The graph of an inequality that includes either the  or  symbol is the graph of the corresponding equation along with all of the points on either side of it. We are given both the line and the circle, so for each inequality, it remains to determine which side of each figure is included. In each case, this can be done by choosing any test point on either side of the figure, substituting its coordinates in the inequality, and determining whether the inequality is true or not. The easiest test point is .

This is true; select the side of this line that includes the origin.

This is true; select the side of this circle that includes the origin - the inside.

The solution sets of the individual inequalities are below:

The graph of the system is the intersection of the two sets, shown below:

The slope of a line determines its steepness.

Since slope is rise over run or 

,

each of the slopes can be compared based on the ratio of rise to run.

A greater rise than run means a steeper line, or in our case, a hill. So for the "steepest" line  , one must rise 5 units and move horizontally 1 unit. 

Compare this with the line  where one must rise one unit and move 8 units horizontally.

As indicated by the solid line, the graph of the inequality at right includes the line of the equation, so the inequality graphed is either 

or

To determine which one, we can select a test point and substitute its coordinates in either inequality, testing whether it is true for those values. The easiest test point is ; it is part of the solution region, so we want the inequality that it makes true. Let us select the first inequality:

 makes this inequality true, so the graph of the inequality  is the one that includes the origin. This is the correct choice.

(Note that if you select the second inequality, substitution will yield a false statement; this will allow you to draw the same conclusion.)

The general form of the graph of a conic section is

where , , , , and  are real coefficients, and  and  are not both 0. 

Here, since the  term is missing, . This indicates that the graph of the equation is a parabola.

The general form of the graph of a conic section is

where , , , , and  are real coefficients, and  and  are not both 0. 

Here,  and .

 and  are of unlike sign. This indicates that the graph of the equation is a hyperbola.

The general form of the graph of a conic section is

where , , , , and  are real coefficients, and  and  are not both 0. 

Here,  and .

 and  are of unlike sign. This indicates that the graph of the equation is a hyperbola.

As indicated by the dashed line, the graph of the inequality at right does not include the line of the equation, so the inequality graphed is either 

or

To determine which one, we can select a test point and substitute its coordinates in either inequality, testing whether it is true for those values. The easiest test point is ; it is not part of the solution region, so we want the inequality that it makes false. Let us select the first inequality:

 makes this inequality false, so the graph of the inequality  is the one that does not include the origin. This is the correct choice.

(Note that if the second inequality had been selected,  would have made it true, so that would not have been the correct choice; we would have again selected the first.)

The general form of the graph of a conic section is

where , , , , and  are real coefficients, and  and  are not both 0. 

Here,  and .

 and  and  are of like sign. This indicates that the graph of the equation is an ellipse.

The circle with center and radius has as its equation, in standard form,

,

so rewrite the equation in this form.

First, rearrange and group the terms as follows:

Complete both perfect square trinomials as follows:

Rewrite the trinomials as the squares of binomials:

Thus, , and the center of the circle is at the point .