Question

How do you solve the system of equations `x-y+z=6`, `4y+1=z`, `3y=2z`?

Answer 1

`x=6 1/5`, `y=-2/5`, and `z=-3/5`

Explanation:

`x-y+z=6`
`4y+1=z`
`3y=2z`

From the 2nd and 3rd equations, we can calculate the values of `y` and `z`.

In the third equation, substitute `z` with `(color(red)(4y+1))`, the value from the second equation.

`3y=2z`

`3y=2(color(red)(4y+1))`

Open the brackets and simplify.

`3y=8y+2`

Subtract `8y` from both sides.

`-5y=2`

Divide both sides by `5`.

`-y=2/5` or `y=-2/5` (multiplying both sides by `-1`)

In the third equation, substitute `y` with `(color(blue)(-2/5))`.

`3y=2z`

`3(color(blue)(-2/5))=2z`

Open the brackets and simplify.

`-6/5=2z`

Divide both sides by `2`.

`-3/5=z` or `z=-3/5`

Now, in the first equation, substitute `y` with `(color(blue)(-2/5))` and `z` with `(color(green)(-3/5))`.

`x-y+z=6`

`x-(color(blue)(-2/5))+(color(green)(-3/5))=6`

Open the brackets and simplify. The product of two negatives is a positive and the product of a positive and a negative is a negative.

`x+color(blue)(2/5)-color(green)(3/5)=6`

Multiply all terms by `5`.

`5x+2-3=30`

`5x-1=30`

Add `1` to each side.

`5x=31`

Divide both sides by `5`.

`x=31/5=6 1/5`