Exercise 40 Page 262

The solution to a system of inequalities is the intersection of the solution sets of the individual inequalities. If a system has no solution, it must be true that there is no intersection and there are no common values. Consider a system in which the boundary lines are parallel. Depending on the inequality symbols in both equations, there are four different possibilities for the solution to the system. The combinations are shown in the table. Notice that we will only consider strict inequalities.

Possibility	System
$1$	$\{\begin{array}{ll}y>x+2& \text{(I)}\\ y<x-4& \text{(II)}\end{array}$
$2$	$\{\begin{array}{ll}y>x+2& \text{(I)}\\ y>x-4& \text{(II)}\end{array}$
$3$	$\{\begin{array}{ll}y<x+2& \text{(I)}\\ y>x-4& \text{(II)}\end{array}$
$4$	$\{\begin{array}{ll}y<x+2& \text{(I)}\\ y<x-4& \text{(II)}\end{array}$

Graphing the systems from the table will allow us to see their solutions.

Possibility $1$

Let's begin with the following system. If you would like a detailed explanation about how to graph a system of linear inequalities, please check out this page.

Notice that the shaded regions do not overlap. This system has no solution.

Possibility $2$

Now we will continue with the second system. Let's graph the system as we did with the first inequality.

The overlap shows the solution to the system.

Possibility $3$

Then, we will continue with the third system. Let's graph the system as we did with the above inequalities.

The overlap shows the solution to the system.

Possibility $4$

Finally, we will graph the fourth system. Let's do it one more time, as we did with the above inequalities.

The overlap shows the solution to the system.

Conclusion

From the graphs, we can see that only Possibility $1$ does not have a solution. Just because the boundary lines of a system are parallel it is not necessarily true that it will have no solution. Our friend is incorrect.