How do you solve the following system?: `1/2x=y-1 , 1/2x+1/3y=5`

Notice that both equations coefficient of `x" is "+1/2`

This makes life easy as we can 'get rid' of one for the variables.

To solve an equation easily you need just one variable and a relationship to some constant.
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Given:
`1/2x=y-1`............................(1)
`1/2x+1/3y=5`......................(2)

It is common practice to convert them so that they are of form
`y=mx+c`. This not strictly necessary in this case as the process will sort this out any way.

Equation (2) has the lesser value of `y` so I am opting for:

`color(brown)("Subtract equation (2) from equation (1) giving:")`

`color(green)("Notice how I have used 0 as a place keeper")`

`" "1/2x+0ycolor(white)(.)=y-1`
`" "underline(1/2x+1/3y=0y+5 )->"subtract"`
`" "color(green)(0" " -1/3y=y-6)`

Add `color(magenta)(1/3y)` to both sides

`" "color(green)(-1/3ycolor(white)(.)color(magenta)(+1/3y)=ycolor(white)(.)color(magenta)(+1/3y)-6)`

`" "0=4/3y-6`

Add 6 to both sides giving:

`6=4/3y`

`color(blue)(y=(6xx3)/4 = 9/2)`
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I am choosing equation (1) for substitution of y

`1/2x=y-1" "->" "1/2x=9/2-1`

`color(blue)(x=7)`
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Check

`1/2x+1/3y=5 " becomes "1/2(7)+1/3(9/2) = 5` Confirmed!