Solve the following system of equations graphically on the set of axes below.

Question

y=x-2

y=-1/3x+6

the -1/3 is a fraction

Answer 1

Hi Gerson.

In this problem you're given a system of 2 linear equations and asked to solve it graphically.

We need to graph both lines and find the point where they intersect.

The pair of coordinates of that point is the solution.

Remember, the solution includes a value for x AND a value for y.

The fact that the equations are in the slope-intercept form makes thing easier.

Just in case you need to review, the slope-intercept form is written as y = mx + b.

The m is the slope which describes how steep the line is, and b is y-intercept.

In the 1st equation; y = x - 2, the slope is 1 and the y-intercept is -2.

(When you multiply by 1, the one doesn't need to be written.)

Consider the y-intercept as your FREE point and plot it at (0, -2).

From that point go right one and up one and plot another point.

Do one or more points and then connect them with a straight line.

Extend the line in BOTH directions.

For the 2nd line, the slope is -1/3 and the y-intercept is 6.

Plot the FREE point at (0, 6)

When the slope is a fraction, we can put that fraction straight to work.

Remember that the slope (m) ca be written as rise/run.

Where the rise is how far up or down, and the run is how far over.

In this case the rise is -1 and the run is 3.

From the y-intercept go to the right 3 and down 1 and plot a point.

Go right 3 and down 1 and plot another point.

Repeat, right 3, down 1 and plot as many as you think your teacher would want.

Connect all the point with a straight line in both directions.

The coordinates, x AND y, make up the solution.

I hope this helps.

Answer 2

This is a great problem to help you understand the connection between algebra and coordinate geometry. The graphing of the system of equations will produce two straight lines that intersect at a point (x,y) that is the solution of the equations.

I cheat and use algebra to get the values of the intersection point, then do the graphing. I set the right sides of the two equations equal to each other and solve for x, then substitute and find y.

Since both equations are in the slope/intercept format (y = mx + b), I plot the y-intercept (0,b) of both and use the slope of each (change in vertical over change in horizontal) to plot the points needed to draw each line -- remembering that a positive slope slants to the right and a negative slope slants to the left. If the graphed lines intersect at a point that has the same values as I got algebraically, then I know I did it right.