, z=14/3−x. Substitute into second: 2x+3(14/3−x)=46/3⟹2x+14−3x=46/3⟹−x=46/3−42/3=4/3.
x=−34,z=6
A is correct because substituting x=−34, y=34, and z=6 satisfies all three given linear equations.
This system-of-equations approach is identical to the method used to find intersection points of three planes in 3D Coordinate Geometry.
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