Question 3.6: Solve the equations 2x + y − z = 4 (1) x + y − z = 3 (2) 2x ......

The simplest approach is to add equation (3) to equation (1), and hence eliminate z, giving an equation in x and y. Then add equation (3) to equation (2), eliminating z again, giving another equation in x and y.

2x + y − z =4          (1)

2x + 2y + z = 12     (3)
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4x + 3y + 0 = 16    (4) adding equations (1) and (2)

x + y − z =3            (2)

2x + 2y + z = 12     (3)
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3x + 3y + 0 = 15     (5) adding equations (2) and (3)

Equations (4) and (5) are the usual two equations in two unknowns, so solve for x and y. Then solve for z later.

4x + 3y = 16            (4)

−3x − 3y = −15       (6) equation (5) multiplied by −1
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x = 1                          adding equations (4) and (6)

So, x = 1. Substitute x = 1 into equations (4), (5) or (6) to solve for y.
Substituting x = 1 into equation (4) gives 4(1) + 3y = 16→y = 4.
Finally, find z by substituting x = 1, y = 4 into any of the equations (1), (2) or (3). For example, substituting into equation (2),

1 + 4 − z = 3 → z = 2

Therefore, the values which satisfy all three equations (1), (2) and (3) are x = 1, y = 4, z = 2.