a The third quadrant is the section of the coordinate plane where x and y take only negative values.
B
b No solutions require no intersection for the individual solution sets, so you can start by looking for non-intersecting boundaries.
C
c Start by thinking about boundaries that would overlap along a line.
D
d Start by thinking about boundaries that would intersect at just one point.
A
aExample System:
x<0 & (I) y<0 & (II)
B
bExample System:
y ≤ 2x +1 & (I) y ≥ 2x +5 & (II)
C
cExample System:
y ≤ 2x +4 & (I) y ≥ 2x +1 & (II)
D
dExample System:
y ≥ |x| & (I) y ≤ 0 & (II)
Practice makes perfect
a We need our system of inequalities to lie on the third quadrant. Let's start by reviewing the quadrants. The coordinate plane is divided into four equal quarters called quadrants. The first quadrant is the one where x and y are positive and the numbers increase counterclockwise.
Notice that the quadrant we are interested in is that where x and y take negative values. We can use the lines x=0 and y=0 which represent the y- and x-axis respectively and restrict them to the third quadrant using < signs.
x<0 & (I) y<0 & (II)
Notice that this is not the only solution. In general, we could use any pair of lines intersecting in the third quadrant with an angle not greater than 90° so that the solution set is always confined to the third quadrant.
b We can set a up a system of inequalities without a solution by using two parallel lines. Since parallel lines do not intersect, by choosing the appropriate inequality symbols for them their solution sets will not intersect either. Let's start by reviewing the slope-intercept form.
y = mx + bIn this equation m is the slope of the line and b is the y-intercept. Recall that parallel lines have the same slope, so we can choose a value for m and use it in both lines while using a different b value. Let's try an example.
y = 2x +1 & (I) y = 2x +5 & (II)
Now we can use the symbol ≤ for Equation (I). This will require us to shade the region below the boundary line. By using the symbol ≥ for Equation (II), it will require us to shade the region above the boundary line.
y ≤ 2x +1 & (I) y ≥ 2x +5 & (II)
As the solutions sets do not intersect, the inequality system has no solution. Notice, however, that this is not the only possibility. We could have used other values for the slopes or y-intercepts. Therefore, there are infinitely many solutions and this is just an example.
c We can use the system from Part B and modify it to meet these requirements. In particular, notice that if we choose the boundaries for our system to be the same line, the solution sets will only intercept at the line itself.
y ≤ 2x +1 & (I) y ≥ 2x +1 & (II)
Again, this is only an example solution, as we could do the same using other lines.
d Notice that, while there is a way to have a system of linear equations intersecting at just one point, the inequalities relating to them will necessarily overlap at a certain region.
However, we can also form a system with a absolute value inequality and a linear inequality. We know that the parent function of an absolute value function is y = |x|, which is V-shaped and starts at the origin. Then we can use a horizontal line passing through this same point. This is y = 0.
We need the absolute value inequality to be shaded upwards and the linear inequality to be shaded downwards. This way the intersection of the solution sets will only be the origin. For this, we can use the inequality symbol ≥ for the absolute value and ≤ for the horizontal line.
Notice that it's important that the inequalities include the boundaries, since the intersection happens at them. This solution is just an example. If we apply the same translation to both functions, the solution set would still be one point.
