Solve the following system of equations graphically.

Jasmine B.

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May 20, 2014
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6
please help me i dont understand how to do the formula and get the graph and how do you make a new thread on here
Solve the following system of equations graphically. Click on the graph until the correct solution of the system appears.

2x + y = 2
x - y = 1
systems5.gif
systems4.gif
systems6.gif
 
The graph of a "linear function" (x and y appear only to first power and not in any functions like "sin(x)" or "log(y)") is a straight line. A straight line is determined by two points. So determine two points by choosing a value of x or y and solving for the other. Mark those two points on the graph and draw the line thorough them.

Here, one function is 2x+ y= 2. If we choose x= 0 (just because it is easy) we have 2(0)+ y= y= 2. So one point is (0, 2). If we choose y= 0, we have 2x+ 0= 2x= 2 so x= 1. Another point is (1, 0). Mark the points (0, 2) and (1, 0) and draw the line through them. (One of the three graphs you show has a line that passes through both (0, 2) and (1, 0).)

The other is x- y= 1. If we choose x= 0 we have -y= 1 so y= -1. One point on the graph is (0, -1). If y= 0 we have x- 0= x= 1. A second point on the graph is (1, 0). Mark those points on the graph and draw the line between them. It should be obvious where the two lines cross and what x, y values satisfy both equations.
 
Solve the following system of equations graphically.

Click on the graph until the correct solution … appears.



systems5.gif
systems4.gif
systems6.gif

Hi Jasmine.

This endeavor seems weird, as an exercise. Did this come from some type of on-line lesson, instead?

That is, did these three graphs come with the system of equations, and you're simply supposed to click intersection points until the machine says, "Correct"?

If the graphs were in fact provided, and this is actually a homework assignment (versus a lesson presentation), then I don't understand the point, but we can answer the question by inspection.

One of the equations in the system is x - y = 1. It's easy to think of this in Slope-Intercept Form.

y = x - 1

This tells us (by inspection) that the slope of the line is 1 and the y-intercept is (0,-1).

There is only one graph above containing a line with slope 1 passing through the point (0,-1).

Cheers :)
 
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