Question

How do you find the LCD for \[\dfrac{5}{ab},\dfrac{4}{b}\]?

Answer 1

Hint: As above given in the question we have to find the LCD (Lowest common denominator or least common denominator. First of all we have to do the multiplication tables of the number for which we have to find the LCD. After that we have to sort out the multiple of numbers which are common and also with that we have determined the least with the common multiples.

Complete step by step solution:
The two fraction gives are\[:-\dfrac{5}{ab}\] and \[\dfrac{4}{b}\]
Here denominator for the first fraction is \['ab'\] and second is \['b'\].
We have determined the LCD which means lowest common denominator. We have to find the common denominator for the two fraction, here such that the fraction remains the same and at the same time and also the common denominator is the natural number which is the lowest possible.
Now, let us take look on the given example
Here, \[\dfrac{-5}{ab},\dfrac{4}{b}\]
Therefore, LCD for the \[ab,b=ab\]
By solving above we can see that the common multiples \[ab\]. But \[b\] is common in both the terms of the denominator.
By satisfying all the required condition we conclude that the LCD for \[\dfrac{5}{ab}\] and \[\dfrac{4}{b}\] is \['ab'.\]

Note: While solving the problem for LCD (least common denominator) first we need to list the multiple of each of the denominators.
As an example we take \[\left( i \right)\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}\]
Multiple of \[2\] are \[\therefore 2\times 1=2,2\times 2=4\] etc.
Multiples of \[3\] are \[\therefore 3\times 1=3,3\times 2=6,3\times 3=9\] etc.
Multiples of \[5\] are \[\therefore 5\times 1=5,5\times 2=10,5\times 3=15\] etc.
After that we have identified the LCM ( Lowest common multiple). If there is not any common multiple present then you may continue to write the multiple until we can across a should multiple.
As an example use take \[30,\] the denominator can share only one multiple.
\[\therefore 2\times 15=30,3\times 10=30,5\times 6=30\]
\[\therefore \] The LCD is \[30.\]
Then we have to rewrite the original equation.
For \[:-\left( \dfrac{15}{15} \right)\times \left( \dfrac{1}{2} \right):\left( \dfrac{10}{10} \right)\times \left( \dfrac{1}{3} \right):\left( \dfrac{6}{6} \right)\times \left( \dfrac{1}{5} \right)\]
\[\therefore \] The new equation is \[\dfrac{15}{18}+\dfrac{10}{80}+\dfrac{6}{80}.\]
At least we have to solve the equation which is written for the required answer.