Question

How do you solve `1/2x+1/3y=-1/6` and `1/2x+2y=13/2` using substitution?

Answer 1

`(x,y)=color(blue)(""(-3,4))`

Explanation:

To simplify this process multiply each equation by appropriate constants to clear the fractions:
[1a]`color(white)("XXX") 1/2x+1/3y=-1/6`
`color(white)("XXXXXXXXXXXXX")rarr`[1b]`color(white)("XXX") 3x+2y=-1`

[2a]`color(white)("XXX") 1/2x+2y=13/2`
`color(white)("XXXXXXXXXXXXX") rarr`[2b]`color(white)("XXX") x+4y=13`

[2b} appears to be the easiest one to convert into a form for substitution:
[2c]`color(white)("XXXXXXXXXXX") x=13-4y`

Substituting this back into [1b]
`color(white)("XXXXX") 3(13-4y)+2y=-1`

`color(white)("XXXXX") 39-10y=-1`

`color(white)("XXXXX") 10y=40`

`color(white)("XXXXX") y=4`

We can now substitute `4` for `y` back in [2c] to get
`color(white)("XXXXX") x=-3`

[as always, you should substitute these values back into the original equations to verify the results... assume that I've done this.]