Math CentralQuandaries & Queries


Question from Diem, a student:

Graph 2x-3y+15=0

Hi Diem,

The equation 2x-3y+15=0 is a linear function (because the degrees of both variables is one).

There are a few different techniques one could use to graph a linear function, but it is usually helpful to rearrange the equation to isolate the y variable and have the equation in y=mx+b form, where m tells us the slope of the line (rise/run) and b is the value of the y-intercept (where the line crosses the y-axis).

I will use the equation 3x - 4y + 8 = 0 to illustrate the technique of isolating y:
-4y = -3x - 8 after subtracting 3x + 8 from both sides
y = 3x/4 + 2 after dividing both sides by -4

The slope and y-intercept information is sufficient to graph the line:
- plot the y-intercept point in the coordinate plane
- use the rise/run values from the slope to move away from the y-intercept and plot another point where you end up

For example, given y=3x/4 +2 as the equation, you would plot a point at 2 on the y-axis (0, 2) and then move 3up and 4 to the right and plot the second point there (4, 5).

- finally, join the two points to draw the graph of the line

Alternatively one could create a table of values where you choose about 3 different x values (any x values) and work out their corresponding y values. These x,y pairs are points on the line. You plot these points and join them to draw the graph of the line.

Hope this helps.


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