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Lesson 2A : Algebra

Lesson Outline:

1. Order of Operations

2. Equations

   a. Variables

   b. Factoring

   c. Solving 

5. Linear equations

   a. Slope-intercept form

   b. Standard form


In this lesson we will look at the basic algebra you will need to know for the GED® test. The algebra on the GED® math test covers variables and equations. The equations may contain combinations of arithmetic expressions. You will need to do arithmetic operations on some equations to find solutions

ORDER OF OPERATIONS

There is a specific order in which you carry out your arithmetic operations on any equation or expression:

  1. First carry out the operations contained in Parentheses
  2. Then any Exponents (powers)
  3. Then Multiplication/Division
  4. Then Addition/Subtraction

Problem:

          Evaluate (2x + 7) + 5(x 2 - 2) when x = 2

  • First we look at what is contained in the parentheses. In this case, there are two sets.

         In the first set we have ( 2x + 7 ) + 5 (x 2 - 2)

         When we plug in 2 for x, we get. = ( 2 * 2 + 7 ) + 5(x 2 - 2)

         We do the multiplication first = ( 4 + 7 ) + 5(x 2 - 2)

         Then the addition = ( 11 ) + 5(x 2 - 2)

  • Now we look at the second set of parentheses. = (11) + 5( x 2 - 3 )

         There is an exponent and a minus sign. We work out the exponent first = (11) + 5( 22 - 2 )

         then we subtract = (11) + 5( 4 - 2 )

  • Now our problem is simply = (11) + 5( 2 )

         Once again, following the order of operations, we multiply 5 and 2 first = 11 + 5 * 2

         And then add to get our answer = 11 + 10 = 21

EQUATIONS

a. Variables

The ‘x’ we used in the previous problem is known as a variable.

Variables are the letters in equations that represent numbers of an unknown value

Such as in x - 4 = 6 or 3y = 21

b. Factoring

Factoring involves breaking a number or expression down into its factors. For example, 3 and 5 are factors of 15, since 3 x 5= 15.

  • We apply factoring to an expression by noticing terms that have a common factor that can be pulled out to help simplify the equation

            2x - 4

  • Here we notice that 2x and 4 have 2 as a common factor

            2x – 4

            = ( 2 * x) – ( 2 * 2)

            = 2 * x – 2 * 2

            = 2 (x - 2)

  • Variables can also be factored out of an expression.

            x 2 + x

           = (x * x) + (x * 1)

           = x * x + x * 1

           = x (x + 1)

Warning:

Be careful how you apply factoring.

  • You could, for example, not factor an x out of the entire expression below because the last term does not have x as a factor:

            (x 2 + x + 2)

  • What you can do is factor x out of the first two terms

            (x 2 + x) + 2

  • putting the parenthesis around them means you are considering them together. And then

            x(x +1) + 2

Be familiar with the factors of individual numbers, the factors of 1-200, for example. This will help you pick out common factors when solving equations.

c. Solving

To solve for a variable, you need to use arithmetic operations to isolate it. As you do that, whatever change you make to one side of the equation, you must also make on the other side.

  • For example, if you add 4 to one side of an equation, you must also add 4 to the other side.

           2x – 4 = 6

           4 +  4

          2x + 0   = 10

          2x = 10

  • Similarly, if you multiply one side by ½ , you must multiply the other side by ½ also.

          2x – 4 = 6

          ½ (2x – 4) = 6 * ½

           x – 2 = 3

And that goes for all the arithmetic operations:

  1. Addition
  2. Subtraction (which is really the adding of a negative)
  3. Multiplication
  4. Division (which is really the multiplying of a fraction)
  5. Taking to a power
  6. Taking a root

This way we still have the same values in the equation even after we have manipulated it to work towards a solution. We will apply this more in discussions about solving different types of equations.

  • Now let us isolate a variable completely and solve for its value

           2x – 4 = 6

  • First move the 4 to the other side.

           2x – 4 = 6

           + 4 + 4

            2x = 10

  • Then get rid of the 2 in front of the x by dividing each side by 2.

           2x = 10

            2      2

            x = 5

There will be problems on the test that ask you to come up with your own equations based on the information in a short reading passage like the following:  

Questions 1-3 are based on the following information

Tom and Katie have just picked up their prescription medications. Tom takes blue pills and Katie takes red pills. Tom takes 3 a day and Katie takes 4 a day. Tom’s bottle of pills will last him twice as long as Katie’s will.

1) In which equation does x represent the number of red pills if Tom was prescribed 30 days worth of pills?

   A) 3 = 30 * x

   B) 4 * x = 30

   C) 4 + x = 30

   D) x / 3 = 30

   E) 3 + x = 30 

# of pills = # of days times # of pills per day

Which would be: x = 30 * 3

  • That is not one of the answer choices. However, if you move the 3 to the other side

           x / 3 = 30 * 3/3

           you get: x / 3 = 30 Which is answer D.

2) In which equation does x represent the total number of Tom’s blue pills if Katie’s bottle contains 40 red pills?

   A) 40 / 4 =-x

   B) 4 – 3 = x /40

   C) x = 2 (40 /4)

   D) x / 3 = 2 (40/ 4)

   E) 2x * 40 = 4 

We are given the new information that Katie has 40 pills.

  • Dividing it by her daily dosage of 4 pills per day gives the # of days for Katie’s medication: 40 / 4

          Remember that Tom’s days are twice as many as Katie’s.

  • Which means that the # of days for Tom’s medication can be expressed as: 2 (40/ 4 )

          Which is equivalent to Tom's days expressed as the # of pills divided by his daily dosage so we can

           set it equal to that : x / 3 = 2(40 / 4)

          Which is answer C).

3}If Tom counts 120 pills in his bottle, how many does Katie have?

   A) 40

   B) 80

   C) 100

   D) 120

   E) Not enough information to answer this problem

  • If Tom has 120 blue pills and he is supposed to take 3 a day, we divide 120 pills by 3 (per/day)

            120 / 3 = the # of days for Tom

            = 40 pills

  • If Tom’s medication is supposed to last twice as many days as Katie’s, then we take Tom’s number of days and divide it by 2 to get

           40 / 2 = # of days for Katie

           = 20 pills

  • Finally, we multiply this by 4 (pills per day for Katie) to get

           20 * 4 = the # of red pills for Katie

           = 80 red pills

        Which is answer B.

LINEAR EQUATIONS

Linear equations are equations with two variables, usually ‘x’ and ‘y’

a. Slope-Intercept form

The most common and useful form of a linear equation is

y = m*x + b

a. This form of the linear equation is known as slope-intercept form.. It expresses the graph of a line. (which is why it is called a linear equation.) The ‘m’ and the ‘b’ describe certain features the line will have on a graph.

  • This is a function and can be expressed as

            f(x) = mx + b

         “Function ‘f’ of ‘x’ equals ‘m’ times ‘x’ plus ‘b’”

b. Standard form is another way of expressing a linear equation:

           Ax + BY = C

Standard form and slope-intercept form are really saying the same thing but in different ways. Here is an example of how you might convert from one form to another

  • Starting with standard form

           3x + 2y = 6

  • move 3x to the right side

           2y = 6 - 3x

  • divide both sides by 2

           y = 3 - 1.5x

  • make the x term the first term on the right side of the equals sign

           y = -1.5x + 3

  • and you have your linear equation in slope-intercept form.

Why do we have slope-intercept form? We will answer this question more thoroughly in our discussion of graphing. For now, know that we can plug in many values for x and get a corresponding value for y. This is what a function is. You can plug in a set of inputs and get a set of outputs.

 

Click on the link below to move on to lesson 2B.

Back: Math Lesson 1B | Next: Math Lesson 2B


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