Question 693930: Graph the system of linear inequality below:
y>x+1
y≥-1/2x-3
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! NOTE: It may help you not to shade when I first talk about shading. You want to keep what part we need to shade in mind until the last part. This may help you save time and it will make the graph look less sloppy.
When we start graphing inequalities, we'll immediately see a difference from regular linear equation graphing. Perhaps the biggest difference is that we have to shade rather than draw lines only. This is because the portions that are shaded represent all of the values that will work for the inequalities. Inequalities deal with ranges of values rather than a single value. That is why we shade. However, to know how to shade we need to start by graphing the line like normal. This is where the second difference comes into play. If the inequality we are dealing with has a sign of greater than or less than (> and < respectively), we draw a dashed line. This can be explained as so: If we have an inequality like y > 12, we know by experience that y cannot equal 12. Why? Well, 12 is not greater than 12; it is equal to it. Therefore, 12 would not be included in the range of values of this inequality. A dashed line in inequality graphing represents the values that are equaled to the number that is under the conditions of greater than or less than. If y > 12 was graphed, it would have a dashed line going through 12 on the y-axis. It would also have a shaded portion above the line representing all the values that are greater than 12. Even though the line goes through 12, it does not mean 12 is an answer because the line is dashed. This is why we use a dashed line when dealing with > or <. If we are dealing with ≥ or ≤ (greater than or equal to), we would draw a solid line because the values that lie on the line would be answers to the inequality. y ≥ 12 can have an answer or y = 12 because 12 = 12. So, let us start with the first problem.
1. y>x+1
More complicated inequalities that need to be graphed would look a little different. The difference would most likely lie in the fact that we would have to find the equation of the line in slope-intercept form. However, this problem already gives us the line in slope-intercept form. The equation of the line in this form would be y=x+1, where the slope is 1 (implied) and b (y-intercept) is 1. We need to be mindful of the sign we are dealing with. The first step in this case is to graph the line with slope 1 and y-intercept at (0,1). Draw the dashed line as far as you can. Now is the time to shade.
Shading isn't a difficult process. Looking at our dashed line, we see that there are two side to it (if we've drawn it as far as we can on the graph). There is the portion to the left of the line and the portion to the right of the line. The portion we shade is based on what points work for the inequality. If the points to the left of the line work, we shade that side. If the points to the right of the inequality work, we shade that portion. So, let's begin with a simple point. I always like to use the origin (0, 0) right off the bat. Unless the line goes through the origin, this point is always handy to use. So, plugging 0 in for x and 0 in for y, does the inequality y>x+1 remain true?
y>x+1
0>0+1
0>1.
No it doesn't, so the point (0, 0) does not work for this graph. Therefore, we do not shade the portion where (0, 0) lies. We would shade the left side of the line. Just to make sure let us use a point on that side. Let us use a point in the second quadrant. Will the point (-3, 6) work?
y>x+1
6>(-3)+1
6>-2.
Yes it does, so we shade the part of the graph to the left of the line. (-3, 6) will be represented in this shading as well as all the points that lie on this graph and that make the inequality y>x+1 true. Note that all points on the dashed line will not work. If we picked (0, 1), or the y-intercept, we would not get a true inequality:
y>x+1
1>0+1
1>1.
1 is equal to 1, not greater than 1.
2. y≥(-1/2)x-3
Now we move on to a greater than or equal to sign. Again, we need to graph the line first. The equation of the line can easily be identified in slope-intercept. The slope should be -1/2 and the y-intercept should be (0, -3). Draw this line with the inequality sign in mind. Since it has "equal to" to the sign's meaning, we would draw a solid line (the same type of line we would use to graph a linear equation). Now we need to shade.
The same rules apply. We have the portion of the graph that lies above the line and below the line. Let us pick the origin point again. Will (0, 0) work in this inequality?
y≥(-1/2)x-3
0≥-1/2(0)-3
0≥-3.
0 is greater than -3, so the point (0, 0) works. The portion above the line that contains the point (0, 0) should be shaded. We could pick a point on the line and the inequality would be true, but it wouldn't necessary tell us which portion we should shade. Now we move on the to ambiguous part.
3. Finding the solution of the system of inequalities
We have graphed both inequalities as though they were by themselves. However, since they are in a system, the shading gets more complicated. If you haven't shaded anything to this part, it may help you save erasing time. The technical way of finding the solution to a system of inequalities is shading the portion of the graph that works for BOTH inequalities. This can be found easily if you have started shading both areas that need to be shaded for both inequalities. The answer is the part of the graph where the shading of the first graph and the shading of the second graph has met. It is dependent on your teacher which method of shading you take. You can shade both inequalities and then shade the portion that is the answer darker than the other parts. You could also hold off the shading entirely and shade only the part that represent the answer for both inequalities. The shading that represents the answer to both inequalities should be inside quadrants 1 and 2. Looking at both lines that we have graphed on the same graph paper, the portion that needs to be shaded darkest (or only if you chose to not shade until this part) is the "V-shaped" portion at the top of the graph. If you plug any values in this area into both equations, both inequalities would be true (any point would work in this region). Again, the solutions to a system of inequalities lie in the portion where the shading of both graphs meet.
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