Learn basic math on line

 
Exponents

Single Variable Equations 

• Simple equations (1 variable, basic operations, whole numbers) 
• Complex equations (multiple variables, decimals, fractions, like terms) 
• Equations consisting of fractions
• Quadratic equations

Multi-Variable Equations


Word Problems


Factoring

• Greatest Common Factor (GCF) 
• Process of factoring 
• Factoring trinomials

Fractions

• Multiplication of fractions
• Lowest Common Multiple (LCM) 
• Addition of fractions
• Complex fractions

Ratios

• Proportions 
• Advanced operations involving ratios

Graphing on Number Lines

• Inequalities 
• Conjunctions 
• Domain

Graphing on a Coordinate Plane

• Graphing points 
• Graphing lines 
• Topics related to graphing on a coordinate plane

Square Roots

• Simplification of roots
• Addition and Subtraction of roots and complex expressions
• Multiplicationi of roots

If you know nothing about finding a greatest common factor, a good place to start would be at the Pre-Algebra page on fractions and finding a GCF.
In Algebra, the greatest common factor is found a little bit differently than it is in Pre-Algebra.  In Algebra, only prime factors of numbers are used, and in many cases, you will be asked to find the GCF of algebraic terms.  Following are two examples:

     2 * 3 * 5 * 7 = 210
     210xy2z3 = 2 * 3 * 5 * 7 * x * y * y * z * z * z

As mentioned above, only prime numbers and literal factors, the letters, are used in this factoring process.  Because only the prime and literal factors are used, the GCF is defined as follows:  The GCF of two or more terms is the product of all prime algebraic factors common to every term, each to the highest power that it occurs in all the terms.  Putting that in more reasonable terms tells us that the GCF has to be made of factors that are present in all the terms for which you are finding the GCF.  Examples:
The expression 6x²y²m² + 3xy³m² + 3x³y² can be rewritten as a product of prime and literal factors -

2 * 3 * x * x * y * y * m * m + 3 * x * y * y * y * m * m + 3 * x * x * x * y * y.

Since the first term is the only term with 2 as a factor, 2 is not a factor of the GCF.  Each term has 3 as a factor at least once, so 3 is a factor of the GCF.

3

Each term also has x as a factor at least once, so x is a factor of the GCF.

3x

y is a factor of each term twice, and m is not a factor of all the terms, so it is not a part of the GCF.

3xy² is the GCF.
 

---

Using the distributive property lets us change an expression from a product to a sum.  For example, an expression such as 3a(x-c) tells you to multiply 3a by x-c.  When you do that, you get the sum 3ax - 3ac.  When you do that in reverse, by writing 3ax - 3ac as the product of the two factors 3a and x-c, you are factoring.

 1. Factor:   4a3b4z3 + 2a2bz4

   
  Solution:   Write out the terms as products of their prime 
              and literal factors.
               
              2*2*a*a*a*b*b*b*b*z*z*z + 2*a*a*b*z*z*z*z
       
              Each term has at least one 2, two a's, one 
              b, and three z's as factors.  Therefore, the GCF
              is 2a2bz3.  
  
              (2a2bz3)(     )
  
              Now that you've got the GCF factored out, 
              you can rewrite the two terms without the 
              factors in the GCF.
  
              2 * a * b * b * b + z
  
              The second pair of parentheses can now be
              filled in with the rewritten terms.
      
              (2a2bz3)(2ab3 + z) is the answer.  

 

---

In this section, we will only help you better understand how to factor quadratic trinomials, or trinomials whose highest power is two.  Also, we assume you know how to multiply binomials (we use the "FOIL" method).
Using a multiplication problem consisting of two binomials, we will show some important things to remember when factoring trinomials, which is the reverse of multiplying two binomials.  Example:
(x - 6)(x + 3) = x² - 6x + 3x - 18 = x² - 3x - 18

1. The first term of the trinomial is the product of the first terms of the binomials.

2. The last term of the trinomial is the product of the last terms of the binomials.

3. The coefficient of the middle term of the trinomial is the sum of the last terms of the binomials.

4. If all the signs in the trinomial are positive, all signs in both binomials are positive.
Keeping these important things in mind, you can factor trinomials.

1. Factor:   x2 - 14x - 15
        
   Solution: First, write down two sets of parentheses to 
             indicate the product.
  
             (     )(     )
   
             Since the first term in the trinomial is 
             the product of the first terms of the binomials, 
             you enter x as the first term of each binomial.  
      
             (x    )(x    )
  
             The product of the last terms of the binomials must 
             equal -15, and their sum must equal -14, 
             and one of the binomials' terms has to be negative.  
             Four different pairs of factors have a product 
             that equals -15.
  
             (3)(-5) = -15     (-15)(1) = -15
             (-3)(5) = -15     (15)(-1) = -15
  
             However, only one of those pairs has a sum of -14.
  
             (-15) + (1) = -14
  
             Therefore, the second terms in the binomial are -15 
             and 1 because these are the only two factors whose 
             product is -15 (the last term of the trinomial) and
             whose sum is -14 (the coefficient of the 
             middle term in the trinomial).
  
             (x - 15)(x + 1) is the answer.
  

In this section we'll help you understand how to deal with ratios.
A fraction is also known as a ratio.  For example, 3/4 is also the ratio of 3 to 4.  Any statement (or equation) that says two ratios are equal is called a proportion.  An important thing to remember when dealing with equal ratios is illustrated below.

3   15                   4 * 15
- = --  Cross multiply.  
4   20                   3 * 20


4 * 15 = 3 * 20
    60 = 60

You can solve for unknowns in proportions by using that process.

             4   21
1.   Solve:  - = --
             m   5
         
  Solution: Set the cross products equal.


            4 * 5 = 21m


            Finish the problem by dividing both sides
            by 21.

 
            20
            -- = m
            21

 

---

Many times, you use ratios to make solving word problems easier.  Many times, you are expected to gather information from the problem that is not directly stated in words.  Gathering this information is called inferring.  This section will help you better understand how to infer information from a word problem and solve using ratios.

If you are told there is a ratio of red marbles to blue marbles that equals 5:7 you are also told about two other ratios!  You know there are 12 total marbles (5 + 7 = 12), so you know the ratio of red marbles to total marbles must be 5:12 and the ratio of blue marbles to total marbles must be 7:12.  Also, it is usually helpful to write all these statements and then reread the problem to find any more information and to better understand the problem.

1. Problem:   The ratio of red marbles to blue marbles is 5 to 7.
              If there are 156 marbles total, how many red marbles
              are there?
      

  Solution:   Write down the information given and inferred.


              R = 5
              B = 7
              T = 12


              With this information, there are three different
              proportions you can write.


              R   5    R   5     B   7
              - = -    - = --    - = --
              B   7    T   12    T   12


              The problem asks about red marbles and total marbles,
              so we will use the second proportion and replace the
              variable T with 156.


                R    5
               --- = -- -> 12 * R = 5 * 156 -> R = 65
               156   12