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

Substitute `y= -x + 9` in equation 1: `3y= -0.5x + 2`

`3(-x + 9) = -0.5x + 2`

`-3x + 27 = -0.5x + 2`

`-3 + 0.5 = 2 - 27`

`-2.5x = -25`

The minus signs cancel each other

`2.5x = 25`

`x = 10`

Now substitute `x = 10` in equation 2: `y = -x + 9`

`y = -10 + 9`

`y = -1`

Substituting a formula into another essentially means, to set one formula equal to a variable and then inserting the formula into the other formula. Though this might seem complicated, this can be done easily with these equations:
`3y = ((-1)/2)x + 2`
`y = -x + 9`
`3(-x+9) = ((-1)/2)x + 2`
`-3x + 27 = ((-1)/2)x + 2`

Now rearrange the equation to collect like terms on either sides.
1) multiply both sides of the equation by 2 to simplify the situation.
`-6x + 54 = -1x + 4`

2) add `1x` to both sides, to eliminate the x on the right side
`-6x (+ x) + 54 = -1x (+ x) + 4`
`-5x + 54 = 4`

3) subtract `54` from both sides
`-5x + 54 (- 54) = 4 (- 54)`
`-5x = -50`

4) now divide both sides of the equation by `-5`
`(-5x)/-5 = (-50)/-5`
`x = 10`

Now just use this value in one of the initial equations.
`y = -x + 9`
`y = -(10) + 9`
`y = -1`