How do you solve `x + 2y = 6` and `x - 4y = 8` using substitution?

If `x-4y=8`
then `x=4y+8`

Therefore we can substitute `color(blue)(4y+8)` for `x`
in the other given equation: `x+2y=6`
giving:
`color(white)("XXX")color(blue)(4y+8)+2y=6`
which simplifies as:
`color(white)("XXX")6y=-2`
or
`color(white)("XXX")y=color(red)(-1/3)`

We can then substitute `color(red)(-1/3)` for `y`
in one of the given equations, say: `x-4y=8`
giving:
`color(white)("XXX")x-4(-1/3)=8`

`color(white)("XXX")x+4/3=8`

`color(white)("XXX")x=6 2/3`

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Verification is always a good idea.
So testing the above solution `(x,y)=(6 2/3,-1/3)` in the given equation not used to evaluate `x`
`color(white)("XXX")x+2y=6` (????)

`color(white)("XXX")6 2/3+2xx(-1/3) = 6` (????)

`color(white)("XXX")6 2/3 - 2/3 = 6` (Correct!)

x+2y=6 -----------------(1) and x-4y=8------------------(2)

Steps:
~>. Substitute one of the following equations, either (1) OR (2)

substituting (1)

Put x subject of formula,

x= 6-2y--------------(*)

~> Replace the equation (*) in equation (2)

x-4y=8
replacing.....

(6-2y)-4y=8
6-2y-4y=8
6-6y=8

Transpose "+6"on the other side of the equation
-6y=8-6
-6y=2

y= `2/-6`

y=`-(1/3)`

Hence, Replace the value obtained for y in equation (*)

x=6-2y
x=6-(`2*-1/3`)
x=6-(`-2/3`)
x=6+(`2/3`)
x=`20/3`

Therefore: x=`20/3` and y=`-(1/3)`