Question

How do you solve `3x+y=2` and `2x+3y=13` using substitution?

Answer 1

`"Point of intersection "(x,y)" "->" "(-1,5)`

Explanation:

Given:
`3x+y=2`............................(1)
`2x+3y=13`.......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider equation (1)

Subtract `3x` form both sides

`3x-3x+y=2-3x`

`0+y=-3x+2" "..................(1_a)`

Using equation `1_a` substitute for `y` in equation (2)

`2x+3(-3x+2)=13................(2_a)`

`2x-9x+6=13`

`-7x+6=13`

Subtract 6 from both sides

`-7x+6-6=13-6`

`-7x+0=7`

Divide bot sides by -7

`(-7)/(-7)xx x=(+7)/(-7)`

`(+1)xx x=-1`

`color(red)(x=-1)`
'~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute `x=-1` into equation (1)

`3x+y=2" "->" "3(-1)+y=2`

`-3+y=2`

Add 3 to both sides

`-3+3+y=2+3`

`color(red)(y=5)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To prove this:
write equation (1) as: `y=-3x+2`
write equation (2) as: `y=-2/3 x+13/3`