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How to solve the system of equations by graphing




y= -3x -4

y= -1/2x + 1

How do you graph this equation? Please advise. Thanks

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2 Answers

Hi Tanesha,

Let's begin by putting both equations into slope-intercept form (y = mx + b):

3x + y = -4 becomes y = -3x - 4 (slope (m) = -3 and y-intercept (b) = -4)

x + 2y = 2 becomes 2y = -x + 2 and then y= -0.5x + 1 (slope (m) = -0.5 and y-intercept = 1)

We now have 1 point (the y-intercept, the y value when x = 0) for each line:

For line,  y = -3x - 4, point is (0,-4)

For line, y = -0.5x + 1, point is (0,1)

Now let's get another point for each line - the x-intercept (the x value when y = 0) by plugging in 0 for y and solving for x in each equation:

For the line, y = -3x - 4, 0 = -3x - 4 gives a value of -4/3 for x, so point is (-4/3,0)

For the line, y = -0.5x + 1, 0 = -0.5x + 1 gives a value of 2 for x, so point is (2,0)

Now we have 2 points for each line, so we can graph each line by connecting the 2 points (x and y intercepts) for each line.

Using the slope of each line, we can plot additional points for each line and then observe the point at which the lines intersect (the point which is common to both lines):

For line, y = -3x - 4, additional points can be plotted either by going down 3 units and over 1 unit to the right (slope is -3/1) from the previous point or by going up 3 units and over 1 unit to the left (slope is 3/-1) from the previous point.  In either, case we are using using the slope (m) of -3 to plot additional points.

For line, y = -0.5x + 1, additional points can be plotted either by going up 1 unit and over 1 unit to the right (slope is 1/1) from the previous point or by going down 1 unit and over 1 unit to the left (slope is -1/-1) from the previous point.  In either case, we are using a slope (m) of 1 to plot additional points.

The point of intersection of these 2 lines, the common point on both lines, gives us our solution.  The common point (our solution) is the point (-2,2).

We can verify that this is our solution by plugging in -2 for x and seeing if we get a y value of 2 in each line equation:

y = -3x - 4, y = -3(-2) - 4, y = 6 - 4, y = 2

y = -0.5x + 1, y = -0.5(-2) + 1, y = 1 + 1, y = 2

Both equations yield a value of 2 for y when plugging in the value of -2 for x, so our intersection point (our olution) of (-2,2) is correct.  

To summarize our graphing procedure:

    1.  Put both line equations into slope intercept form (y = mx + b)

    2.  Plot the the y-intercept (b) for each line.

    3.  Find the x intercept for each line by setting y to zero and solving for x in the slop-intercept equation for each line.

    4.  Connect the 2 points (x and y intercept points) for each line.

    5.  Plot additional points for each line as needed (using slope (m) of each line) to determine at which point the lines intersect.  The intersection point is the solution.

    6.  Check the solution by plugging in the x value of the intersection point in each line equation (slope-intercept form) and confirming the y value of the intersection point in each case.

I trust the above procedure will clarify for you how to graphically solve any system of 2 line equations.

Thanks for submitting this question.

Regards, Jordan.   

y = -3x - 4

pick a number as close as possible to the origin. let's pick 0

plug in 0 to x:

y = -3(0) - 4 = -4

so, we have our first point (0, -4)

we can find another point by picking another number to be plugged in to x or we can use the given slope of the equation.

Since the slope is -3, we can represent this as -3/1. We know the slope is the change in y over the change in x. So, from the point (0, -4) move your pencil to the right one unit and down three units. This location will be another point. Now, just use a ruler to connect these points.

Another way to get another point by using the slope is adding the change in y to the y coordinate of the first point and adding the change in x to the x coordinate of the first point. So, we have (0 + 1, -4 + (-3)) = (1, -7), this is the second point.

Do the same for the other equation.