View Full Version : help with simplifying (3/8y + 5/6x) / 5/12xy

aajlh

06-03-2009, 03:37 AM

Just joined this forum and I'm hoping to get some help. I am home educating three boys and maths is definitely not my forte. In fact, I was terrible at maths at school (always wanted to know why things were done in a certain way rather than just do it because that was the formula. Consequently I got completely left behind still trying to work out 'why' (plus spending most of my childhood in hospital and going to 13 different schools didn't help). So, now I am learning maths with my boys and actually loving it. I don't have anyone to help me with problems that we have difficulty with so I hope people here might help. Anyway enough of the 'sob story'. Here are a few problems we have been struggling with.

(3/8y + 5/6x) / 5/12xy

I'm not sure how to write maths so perhaps this makes more sense:

(3/8y + 5/6x) divided by 5/12xy

The answer in the book is: 9x + 20y divided by 10.

If someone could show me the workings for how this answer is obtained that would be great.

Another problem we have is:

(27x^2)^(1/3) divided by 1/3(x^3)^1/2. In other words 27 'x' squared all raised to the power of a third. That is then divided by a third times (x cubed) raised to the power of a half. The answer in the book is given as 9/x^5/6 (ie, 9 divided by (x to the power of 5 sixths). I am completely lost with this one and if someone can show me how to do it I might have a chance of doing the other similar ones in the book.

Thanks

Anne

stapel

06-03-2009, 07:17 AM

I am home educating three boys and maths is definitely not my forte.

Unfortunately, attempting to tutor through a "translator" who "doesn't speak the language" does not have a history of success. It would be better if we could speak with the students directly. :oops:

(3/8y + 5/6x) / 5/12xy

Before starting the division, first simplify the fraction sum on top. I am guessing the fractions to be as follows:

. . . . .\frac{3}{8y}\, +\, \frac{5}{6x}

The common denominator is 24xy, so convert:

. . . . .\frac{9x}{24xy}\, +\, \frac{20y}{24xy}

Then combine the fractions. Once you have (fraction)/(fraction), you can flip and multiply.

(27x^2)^(1/3) divided by 1/3(x^3)^1/2.

I believe the expression is as follows:

. . . . .\frac{\left(27x^2)^{\frac{1}{3}}}{\frac{1}{3}\, \left(x^3\right)^{\frac{1}{2}}}

A good first step might be to deal with the fractional exponents (http://www.purplemath.com/modules/exponent5.htm), yielding:

. . . . .\frac{3x^{\frac{2}{3}}}{\left(\frac{x^{\frac{3}{2 }}}{3}\right)}\,=\, \left(\frac{3x^{\frac{2}{3}}}{1}\right)\left(\frac {3}{x^{\frac{3}{2}}}\right)

...and so forth. :wink:

soroban

06-03-2009, 10:46 AM

Hello, aajlh!

Welcome aboard!

\text{Simplify: }\;\frac{\dfrac{3}{8y} + \dfrac{5}{6x}}{\dfrac{5}{12xy}}

If we are given a complex fraction (one with more than two "levels")

. . there is a procedure that converts it to a simple fraction.

Multiply top and bottom by the common denominator of all the denominators.

The denominators are: .8y,\;6x,\;12xy

. . The LCD is: .24xy

\text{Multiply top and bottom by }24xy\!:\quad \frac{24xy\left(\dfrac{3}{8y} + \dfrac{5}{6x}\right)}{24xy\left(\dfrac{5}{12xy}\ri ght)} \;=\;\frac{24xy\left(\dfrac{3}{8y}\right) + 24xy\left(\dfrac{5}{6x}\right)}{24xy\left(\dfrac{5 }{12xy}\right)}

\text{Reduce and simplify: }\;\frac{\rlap{//}24^{^3}x\rlap{/}y\cdot\dfrac{3}{\rlap{//}8y} + \rlap{//}24^{^4}\rlap{/}xy\cdot\dfrac{5}{\rlap{//}6x}}{\rlap{//}24^{^2}\rlap{//}xy\cdot\dfrac{5}{\rlap{////}12xy}} \;=\;\frac{9x+20y}{10}

aajlh

06-03-2009, 07:15 PM

Hi stapel and soroben

Thanks for replying. How did you get your fractions to look like they had been written, including striking out numbers that have been cancelled, indices, large brackets big enough to cover the top and bottom of a fraction, etc? I've managed to do some underlining this time but can't get the numbers under the right fractions on the bottom line.

Soroben I understand you method, thanks.

Stapel I am a bit stuck as to how to keep going after that.

Once I get:

(9x + 20y) * 12xy

24xy 24xy 5

I then get:

9x+20y * 12xy

24xy 5

then???

108x^2*y + 240xy^2

120xy

I suspect that's not right and, even if it is, can you please show me how to continue right to the solution?

Thanks

Anne

aajlh

06-03-2009, 07:25 PM

Hi again Stapel

Can you help me keep going to the solution please?

3x^(2/3) * 3

1 x^(3/2)

then:

9x^(2/3)

x^(3/2)

then:

deal with the indices. As they are divided I should subtract them.

So: 2 - 3

3 2

= 4 - 9

6 6

= -5

6

So the solution is?

9

x^(-5/6)

Which is wrong but getting closer.

Can you show me what I am doing wrong please?

Thanks

Anne

Denis

06-03-2009, 11:26 PM

3/2 - 2/3 = 5/6 (not -5/6)

aajlh

06-03-2009, 11:42 PM

Hi Denis

Aren't I trying to do 2/3 - 3/2 though?

Anne

Denis

06-04-2009, 12:21 AM

Nope Anne; the other way around; examples:

a^2 = 1 / a^(-2)

a^7 / a^13 = 1 / a^(13-7) = 1 / a^6

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