How do solve the following linear system?: `8x+2y=3 , 2x+7=-5y`?

`8x+2y = 3`

`2x+7=-5y`

`* * * * * * * * * * * * * * * * * * *`

Let's isolate for `y` in the second equation:

`(-2x-7)/5 = y`

Now plug that into the first equation for `y`:

`8x+(2(-2x-7))/5 = 3`

`color(gray)(5/5) xx 8x - (4x)/5 - 14/5 = 3`

common denominator

`( 40x - 4x -14 )/5 = 3`

multiply both sides by `5`

`36x =29`

`color(green)(x = 29/36 = 0.806`

`color(white)(..)`

Now let's solve for `y`:

`8(29/36) + 2y = 3 xx color(gray)(36/36)`

`232/36 + 2y = 108/36`

`2y = 108/36 - 232/36`

`2y = -124/36`

`y = -124/72 = -1.722`

`* * * * * * * * * * * * * * * * * * * * * * * *`

Let's check our work by graphing the two equations and seeing where they intersect

Looks right! Good job

Given
[1]`color(white)("XXX")8x+2y=3`
[2]`color(white)("XXX")2x+7=-5y`

Converting [2] into standard form:
[3]`color(white)("XXX")2x+5y=-7`

We note that if we multiply [3] by `4` the coefficient of `x` becomes the same as that of [1]
[4]`color(white)("XXX")8x+20y=-28`

Subtracting [1] from [4] (to get rid of the `x` variable
[5]`color(white)("XXX")18y=-31`

Dividing both sides of [5] by `18`
[6]`color(white)("XXX")y=-31/18`

Substituting `(-31/8)` for `y` in [1]
[7]`color(white)("XXX")8x+2 *(-31/18)=3`

[8]`color(white)("XXX")8xcolor(white)("xxxxxxxxxxxx")=3+31/9=58/9`

[9]`color(white)("XXX")x=29/36`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Verifying by substituting `(-31/18)` for `y` and `29/36` for `x` in [2]
[10]`color(white)("XXX")2 * (29/36)+7 color(white)("xxx")?=?color(white)("xxx") -5 * (-31/18)`

[11]`color(white)("XXX")29/18+126/18color(white)("xxxxx")?=?color(white)("xx")155/18`

Yes: results verified!