Question

How do you solve using elimination of `x-15y=3` and `3x+y=21`?

Answer 1

`(x,y)=(159/23, 6/23)`

Explanation:

Given
[1]`color(white)("XXX")x-15y=3`
[2]`color(white)("XXX")3x+y=21`

Eliminate the `x` terms by subtracting `3xx`[1] from [2]

#{: ([2],color(white)("XXX"),,"(",3x,+y,"=",21,")"), (3xx[1],color(white)("XXX"),-,"(",3x,-45y,"=",9,")") :}#

[4]`color(white)("XXX")46y=12`
[5]`color(white)("XXX")y=6/23`

Eliminate the `y` terms by adding `15xx[2]` to [1]

#{: ([1],color(white)("XXX"),,"(",x,-15y,"=",3,")"), (15xx[2],color(white)("XXX"),+,"(",45x,+15y,"=",315,")") :}#

[6]`color(white)("XXX")46x=318`
[7]`color(white)("XXX")x=159/23`

Answer 2

I found:
`x=159/23`
`y=6/23`

Explanation:

We can multiply the first equation by `-3` and add (in column) to the second:
`{(-3xx[x-15y=3]),(3x+y=21):}`
`{(-3x+45y=-9),(3x+y=21):}` add:
`0+46y=12`
So:
`y=12/46=6/23`
use this into the first equation:
`x-15*color(red)(6/23)=3`
`x=90/23+3=159/23`