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Mathematics 19 Online
OpenStudy (anonymous):

simplify (5/4x^2y)-(y/14xz)

OpenStudy (asnaseer):

ok, this one is more tricky. I assume this is the question?\[\frac{5}{4x^2y}-\frac{y}{14xz}\]so we have 'x', 'y' and 'z' as variables here?

OpenStudy (asnaseer):

?

OpenStudy (asnaseer):

can you please confirm whether I have the question correct?

OpenStudy (anonymous):

yess .

OpenStudy (asnaseer):

ok, so first step is to get a common denominator - do you know how to do that?

OpenStudy (anonymous):

yes it would be 14xz(4x^2y) for the first fraction and 4x^2y(14xz) for the second

OpenStudy (asnaseer):

yes - you can do that but have you been taught how to work out the "lowest common multiple"?

OpenStudy (anonymous):

uhmmm yea . but im not sure that would be used here

OpenStudy (asnaseer):

working out the LCM means you get simpler equations to work with.

OpenStudy (asnaseer):

we can do it your way first and then I can show you how to use LCM method later

OpenStudy (anonymous):

okayy then

OpenStudy (asnaseer):

so, remember that any number multiplied by one is the same number. therefore:\[\frac{5}{4x^2y}=\frac{5}{4x^2y}\times\frac{14xz}{14xz}\]agreed?

OpenStudy (anonymous):

where did the y go?

OpenStudy (asnaseer):

I am just looking at the first term here:\[\frac{5}{4x^2y}=\frac{5}{4x^2y}\times1=\frac{5}{4x^2y}\times\frac{14xz}{14xz}\]since:\[\frac{14xz}{14xz}=1\]

OpenStudy (anonymous):

okayy .

OpenStudy (asnaseer):

so that then gives:\[\frac{5}{4x^2y}=\frac{5}{4x^2y}\times1=\frac{5}{4x^2y}\times\frac{14xz}{14xz}=\frac{70xz}{56x^3yz}\]ok?

OpenStudy (anonymous):

u do have x ^ 3 right ? and z ^ 2 ?

OpenStudy (anonymous):

not z 6 2 , scratch that

OpenStudy (anonymous):

z ^ 2 *

OpenStudy (asnaseer):

ok, now lets look at the next term:\[\frac{y}{14xz}=\frac{y}{14xz}\times1=\frac{y}{14xz}\times\frac{4x^2y}{4x^2y}=\frac{4x^2y^2}{56x^3yz}\]agreed?

OpenStudy (anonymous):

absolutely .

OpenStudy (asnaseer):

good, so now we can say that:\[\frac{5}{4x^2y}-\frac{y}{14xz}=\frac{70xz}{56x^3yz}-\frac{4x^2y^2}{56x^3yz}\]make sense?

OpenStudy (anonymous):

yess . those r the equations you just got right ?

OpenStudy (asnaseer):

yes

OpenStudy (anonymous):

okay then .

OpenStudy (asnaseer):

so now we have two fractions that have the same denominator, which means we can safely combine the numerators to get:\[\frac{5}{4x^2y}-\frac{y}{14xz}=\frac{70xz}{56x^3yz}-\frac{4x^2y^2}{56x^3yz}=\frac{70xz-4x^2y^2}{56x^3yz}\]

OpenStudy (anonymous):

okay . now do we add/subtract common terms?

OpenStudy (asnaseer):

you can't actually add/subtract common terms here as there are none. what you need to do next is to factorise the numerator.

OpenStudy (anonymous):

im not sure how to do that .

OpenStudy (asnaseer):

you can see the numerator has 'x' in both terms and also that 70 and 4 are both divisible by 2

OpenStudy (anonymous):

so then the numerator will look like ... 2x(35z-2xy^2)

OpenStudy (asnaseer):

exactly! so we get:\[\frac{5}{4x^2y}-\frac{y}{14xz}=\frac{70xz}{56x^3yz}-\frac{4x^2y^2}{56x^3yz}=\frac{70xz-4x^2y^2}{56x^3yz}=\frac{2x(35z-2xy^2)}{56x^3yz}\]

OpenStudy (asnaseer):

now you should be able to cancel some common terms between the numerator and the denominator

OpenStudy (anonymous):

alright , let me try and work this out .

OpenStudy (asnaseer):

great! :)

OpenStudy (anonymous):

just kidding ! im lost already /: ugh

OpenStudy (asnaseer):

just look at what cancels out between the numerator and the denominator

OpenStudy (asnaseer):

e.g. there is a 2 in the numerator and a 56 in the denominator - so you should be able to divide numerator and denominator by 2

OpenStudy (anonymous):

then i wwill get 28x^3yz in the denom?

OpenStudy (asnaseer):

yes

OpenStudy (asnaseer):

now you should also be able to divide numerator and denominator by 'x'

OpenStudy (anonymous):

28x^2yz?

OpenStudy (asnaseer):

that is the correct denominator - what do you get for the numerator now?

OpenStudy (anonymous):

(35z−2xy2)

OpenStudy (asnaseer):

perfect! you've got the right answer. :)

OpenStudy (anonymous):

yayyyyy !!!

OpenStudy (asnaseer):

if we were to find the LCM at the start, then you wouldn't need to simplify at the end as we id here as it would naturally come out simplified. however, I suggest you first get used to simplifying these using the method above and, once you are confident with this method, then I can show you how to make use of the LCM.

OpenStudy (anonymous):

its such an in depth question , which is what makes it so hard , but thank you for your help! and i will want to learn lcm eventually ...

OpenStudy (asnaseer):

thats good to hear. :)

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