All Algebra 1 Resources
Example Questions
Example Question #1 : Graphing Inequalities
Graph the compound inequality:
and
The compund inequality requires a graph in which the values of are greater than 5 AND less than or equal to 9. That is, is all the real numbers between 5, not included, and 9, included. Since 5 is not included, there is no "or equal to" sign on the greater than inequality, we place and open circle above 5. Since 9 is included, we place a closed circle above 9.
Example Question #1 : Graphing Inequalities
Which of the following graphs depicts the inequality:
First, graph the line of the equation. It is written in slope-intercept from; therefore the slope is and the y-intercept is .
When an inequality is written as less than or greater than, a dashed line is used. When an inequality is written as less than or equal to, or greater than or equal to, a solid line is used. Therefore, a dashed line should be used, eliminating two of the answer choices.
Next, use a test point to determine which regions should be shaded. A test point can be any point not on the line; the origin is generally a good choice. If the origin, , is subsituted into the question and the statement is TRUE, the graph should be shaded on the side of the line that contains the origin. If the statement is false, the other side should be shaded.
This statement is TRUE; the section containing the origin should be shaded.
Example Question #1 : Solving Inequalities
Solve the compound inequality and express answer in interval notation:
or
(no solution)
For a compound inequality, we solve each inequality individually. Thus, for the first inequality, , we obtain the solution and for the second inequality, , we obtain the solution . In interval notation, the solutions are and , respectively. Because our compound inequality has the word "or", this means we union the two solutions to obatin .
Example Question #1 : Graphing Inequalities
Which inequality describes the graph?
First, we find the equation of the boundary line using the two intercepts. The slope is
.
The -intercept is .
The slope-intercept form of the equation is therefore
.
Put this in standard form:
The inequality is therefore either or . To determine which, test a point that falls in the shaded region. The easiest is :
This inequality holds, so the answer is .
Example Question #2 : Graphing Inequalities
Refer to the above diagram. which of the following compound inequality statements has this set of points as its graph?
A horizontal line has equation for some value of ; since the line goes through a point with -coordinate 3, the line is . Also, since the line is solid and the region above this line is shaded in, the corresponding inequality is .
A vertical line has equation for some value of ; since the line goes through a point with -coordinate 4, the line is . Also, since the line is solid and the region right of this line is shaded in, the corresponding inequality is .
Since only the region belonging to both sets is shaded - that is, their intersection is shaded - the statements are connected with "and". The correct choice is .
Example Question #1 : Graphing Inequalities
Which of the following inequalities is graphed above?
First, we determine the equation of the boundary line. This line includes points and , so the slope can be calculated as follows:
Since we also know the -intercept is , we can substitute in the slope-intercept form to obtain the equation of the boundary line:
The boundary is included, as is indicated by the line being solid, so the equality symbol is replaced by either or . To find out which one, we can test a point in the solution set - for ease, we will choose :
_____
_____
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0 is less than 3 so the correct symbol is .
The inequality is .
Example Question #3 : Graphing Inequalities
Which of the following inequalities is graphed above?
First, we determine the equation of the boundary line. This line includes points and , so the slope can be calculated as follows:
Since we also know the -intercept is , we can substitute in the slope-intercept form to obtain equation of the boundary:
The boundary is included, as is indicated by the line being solid, so the equality symbol is replaced by either or . To find out which one, we can test a point in the solution set - for ease, we will choose :
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_____
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0 is less than 7 so the correct symbol is .
The correct choice is .
Example Question #2 : Graphing Inequalities
Which of the following inequalities is graphed above?
First, we determine the equation of the boundary line. This line includes points and , so the slope can be calculated as follows:
Since we also know the -intercept is , we can substitute in the slope-intercept form to obtain equation of the boundary:
The boundary is excluded, as is indicated by the line being dashed, so the equality symbol is replaced by either or . To find out which one, we can test a point in the solution set - we will choose :
_____
_____
_____
_____
1 is greater than 0 so the correct symbol is
The inequality is
Example Question #6 : Graphing Inequalities
The above graph depicts which of the following equations or inequalities?
Given the above graph, we can initially deduce that , , and are not the correct answer; the dashed line in the graph indicates that no point on the line is a solution to the inequality. Thus, we're left with and .
We can use a test point to determine which of the remaining inequalities is the correct answer. The test point can be any point that is not on the line, so let's select in this case. Plugging into yields . Since this is true, we know that every point on the same side of the line as will yield a true result, and that our graph represents .
Example Question #7 : Graphing Inequalities
The above graph depicts which of the following equations or inequalities?
Given the above graph, we can initially deduce that , , and are not the correct answer; the dashed line in the graph indicates that no point on the line is a solution to the inequality. Thus, we're left with and .
We can use a test point to determine which of the remaining inequalities is the correct answer. The test point can be any point that is not on the line, so let's select in this case. Plugging into yields . Since this is true, we know that every point on the same side of the line as will yield a true result, and that our graph represents .