Question

How do you solve the system `x= 175+15y`, `.196x= 10.4y`, `z=10*y`?

Answer 1

`x=53.05`
`y=-8.13`
`z=-81.3`

Explanation:

To begin, plug in a equation for a variable that is already isolated. There are two variables that are already isolated, and those are `x and z`. You could choose either to work with, doesn't matter, but I'm going to use `x`

So, plug the equation for `x` into another equation that has that same variable in it. Which would be:

`.196x=10.4y`

To begin, plug `x` equation into `.196x=10.4y`

`.196(175+15y)=10.4y`

Distribute `.196` throughout the set of parenthesis

`34.3+2.94y=10.4y`

Begin to isolate `y` by subtracting `34.3` on both sides of the equation

`2.94y=-23.9`

Isolate y by dividing `2.94` on both sides of the equation

`y=-8.13`

Now, we have solved for one of the variables. The next thing to do is plug `y` into another equation that contains `y` in it. An easy one to use would be `x=175+15y` because `x` is already isolated

So, plug `y=-8.13` into `x=175+15y`

`x=175+15(-8.13)`

Distribute `15` throughout the set of parenthesis

`x=175-121.95`

Subtract

`x=53.05`

And now we have solved for `y` and `x`. The only variable left is `z`

To solve for `z`, you have to plug `y=-8.13` into the equation

`z=10(-8.13)`

Multiply

`z=-81.3`

`x=53.05`
`y=-8.13`
`z=-81.3`