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Add Fractions - Please help me to answer

by Sean
(Taguig City Philippines)











































how to answer 2 1/4+3/8=????? and reduce it to lowest terms

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Aug 15, 2011
Add Fractions
by: Staff


The question:

by Sean
(Taguig City Philippines)


how to answer 2 1/4+3/8=????? and reduce it to lowest terms


The answer:

2 ¼ + 3/8=?????

There are three widely used methods of adding these fractions: (1) add the whole numbers and the fractional components separately (2 + 0) + (¼ + 3/8); (2) convert the two numbers to their decimal equivalents, and then add the decimal equivalents (2.25 + 0.375); (3) convert any mixed numbers to improper fractions, and then add the fractions.


I’m going to demonstrate the third method.

Begin by converting the mixed number 2 ¼ into an improper fraction. This will make the addition of the two fractions much easier.

(2*4 + 1 = 9, so the improper fraction will be 9/4)

2 ¼ + 3/8

= 9/4 + 3/8


Fractions cannot be added directly unless the denominators of both fractions are the same.

Convert the fraction 9/4 to a fraction with a denominator of 8 (the same as the denominator of the fraction 3/8)

To accomplish this, multiply the fraction 9/4 by another fraction: 2/2. Note that 2/2 = 1, so we are actually multiplying by 1. The decimal value of the fraction 9/4 will not change.

= (9/4)*(2/2) + 3/8

= (9*2)/(4*2) + 3/8

= (18)/(8) + 3/8

= 18/8 + 3/8

Since the two fractions now have the same denominator, they can be easily added. (Remember, only the numerators are added. The denominator of 8 remains the same)

= (18 + 3)/8

= 21/8

This fraction cannot be reduced, but it can be converted into a mixed number.

= 21/8

To convert 21/8 to a mixed number, divide 21 by 8

= 21÷8

= 2, remainder 5

= 2 5/8


The final answer is: 2 5/8 (or 21/8 or 2.625)


Check the answer with a calculator:

2 ¼ + 3/8 =?????

= 2.25 + 0.375

= 2.625

Since 2.625 is the same answer obtained by converting 2 5/8 to its decimal equivalent, the answer of 2 5/8 is correct.



Thanks for writing.

Staff
www.solving-math-problems.com

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