Question

Solve the simultaneous equation

2x+2y+4z=24
6x+3y=15
y+2z=11 (5 marks).

Answer 1

First of all, we can simply the first and second equations by
`2` and
`3`, respectively, to get

`x+y+2z=12`

`2x+y=5`

`y+2z=11`

If we subtract the third equation from the first, we already figure out that
`x=1`. The system becomes

`y+2z=11`

`y=3`

`y+2z=11`

so we know
`y=3` as well. Let's plug the values for
`x` and
`y` in any of the three equations to get

`y+2z=11 \iff 3+2z=11 \iff 2z=8 \iff z=4`

So, the solution is
`x=1`,
`y=3`,
`z=4`