Question

How do you solve `4x + y = 4` and `2x + 8y = 0` using substitution?

Answer 1

`(x,y)=(16/15,-4/15)`
(see below for solution method using substitution)

Explanation:

Given: `4x+y=4`

This implies: `y=4-4x`
So we can substitute `(4-4x)` for `y`
in the second given equation: `2x+8y=0`
to get:
`color(white)("XXX")2x+8(4-4x)=0`

Simplifying:
`color(white)("XXX")2x+32-32x=0`

`color(white)("XXX")30x=32`

`color(white)("XXX")x=16/15`

The we can substitute `16/15` for `x`
in the first given equation: `4x+y=4`
to get:
`color(white)("XXX")4*(16/15)+y=4`

`color(white)("XXX")y=4-64/15 = -4/15`

Answer 2

`(x,y)=(16/15,-4/15)`

Explanation:

Solve by substitution and elimination

`color(blue)(4x+y=4`

`color(blue)(2x+8y=0`

We can eliminate `4x` from the first equation by `2x` in the second equation (whole equation) with `-2` to get `-4x`

`rarr-2(2x+8y=0)`

Use the distributive property

`color(brown)(a(b+c=x)=ab+ac=ax`

`rarr-4x-16y=0`

Now, add the above equation with the first equation to eliminate `4x`

`rarr(4x+y=4)+(-4x-16y=0)`

`rarr-15y=4`

Divide both sides by `-15`

`rarr(cancel(-15)y)/cancel(-15)=4/-15`

`color(green)(rArry=-4/15`

Now,substitute the value of `y` to the first equation

`rarr4x+(-4/15)=4`

`rarr4x-4/15=4`

Add `4/15` both sides

`rarr4x-4/15+4/15=4+4/15`

`rarr4x=64/15`

Divide both sides by `4`

`rarr(cancel4x)/cancel4=64/15-:4`

(I have written `64/15/4` as `64/15-:4`)

Take the reciprocal and multiply

`rarrx=cancel64^16/15*1/cancel4^1`

`color(green)(rArrx=16/15`

`color(blue)(ul bar |(x,y)=(16/15,-4/15)|`