Question

How do you solve `x-3y=-9` and `x+3y=3` by graphing?

Answer 1

Solution is `(-3,2)`

Draw the two graphs `x-3y=-9` and `x+3y=3` and find the coordinates `(x, y)` of where they intersect.

Explanation:

You have two equations, `x-3y=-9` and `x+3y=3`

To plot this on a graph, do some transformation so that you have just `y` on one side (`y=...`)

Part 1

You need to get those equations into the form `y=...`

For the first equation,

`x-3y=-9`

Add 9 to both sides.

`x-3y+9=0`

Add 3y to both sides.

`x+9=3y`

Dividing both sides by 3, you get

`y=(x+9)/3`

For the second one, you do the same thing

`x+3y=3`

Subtracting x from both sides,

`3y=3-x`

Then divide both sides by 3

`y=(3-x)/2`

Part 2

Plot these two equations on the same graph

`y=(x+9)/3` and `y=(3-x)/2`

graph{(x-3y+9)(x+3y-3)=0 [-10, 10, -5, 5]}

The coordinates of where those two lines cross will be your answer. Which will be `(-3,2)`

Here's why the coordinates of where the two lines cross is the answer:

What the line for `x-3y=-9` means is that every point on that line will have coordinates `(x,y)` that satisfy `x-3y=-9`

So for instance, if you find the point `(-9,0)` is on that line, that means that `x=-9, y=0` is a pair of numbers that obeys `x-3y=-9`:

`(-9)-3(0)=-9`

So if you want to find a pair of numbers `(x,y)` which fit `x-3y=-9`, then just take the coordinates of any point on that line.

The same thing applies for more than one line. Since a point that lies on some line `y=(...)x+(...)` will satisfy that equation, then a point that lies on two different lines at the same time will satisfy both equations represented by those lines. So any point that lies on two lines (is the intersection) will have coordinates that fit the equations of both lines.