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Perform the indicated operation. 15/k-5 + -3k/k-5 3 -3 0 -5
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Asked 5/19/2014 6:34:08 AM
Updated 5/20/2014 8:07:33 PM
1 Answer/Comment
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Flagged by yeswey [5/20/2014 8:05:07 PM]
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User: Perform the indicated operation. 15/k-5 + -3k/k-5 3 -3 0 -5

Weegy: (15)/(k)-5+(-(3k)/(k-5)) Remove the parentheses around the expression -(3k)/((k-5)). (15)/(k)-5-(3k)/(k-5) To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). [ In this case, the LCD is k. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (15)/(k)-5*(k)/(k)-(3k)/(k-5) Multiply -5 by k to get -5k. (15)/(k)-(5k)/(k)-(3k)/(k-5) Combine the numerators of all expressions that have common denominators. (15-5k)/(k)-(3k)/(k-5) Factor out the GCF of 5 from the expression 15. (5(3)-5k)/(k)-(3k)/(k-5) Factor out the GCF of 5 from the expression -5k. (5(3)+5(-k))/(k)-(3k)/(k-5) Factor out the GCF of 5 from 15-5k. (5(3-k))/(k)-(3k)/(k-5) Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of k(k-5). The (5(3-k))/(k) expression needs to be multiplied by ((k-5))/((k-5)) to make the denominator k(k-5). The -(3k)/((k-5)) expression needs to be multiplied by ((k))/((k)) to make the denominator k(k-5). (5(3-k))/(k)*(k-5)/(k-5)-(3k)/(k-5)*(k)/(k) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of k(k-5). (5(3-k)(k-5))/(k(k-5))-(3k)/(k-5)*(k)/(k) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group. (5(3*k+3*-5-k*k-k*-5))/(k(k-5))-(3k)/(k-5)*(k)/(k) Multiply 3 by k to get 3k. (5(3k+3*-5-k*k-k*-5))/(k(k-5))-(3k)/(k-5)*(k)/(k) Multiply 3 by -5 to get -15. (5(3k-15-k*k-k*-5))/(k(k-5))-(3k)/(k-5)*(k)/(k) Multiply -k by k to get -k^(2). (5(3k-15-k^(2)-k*-5))/(k(k-5))-(3k)/(k-5)*(k)/(k) Multiply -k by -5 to get 5k. (5(3k-15-k^(2)+5k))/(k(k-5))-(3k)/(k-5)*(k)/(k) According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. ]
thederby|Points 1530|

Question
Asked 5/19/2014 6:34:08 AM
Updated 5/20/2014 8:07:33 PM
1 Answer/Comment
This conversation has been flagged as incorrect.
Flagged by yeswey [5/20/2014 8:05:07 PM]
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15/(k-5) + [-3k/(k-5)]
= 15/(k-5) - 3k/(k-5)
= (15 - 3k)/(k - 5)
= -3(k - 5)/(k - 5)
= -3
Added 5/20/2014 8:07:33 PM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [5/20/2014 8:09:30 PM]
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