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Beginning Algebra
Answer/Discussion to Practice Problems
Tutorial 7: Multiplying and Dividing Real Numbers


checkAnswer/Discussion to 1a


(-2)(-25) = 50.

The product of the absolute values 2 and 25 is 50 and they have the same sign, so that is how we get the answer 50.




checkAnswer/Discussion to 1b


(0)(-100) = 0.




checkAnswer/Discussion to 1c


Working it left to right we get:

(-2)(3)(5) = 
(-6)(5) =




checkAnswer/Discussion to 2a

problem 2a

(-25)/(5) = -5

The quotient of the absolute values (25)/(5) = 5 and they have opposite signs, so that is how we get the answer -5.




checkAnswer/Discussion to 2b

problem 2b

7/0 = undefined.




checkAnswer/Discussion to 2c

problem 2c

*Div. is the same as mult. by reciprocal
*Mult. num. together
*Mult. den. together
*(-)(-) = -
*Reduce fraction




checkAnswer/Discussion to 3a

problem 3a

Since we have several operations going on in this problem, we will have to use the order of operations to make sure that we get the correct answer. 

If you need to review the order of operations go to Tutorial 4: Operations of Real Numbers.

*Evaluate inside the absolute values



*Reduce fraction




checkAnswer/Discussion to 4a

problem 4a


Plugging 5 for x and -5 for y and simplifying we get:


*Plug in 5 for x and -5 for y



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Last revised on July 25, 2011 by Kim Seward.
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