Question

How do you solve `x= 3y-1` and `x+2y=9` using substitution?

Answer 1

`(5,2)`

Explanation:

You know the value of the variable `x`, so you can substitute that into the equation.
`overbrace((3y - 1))^(x) + 2y = 9`

Remove the parentheses and solve.
`3y - 1 + 2y = 9`

`=> 5y - 1 = 9`

`=> 5y = 10`

`=> y = 2`

Plug `y` into either equation to find `x`.
`x = 3overbrace((2))^(y) - 1`

`=> x = 6 - 1`

`=> x = 5`

`(x,y) => (5,2)`

Answer 2

`x=5, y=2`

Explanation:

Given `x=3y-1 and x+2y=9`

Substitute `x=3y-1` into `x+2y=9`,

`(3y-1)+2y=9`
`5y-1=9`
`5y=10`
`y=2`

Substitute y=2 into the first equation,
`x=3(2)-1`
`x=5`

Answer 3

`x = 5`
`y = 2`

Explanation:

If

`x = 3y -1`

then use that equation in the second equation. This means that

`(3y - 1) + 2y = 9`

`5y - 1 = 9`

`5y - 1 + 1 = 9 + 1`

`5y = 10`

`(5y)/5 = 10/5`

`y = 2`

Having said this, just replace the `y` in the first equation in order to get the `x`.

`x = 3(2) -1`

`x = 6 -1`

`x = 5`

After that, just check that the values make sense:

`x = 3y - 1`

`5 = 3(2) -1`

`5 = 6 - 1`

`5 = 5`

And for the second one:

`x + 2y = 9`

`5 + 2(2) = 9`

`5 + 4 = 9`

`9 = 9`

Both answers satisfy both equations, which makes them correct.