Module Five:

Miscellaneous Topics

 

 

In this module, we will review radical expressions, graphing linear equations, and solving systems of equations.

 

Radical Expressions

 

Example 1.  

                       

 

Method A.

 

Step 1.           

Step 2.           

Step 3.           

 

Method B.

 

Step 1.           

Step 2.           

Step 3.           

 

 

 

 

Example 2.  

                        Simplify:      

 

Step 1.           

Step 2.           

Step 3.           

Step 4.            25a2 · 3b

Step 5.           

 

 

Example 3.  

                        Perform the indicated operations:

                                               

 

Step 1.           

Step 2.           

Step 3.           

Step 4.           

 

 

Interpreting Graphs of Linear Equations

 

The next topic is interpreting the graph of a linear equation.  In these examples, a graph will be given.  We will ask questions whose answers require interpretation of the graph.

 

 

Example 4.  

                        The graph of y + 2x = 4 is shown below.


 


                        A. What are the coordinates of the x-intercept?

 

                        Step 1.            x = 2 and y = 0

                        Step 2.            (2, 0)

 

                        B. What are the coordinates of the y-intercept?

 

                        Step 1.            x = 0 and y = 4

                        Step 2.            (0, 4)

 

                        C. Does the line contain the origin?

 

Method A.

 

                        Step 1.            The origin is the point (0, 0).

                        Step 2.            Using substitution where x = 0 and y = 0 we obtain:

    y + 2x = 4

 0 + 2 · 0 = 4

            0 = 4

Step 3.            The line does not contain the origin. 

 

Method B.

 

Step 1.            Visual inspection of the graph.  The line does not pass through the point (0, 0).  Therefore, the line does not contain the origin.

 

D. Determine whether the slope of the line is positive, negative, or neither.

 

Method A.

 

Step 1.            In the formula, y = mx + b, the slope is represented by the letter m.

 

Step 2.            Solving y + 2x = 4 for y, we obtain y = – 2x + 4 which is in the form y = mx + b.

 

Step 3.            m = – 2 and the slope is negative.

 

Method B.

 

Step 1.            By visual inspection of the graph, as we view the line from left to right, the y-coordinates decrease in value.

 

Step 2.            The slope is negative.

                       

Example 5.  


                        Which of the following is the graph of 3x – y = 0?

 

 


Step 1.            Solving 3x – y = 0 for y, we obtain y = 3x which is in the form

 y = mx + b.

           

Step 2.            The slope is m = 3, and the slope is positive.

 

Step 3.            Since the slope is positive, graph A is the answer..

 

 

 

 

Systems of Equations

 

 

 

The third topic we will discuss is solving systems of equations

 

 

 

Example 6.  

Solve the following system of equations for x and y using the substitution method.

 

4x – y = 22

        x = 3y

 

Step 1.            Substitute 3y for x in 4x –  y = 22.

 

4(3y) – y = 22

 

Step 2.            Solve for y:

 

                                                12y – y = 22

                                                      11y = 22

                                                          y = 2

 

Step 3.            Determine the value of x by substituting y = 2 into x = 3y.

 

                                                x = 3y

                                                x = 3(2)

                                                x = 6

 

Step 4.            x = 6 , y = 2  written as (6, 2)

 

 

 

Example 7.  

Solve the following system of equations for x and y using the elimination method.

 

 3x + y = 6

2x – 4  = y

 

Step 1.            Rewrite 2x – 4 = y as 2x – y = 4

 

Step 2.            Add 2x – y = 4 and 3x + y = 6

 

                                    2x – y = 4

                               +  3x + y = 6

                                    5x       = 10

 

Step 3.            Solve for x.  

                                    5x = 10

                                      x = 2

 

Step 4.            Determine the value of y by substituting x = 2 into the equation 3x + y = 6.

 

                                                  3x + y  = 6

                                                3(2) + y = 6

                                                     6 + y = 6

                                                           y = 0

 

Step 5.            x = 2, y = 0 written as (2, 0)

 

Example 8.  

Solve the following system of equations for x and y using the elimination method.

 

   x + 11 = 7y

x – 7y  = 9

 

Step 1.            Rewrite x + 11 = 7y as x – 7y = – 11

 

Step 2.            Subtract x – 7y = 9 from x – 7y = – 11

 

                                      x – 7y   = – 11

                                – ( x – 7y) = – (9)

                                                0 = – 20

 

Step 3.            Since 0 = – 20 is a false statement, the system has

no solution and is called an inconsistent system.

Practice problems.

 

Solution to problem 1:

 

 

 

 

Solution to problem 2:

 

 

 

Solution to problem 3:

 

 

4. The graph of 5x – 4y = 20 is shown below.

a)   What are the coordinates of the x-intercept?

b)   What are the coordinates of the y-intercept?

c)   Does the line contain the origin?

d)   Determine whether the slope of the line is positive, negative or neither.

 

 

 

 

 

 

 

 

 

 

Solution to problem 4:

 

a)   (4, 0)

b)   (0, – 5)

c)   no

 

5.   Which of the following is the graph of x + 2y = 0?

 

 

 

 

 

 

Solution to problem 5:

 

A

6.   Solve the following system of equations for x and y using the substitution method.

 

     x – 3y = 1 and 2x + 8 =y

 

Solution to problem 6:

 

Solve x – 3y = 1 for x and obtain

x = 1 + 3y

 

Then substitute into 2x + 8 = y:

   2(1 + 3y) + 8 = y

 

Now solve for y.

 

   2 + 6y + 8 = y

       10 + 6y = y

              10 = – 5y

            – 2  = y

 

Substitute - 2 for y into x = 1 + 3y

and obtain

     x = 1 + 3( – 2)

    x = – 5

 

Hence the solution is the ordered pair ( – 5, – 2)

 

7.   Solve the following system of equations for x and y using the elimination method.

     

      x + 4y = 16 and 6x + y = 27

 

Solution to problem 7:

 

Multiply the first equation by  – 6

 

– 6(x + 4y) = –6 (16) resulting in –6x – 24y = – 96

 

– 6x – 24y = – 96

   6x  +  y  =     27

       – 23y = – 69

              y = 3

 

Substituting 3 for y into x +  4y = 16 we obtain

 

x + 4(3) = 16

  x + 12 = 16

          x = 4

 

the ordered pair is (4, 3)