Hints for solving the problems(8.1)
Multiplying and Dividing
Any rational expression will be undefined if the denominator becomes zero (0). To solve the first four problems take the denominator and set it equal to zero (0) and solve for the variable. The answer you get will be the exclusions, the numbers that you can’t substitute into the denominator for the variable. If you do, you will get zero (0)
Problem No. 2 - For what value (s) of x is the rational expression below undefined?
Set the denominator equal to 0 (x - 3)(x + 5) = 0
Solve x - 3 = 0 & x + 5 = 0 x = 3 & x = -5
The answer to the problem is: {3, -5}. This means that if you substitute 3 or -5 for x in the denominator it will make one of the binomials zero (0) and when you multiply by 0 you get 0 in the denominator. Remember you cannot divide by zero (0). When zero (0) is in the denominator, the fraction is undefined. Try it, take 2 and divide it by 0 on your calculator and see what you get.
Problem No. 6 - Reduce to lowest terms:
Remember when you reduce fractions, you reduce the numbers as you always did and the exponents of the like terms, that are variables, are subtracted. I will set up the above problems in steps.
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the ans.
Remember reduce the numbers by dividing by the GCF, and the letters will have their exponents subtracted, with the difference in the term (numerator or denominator) that has the highest exponent in the original problem.
Problem No. 8 - Reduce to lowest terms:
This problem, you must factor the numerator and denominator completely. Why? Because you cannot reduce any number, variable, or binomial unless the problem is in factored form.
Notice that the numerator and
denominator have a common factor of (x + 3). Reduce it out and the answer is:
Problem No. 13 - Find: 
Here
you multiply the numerator by numerator and denominator by denominator. Of
course if you can reduce by division (cancel) first the problem may be easier.
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the ans.
Notice the 10 and 15 were divided by 5, and the 9 and 21 were divided by 3. The variables were multiplied in each term by adding the exponents of like factors. Then division was done by subtracting the exponents.
Problem No. 18 - Find: 
All division problems must be
changed to multiplication by inverting the term on the right of the division
sign and multiplying.
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the
ans.
Problem No. 25 - Simplify the complex fraction below.
A complex fraction is a division problem. Take the denominator and invert it and multiply it by the numerator.
reducing you get
the ans.
Here is another way. Multiply the numerator by numerator and denominator by denominator and reduce after that.
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the ans.
Adding and Subtracting
Problem No. 30 - Find: 
When
adding or subtracting fractions, the denominators must be the same. When they
are already the same, (as in this problem) just add the numerators, keep the
same denominator and reduce the answer if you can.
The ans. Problem cannot be
reduced.
Problem No. 35 Find: 
Watch
out for this problem, because as you combine the numerators, the minus (-) sign
between the two fraction changes the sign of both terms in the
fraction on the right.
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the ans.
A 2 can be factored out in the numerator, but it would not produce a binomial like the denominator. So there is no need to factor it out. But you must be aware that if you can factor the numerator, (after combining) and get a binomial like one in the denominator, then you must do it in order to reduce further.
Problem No. 42 - Add and reduce your answer to lowest terms.
Same denominators, combine the numerators.
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Factor the numerator and the
denominator 
Oh!, wow, the
numerator and denominator have a common factor in the form of a binomial. Reduce
then out and the answer is: 
.
Problem No. 52 - Subtract and reduce your answer to lowest terms.
Same procedure as last problem, but watch out for
the minus (-) between the two fractions.
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the
ans.