   SEARCH HOME Math Central Quandaries & 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.

Leeanne     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.