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Welcome to our Math lesson on Systems of Linear Inequalities, this is the first lesson of our suite of math lessons covering the topic of Systems of Inequalities, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.
Let's consider the two inequalities in the introduction section. Unfortunately, the first two (analytical) methods used for solving systems of linear equations (elimination and substitution method) cannot be used in systems of linear inequalities, as here we have to find a region of the coordinates plane XOY where both inequalities are true. This region can extend at different orientations, so it is impossible to identify all values that belong to the solution set without referring to a figure that shows the graphs of every single inequality.
Therefore, we can solve graphically the system of linear inequalities
Obviously, as in all inequalities solved in the previous tutorial, when solving such as system only graphically, we must isolate the variable y and solve each inequality in terms of the other variable x. Then, we plot the graphs of each inequality and see where the double shaded area that shows the region where both inequalities are true extends. Therefore, only the last part of the solution reminds us of the existence of a system of linear inequalities; the rest of the solution looks like solving two different exercises containing a single inequality.
However, we can still use the first two methods when checking the solution, especially when the corresponding equations form independent systems, i.e. when they have a single common point. We can check whether this point of convergence is the same as shown in the graph or not. Hence, we can write for the two inequalities in our example:
and
This means the solution set includes the double shaded area below the lines y = 3x/2 -1/2 (without including the line) and y = -x + 4 (including the line itself), as shown in the figure below.
Now, let's check whether points A and B in the "Introduction" section belong to the solution set or not. Let's insert points A(-2, 0) and B(2, 1) into the graph.
From the graph, it is evident that point A does not belong to the solution set of the given system of linear inequalities, as it belongs only to the solution set of y ≤ -x + 4 but not to that of y < 3x/2 -1/2. On the other hand, point B belongs to the solution set of the given system of linear inequalities, as it is inside the double shaded zone.
Let's check our graphical solution by making the numerical proof. We can take either the original inequality or the revised form used to plot the graph, as the result will not change. Thus, substituting the coordinates of point A (x = -2, y = 0) in the first inequality yields
while substituting the same point in the second inequality yields
Now, let's consider the coordinates of point B (x = 2, y = 1). Thus, when substituting them in the first inequality, yields
and when substituting the coordinates of the same point in the second inequality yields
Therefore, only point B belongs to the solution set of the given system of inequalities, as when substituting its coordinates in the system of inequalities, makes both inequalities true.
Remark! The graphing method not only makes us solve systems of linear inequalities, but it also helps us solve easier the corresponding systems of equations when the coefficients are such that they give rational numbers as solutions. Thus, instead of finding a pair of values obtained through complicated calculations, we simply plot the graphs and see what the coordinates of the intercept point are.
Solve the following systems of inequalities.
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