Hints for solving the problems (5.3)
Solving Linear Systems
Problems 1-28
In section 4.3 you graphed one inequality. First the equation was graphed and then one of the regions was shaded in, according to the point you substituted into the original inequality. In this exercise you will graph two inequalities and find out where the cross-shading is.
Problem No. 7
Graph the system of inequalities below.
x + y ≥ 4 x - y < -2
x + y = 4 x - y = -2 Graph the equality first.
y = -x + 4 -y = -x - 2 Use y = mx + b
y = x + 2
m = -1/1 m = 1/1
b = 4 (0, 4) b = (0, 2)
The first equation will start on (0,4). The slope says to move one unit to the right and one unit down to the point (1, 3). When you draw this line it will be a solid line. Substitute (0, 0) into the original inequality 0 + 0 ≥ 4, 0 ≥ 4 False. So the shading is up and to the right. The shading area does not include the test point (0, 0).
The second equation (on the same graph) will start at (0, 2). The slope says to move one unit to the right and one unit up to the point (1,3) (intersecting with the first equation at this point). When you draw this line it will be a broken line. Substitute (0, 0) into the original inequality 0 - 0 < -2, 0 < -2 False. So shading is up and to the left. The shading does not include the test point (0, 0).
The intersection of the two inequalities is where the two shadings cross each other, which is up and to the right of the first equation and up and to the left of the second equation. Put some extra lines or shading to show the Solution.
Problem No. 14 – Graph the system of inequalities below.
y ≥ (1/4)x + 4 y ≤ (1/4)x - 3
y = (1/4)x + 4 y = (1/4)x - 3 Graph the equality first.
m = 1/4, b = 4 (0,4) m = 1/4, b = -3 (0, -3)
The two lines will be parallel when graphed. The slopes are the same. The first equation will be graphed with the y-intercept at (0, 4). The slope says to go four units to the right and one unit up and you land on the point (4, 5). When you draw this line it will be solid. Substitute (0, 0) into the original inequality. A false statement is produced, so you shade up and to the left (the region without the point (0, 0)).
The second equation will be graphed with the y-intercept at (0, -3). The slope says to go four units to the right and one unit up and you land on the point (4, -2). When you draw this line it will be solid. Substitute (0, 0) into the original inequality. A false statement is produced, so you shade down and to the right (the region without the point (0, 0)).
After you have shaded the correct regions, you will notice that there is no cross-shading. One region is shaded up and to the left and the other is shaded down and to the right. No points are common to both inequalities.
Problem No. 25 – Graph the system of inequalities below.
y > x y ≤ 3
y = x y = 3 Graph the equality first.
m = 1/1, b = 0 (0, 0) (0, 3)
The first equation will be graphed with the y-intercept at (0, 0). The slope says to one unit to the right and one unit up and you land on the point (1,1). You will notice when you connect the two points and extend the line it will run through the middle of quadrants 1 & 3. You can’t substitute (0, 0) into the original inequality, because the line runs through that point.
So substitute the point (0, 1) into the original inequality. A true statement is produced, so you shade up and to the right of the broken line you just drew.
The second equation will be graphed with a point at (0, 3). A solid horizontal line is drawn through this point. Substitute (0, 0) into the original inequality and true statement is produced, so you shade down from the horizontal line you just drew.
The intersection of the two inequalities is down from the horizontal line and to the left of the first inequality.