### All Algebra II Resources

## Example Questions

### Example Question #1 : Graphing Systems Of Equations And Inequalities

This graph shows a solution at its intersection, the point (1,-1). This solution could be the solution for which of the following systems of equations?

**Possible Answers:**

**Correct answer:**

Based on the graph, you can see that the purple line has the equation based on the fact that it goes through the points (0,0) and (1,-1). The red line goes through the points (0,-4) and (1,-1). We can see that its slope is and its y-intercept is at -4. Putting this into y=mx+b slope intercept form, we get . Therefore, this graph is the solution to the system of equations

### Example Question #2 : Graphing Systems Of Equations And Inequalities

**Possible Answers:**

2 workouts

5 workouts

10 workouts

8 workouts

**Correct answer:**

8 workouts

A system of equations that models this, where a is the number of workouts Alf has completed and t is the number of workouts Tanner has completed is:

Next, graph each of these equations:

Then, identify where the two lines meet each other. This is at (2,8). Therefore, after they each complete 8 workouts (in 2 weeks time) they will have completed the same number of workouts.

### Example Question #3 : Graphing Systems Of Equations And Inequalities

This graph shows which of the following inequalities?

**Possible Answers:**

**Correct answer:**

The function graphed is . We know that it is strictly less than because the line is dotted (not solid). We also know that y is less than because the portion of the graph below the function is shaded.

### Example Question #4 : Graphing Systems Of Equations And Inequalities

**Possible Answers:**

4 classes

16 classes

4 classes

16 classes

**Correct answer:**

16 classes

To start, we want to write two equations that model this situation. Those equations are:

(where c is the number of classes Miles teaches)

The graph of these two equations is below:

Finally, the intersection of these two lines is associated with the total number of classes Miles needs to teach to break even. We can see that the intersection is at (16, 400). Therefore, after teaching 16 classes, Miles will have made $400 from the gym and broke even on his investment.

### Example Question #5 : Graphing Systems Of Equations And Inequalities

**Possible Answers:**

20 classes

120 classes

500 classes

6 classes

**Correct answer:**

20 classes

To start, we want to write two equations that model this situation. Those equations are:

(where c is the number of classes Emeka teaches)

The graph of these two equations is below:

Finally, the intersection of these two lines is associated with the total number of classes Emeka needs to teach to break even. We can see that the intersection is at (20, 500). Therefore, after teaching 20 classes, Emeka will have made $500 from the gym and broke even on his investment.

### Example Question #6 : Graphing Systems Of Equations And Inequalities

When graphed, the lines represented in the system of equations below will intersect in which quadrant?

3x - 2y = 0

4x+3y = -17

**Possible Answers:**

Quadrant IV

Quadrant I

Quadrant III

The lines will intersect on one of the axes, not in a quadrant.

Quadrant II

**Correct answer:**

Quadrant III

Recognize that the intersection of two lines occurs when the two lines meet at the same point - that is, where their x- and y-coordinates are the same. So you can find the point of intersection by simply solving the system. Here that is likely best done through the Elimination Method. If you stack the equations:

3x - 2y = 0

4x+3y = -17

You should see that you have a negative y term in the first equation and a positive y term in the second. That means that if you can get the coefficients of y to be the same, you can sum the equations and eliminate the y-term altogether. To do that, multiply the top equation by 3 and the bottom equation by 2 so that your y-coefficients are each 6:

3(3x - 2y = 0) --> 9x - 6y = 0

2(4x + 3y = -17) --> 8x + 6y = -34

When you sum the equations, then, you arrive at:

17x = -34

Which means that x = -2.

Then plug x = -2 into one of the equations. If you use the first, you'll get:

3(-2) - 2y = 0

Meaning that:

-6 - 2y = 0

-6=2y

y = -3

Since the point of intersection is (-2, -3), with both coordinates negative, the point will lie in Quadrant III.