Question

How do you solve the system of equations `7x + 5y + 7z = 60`, `15x + 2z = - 195`, and `5z = 75`?

Answer 1

`x=-15`
`y=12`
`z=15`

Explanation:

First solve for `z` using the third equation

`z=75/5=15`

Now that we know `z`, we can plug it into the second equation and solve for `x`

`15x+2(15)=-195`

`15x=-225`

`x=-15`

Now we can find `y` using the values of `x` and `z` and plugging them into the first equation

`7(-15)+5y+7(15)=60`

`5y=60`

`y=12`

Answer 2

`x =-15`
`y = 12`
`z=15`

Explanation:

This question is far easier than it appears at first.

The key idea is that if a a linear equation has `1` variable, there will be one solution, but as soon as there are `2` variable, `2` equations are required and likewise for `3` variables, there must be `3` equations.

We have all of these scenarios presented here.

Solve the `3rd` equation first as it only has `1` variable,

`5z = 75`
`color(blue)(z =15)" "larr` now use this value for `z` in the `2nd` equation:

`15x +2color(blue)(z) = -195`
`15x +2color(blue)((15)) = -195`
`15x +color(blue)(30) = -195`
`15x = -225`
`color(green)(x =-15)" "larr` use this value for `x` in the first equation

`7color(green)(x)+5y+7color(blue)(z)=60`
`7color(green)((-15))+5y+7color(blue)((15))=60`

`-105 +5y +105 = 60`
`5y=60`
`y=12`