Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Find the LCD that would eliminate the fractions in this equation. 6x/11 + 9/2 = 3

OpenStudy (anonymous):

Please help!

OpenStudy (anonymous):

First you need to add: \[ \frac{6x}{11}+\frac{9}{2} \]

OpenStudy (anonymous):

Okay, then what?

OpenStudy (anonymous):

Hard to explain the rest until you add them.

OpenStudy (anonymous):

So 15x/13 ..?

OpenStudy (anonymous):

Nope. First make the denominators the same.

OpenStudy (anonymous):

I don't know how. This is why I posted this, for help. Could you demonstrate..?

OpenStudy (anonymous):

Well, you have denominators \(11\) and \(2\). Multiply them and get \(22\). We do this to both fractions: \[ \frac{6x}{11} \times \frac 22 = \frac{12x}{22}\\ \frac{9}{2}\times \frac{11}{11} = \frac{99}{22} \]

OpenStudy (anonymous):

Then you can add them:\[ \frac{12x}{22}+\frac{99}{22}=\frac{12x+99}{22} \]

OpenStudy (anonymous):

I still don't understand. So how do you get to lowest common denominator?

OpenStudy (anonymous):

Well, to get a common denominator in general, you can multiply the denominators. In this case \(2\times 11=22\)

OpenStudy (anonymous):

There are two ways to get the lowest common denominators. 1) Multiply the numbers together, then divide by their greatest common factor 2) Write out the multiples of each number and find the lowest number that is a multiple of both numbers.

OpenStudy (anonymous):

Method 1: The factors of 2 are 1, 2 The factors of 11 are 1, 11 so the greatest common factor is 1. Thus the least common denominator is \(\frac{2\times 11}{1} = 22\) Method 2: The multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26 The multiple of 11 are 11, 22, 33 The lowest multiple belonging to both is 22.

OpenStudy (anonymous):

For 4 and 6 we can do... Method 1: Factors 4: 1, 2, 4 Factors 6: 1, 2, 3, 6 GCF: 2 LCD: \(\frac{4\times 6}{2}= \frac{24}{2} = 12\) Multiples 4: 4, 8, 12, 16 Mutliples 6: 6, 12, 18, 24 LCD: 12

OpenStudy (anonymous):

Thank you

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!