Question

How do you solve `15x - 20y = 1` and `5y = 1 + 10x` using substitution?

Answer 1

`x,y=-1/5`

Explanation:

`color(blue)(15x-20y=1`

`color(blue)(5y=1+10x`

Use the second equation

`rarr5y=1+10x`

Divide both sides by `5`

`rarr(cancel5y)/cancel5=(1+10x)/5`

`rarry=1/5+2x`

Now substitute this value to the first equation

`rarr15x-20(1/5+2x)=1`

Use the distributive property `color(brown)(a(b+c)=ab+ac`

`rarr15x-20/5-40x=1`

`rarr15x-4-40x=1`

`rarr-25x-4=1`

Add `4` both sides

`rarr-25xcancel(-4+4)=1+4`

`rarr-25x=5`

`color(green)(rArrx=-5/25=-1/5`

Substitute the value of `x` to the second equation

`rarr5y=1+10(-1/5)`

`rarr5y=1-10/5`

`rarr5y=1-2`

`rarr5y=-1`

`color(green)(y=-1/5`

`:. color(indigo)(ul bar |x=y=-1/5|`