Multiplying and dividing fractions in basic arithmetic and in algebra involves a lot of factoring. Factoring will also be helpful as you combine fractions with the last two operations—addition and subtraction.

Let’s begin with an example from arithmetic. You will see that the process used to add simple fractions applies directly to more complex algebraic fractions as well.

Consider the following addition problem:

Because the denominators are equal, this addition problem could be written as a single fraction, with the numerators summed over the denominator:

Because the 7 and the 2 are like terms, they can be added:

This example from basic arithmetic can be extended slightly to become an algebra problem that includes variables:

Again, it can be written as a single fraction and the like terms combined, as shown:

A common denominator is needed when two or more fractions are added.

As long as the fractions being added have equal denominators, you can combine the numerators and create a single fraction, as in the following examples:

11x and 15 have no common factors, so is the simplified answer.

5x and 15 have a common factor of 5, so each term must be divided by 5 to give the simplified answer of

11a² and 10 have no common factors, so is the simplified answer.

In algebra, unlike basic arithmetic, it’s acceptable to leave a fraction with a numerator larger than the denominator. Common factors do need to be divided out, though, as you see in the next example:

5x³ and 4y have no common factors, so is the simplified answer.

Now try these addition problems, checking your answers on pages 143-144.

Recall from Hours 4 and 7 that subtracting is closely related to adding. As you learn how to subtract fractions, you’ll see that the process is almost identical to the addition problems in the earlier section. However, be prepared to watch for changing signs, just as you did when you subtracted polynomials.

Consider the following subtraction problem:

Because the denominators are equal, the two numerators can be combined to form a single fraction: -2 and 7a have no common factors, so is the simplified answer.

Watch the signs carefully if the first term is negative: -15 and 5 have a common factor of 5. Dividing both terms by 5 gives the simplified answer of -3x.

In the following example, the second numerator is a binomial, but the subtraction process is the same:

By changing the signs in the binomial 6b + 5, this subtraction problem becomes an addition problem, as shown here:

Adding the fractions gives the following solution: -3b, -5, and 10 have no common factors, so is the simplified answer.

When both numerators are binomials, changing the signs in the second one allows you to add the fractions: -x and 14x² have a common factor of x. Dividing both terms by x gives the simplified answer of

Now try the following problems, checking your answers on page 144.

In all the addition and subtraction problems so far, the fractions had common denominators. Combining the numerators allowed you to create a single fraction. But what if the denominators are not equal? Your factoring skills will help you get each problem into a new form, and then you can do the adding and subtracting.

When the denominators in fractions are not equal, those fractions cannot be added or subtracted. Consider the following addition problem:

These two fractions cannot be combined into one single fraction—at least, not the way they are now. If you can change one denominator to equal the other, then you can combine them into one fraction.

Think about the two denominators, 2 and 10. If you multiply 2 times 5, you get 10. However, multiplying only the denominator 2 by 5 would change the value of the first fraction. What if you multiply both the numerator and denominator by 5? It would look like this:

Because , you are really just multiplying the original fraction by 1, which doesn’t change its value. The addition problem would become this:

This new addition problem is easily solved:

You began with the denominators of 2 and 10. Knowing that 2 is a factor of 10 allowed you to see that multiplying by the missing factor of 5 would give you the common denominators you need when adding or subtracting. The factoring process makes it all possible.

The following addition problem also has different denominators:

Because 4 is not a factor of 6, you need to find a new number that both 4 and 6 are factors of. There are many ways to do this, but the easiest is to multiply the larger number, 6, by 2, which gives 12. Is 4 a factor of 12? Yes. 4 times 3 is 12.

Multiplying the numerator and denominator of by 2 and the numerator and denominator of by 3, you have found the least common denominator, which is 12. You also have two fractions that can easily be combined into one: 37 and 12 have no common factors, so is the simplified answer.

The factoring process from the previous examples works in all problems that need common denominators, including those with variables, as in the next example:

The denominators already have common factors of x² that are equal, but the coefficients, 5 and 20, are not. 5 times 4 is 20, so 4 is multiplied by the original numerator and denominator in the first fraction: 39 and 20x² have no common factors, so is the simplified answer.

The following problem shows different coefficients and variables in the denominators of a subtraction problem, but finding the common factors helps you find the LCD: 3 and 7 have no common factors. Multiplying the larger number, 7, by 2 doesn’t help, since 3 is not a factor of 14. Multiplying 7 by 3 works, so 21 is part of the LCD. For the variables, b is a factor of b³ because b times b² equals b³. The LCD for this problem is 21b³.

The first fraction’s numerator and denominator need to be multiplied by 7, and the second fraction’s numerator and denominator need to be multiplied by 3b². The work is shown here:

None of the terms in the fraction have common factors, so is the simplified answer.

The previous problem is more complex, but the process works the same as in the easier ones. Understanding and using the algebraic processes is still your goal, no matter how complex the problem is.

Now try the following problems, checking your answers on page 144.

Based on what you’ve learned in Hour 12, do the following problems. Check your answers with the solution key in Appendix B.

Simplify: