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Sum of Two Rational Expressions with Different Denominators - Math










































Give an example of a sum of two rational expressions with different denominators, then perform the operation by showing all the steps, including how you found the common denominator. These rational expressions must have a variable in the denominator, such as (3x+1) / (x^2 – 1). Give another example of a sum, ratio, product, or different of two rational expressions for your classmates to solve. Again, the example must have a variable in the denominator. When might you use rational expressions in real life?


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May 18, 2012
Sum of Two Rational Expressions with Different Denominators
by: Staff


Question:

Give an example of a sum of two rational expressions with different denominators, then perform the operation by showing all the steps, including how you found the common denominator. These rational expressions must have a variable in the denominator, such as (3x+1) / (x^2 – 1). Give another example of a sum, ratio, product, or different of two rational expressions for your classmates to solve. Again, the example must have a variable in the denominator. When might you use rational expressions in real life?



Answer:


The technique for adding rational expressions with different denominators is exactly the same procedure as adding fractions with different denominators.

So . . . let’s briefly review the process as it applies to fractions, and then apply it to rational expressions.


1/3 + 2/5


To add these fractions, they must have a common denominator

1/3 + 2/5

= (1/3)*(5/5) + (2/5)*(3/3)

= (1*5)/(3*5) + (2*3)/(5*3)

= 5/15 + 6/15


Both fractions can now be added because they each have the same denominator

5/15 + 6/15

= (5 + 6)/15

= 11/15


Therefore,

1/3 + 2/5 = 11/15



Now, let’s apply this same procedure to the addition of two rational expressions:

Add

(3x + 1) / (x - 1) and (x) / (4x + 1)


= (3x + 1) / (x - 1) + (x) / (4x + 1)



To add these fractions, they must have a common denominator

The new common denominator will be equal to the two existing denominators multiplied together:

(x - 1) * (4x + 1) = 4x² - 3x - 1

Convert the two fractions to the common denominator (x - 1) * (4x + 1)


(3x + 1) / (x - 1) + (x) / (4x + 1)

= [(3x + 1) / (x - 1)] * [(4x + 1) / (4x + 1)] + [(x) / (4x + 1)] * [(x - 1) / (x - 1)]

= [(3x + 1) * (4x + 1)] / [(x - 1) * (4x + 1)] + [(x) * (x - 1)] / [(4x + 1) * (x - 1)]

= [(3x + 1) * (4x + 1)] / [(x - 1) * (4x + 1)] + [(x) * (x - 1)] / [(x - 1) * (4x + 1)]

= [(3x + 1) * (4x + 1)] / (4x² - 3x – 1) + [(x) * (x - 1)] / (4x² - 3x – 1)

At this point the denominators of each rational expression are the same

= [(3x + 1) * (4x + 1)] / (4x² - 3x – 1) + [(x) * (x - 1)] / (4x² - 3x – 1)

= (12x² + 7x + 1) / (4x² - 3x – 1) + (x² - x) / (4x² - 3x – 1)


Both fractions can now be added because they each have the same denominator

= (12x² + 7x + 1) / (4x² - 3x – 1) + (x² - x) / (4x² - 3x – 1)

= [(12x² + 7x + 1) + (x² - x)] / (4x² - 3x – 1)

= (12x² + 7x + 1 + x² - x) / (4x² - 3x – 1)

= (12x² + x² + 7x - x + 1) / (4x² - 3x – 1)

= (13x² + 6x + 1) / (4x² - 3x – 1)


>>> the final answer for this example is:

(3x + 1) / (x - 1) + (x) / (4x + 1) = (13x² + 6x + 1) / (4x² - 3x – 1)





Thanks for writing.

Staff
www.solving-math-problems.com


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