# Solve linear systems where some of the equations have missing terms

Solve linear systems where some of the equations have missing terms

If a linear system has an equation missing a term or terms, one elimination step can be omitted.

Example

Solve the system

6x – 12y = -5 (1)

8y + z = 0 (2)

9x – z = 12 (3)

Since equation (3) is missing the variable y, a good way to begin the solution is to eliminate y again using equation (1) and (2).

12x – 24y = – 10 Multiply each side of (1) by 2.

24y + 3z = 0 Multiply each side of (2) by 3.

12x + 3z = -10 Add. (4)

Use this result, together with equation (3), 9x – z = 12,  to eliminate z. Multiply equation (3) by 3.

This gives

27x – 3z = 36 Multiply each side of (3) by 3.

12x + 3z = – 10 (4)

39x = 26          Add.

x =   = .

Substituting into equation (3) gives

9x – z = 12 (3)

9() – z = 12 Let x = .

Substituting  -6 for z in equation (2) gives

8y + z = 0 (2)

8y – 6 = 0 Let z = – 6.

8y = 6

y = .

Thus, x = , y = , and z  = -6. Check these values in each of the original equations of the system to verify that the solution set of the system is {(,, -6 )}