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 )}