Question

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

Answer 1

`(x,y)=(1/2,-1/2)`

Explanation:

Given
[1]`color(white)("XXX")3x+5y=-1`
[2]`color(white)("XXX")5x-y=3`

By substitution
Re-write [2] as
[3]`color(white)("XXX")y=5x-3`

Substitute `(5x-3)` for `y` in [1]
[4]`color(white)("XXX")3x+5(5x-3)=-1`

Simplify
[5]`color(white)("XXX")3x+25x-15=-1`

[6]`color(white)("XXX")28x=14`

[7]`color(white)("XXX")x=1/2`

Substitute `1/2` for `x` in [2]
[8]`color(white)("XXX")5(1/2)-y=3`

[9]`color(white)("XXX")y=5/2-3=-1/2`

Answer 2

`(x,y)->(1/2,-1/2)`
`color(brown)("Finding x is explained in detail. Finding y is explained using shortcuts.")`

Explanation:

Given:

`color(brown)(3x+5y=-1)` ........................................(1)

`color(brown)(5x-y=3)`..................................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To find "x)`

From (2) we have `y=5x-3...........(2_a)`

Substitute `(2_a)`into (1) giving:

`3x+5(5x-3)=-1`

`3x+25x-15=-1`

`color(brown)(28x-15=-1)`
'-----------------------------------------------------------
Add `color(blue)(15)` to both sides giving:

`color(brown)(28x-15color(blue)(+15) =-1 color(blue)(+15))`

`28x=14`
'--------------------------------------------------------
Divide both sides by 28 which is the same as `color(blue)(xx1/28)`

`28/28 xx x= 14xx 1/28`

but `28/28=1" and " 14/28 = 1/2` giving:

`color(magenta)(x=1/2)`...........................................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("To find y")` (Using shortcuts)

Substitute (3) into (1) or (2): Using (2) is easier!

`5x-y=3 -> 5(1/2)-y=3`
`color(magenta)(y=2 1/2 -3 =-1/2)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`(x,y)->(1/2,-1/2)`