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I have the following equation:

$$\frac{1000−6n}{3+10n}=k$$

I know it can be rearranged into this, but I do not quite understand how:

$$(5k+3)(10n+3)=5009$$

I know the first thing I can do is multiple both sides by $3 + 10n$ to get:

$$1000 - 6n = 3k + 10kn$$

Then I can add $6n$ to both sides to get:

$$ 1000 = 3k + 10kn + 6n$$

This is where my math knowledge now dies. I am unsure of what the next steps are. Could someone please explain in an idiot proof way of what needs to be done next to get the answer I am looking for?

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You have an error:

$$1000=3k+10kn \color{red}{+}6n$$

From this, multiply both sides by $5$ and add $9$ to obtain: $$5009=50kn+15k+30n+9$$

and factorise. Use:

$$(an+b)(ck+d)=(ac)kn+(bc)k+(ad)n+(bd)$$ as a template to help with this.


Clarification on how this is achieved:

We take $$3k+10kn+6n=1000$$ and rewrite as $$(an+b)(ck+d)=1000+\lambda \tag i$$ where $$ac=10\tag1$$ $$ad=6\tag2$$ $$bc=3\tag3$$ $$bd=\lambda\tag 4$$

Observe $(1)/(3)=\frac ab=(2)/(4)$, so $$\frac{10}{3}=\frac{6}{\lambda}\implies\lambda =\frac95$$ This means that $(i)$ becomes $$(an+b)(ck+d)=1000+\frac 95$$ Hence we multiply the whole system by $5$ to keep it in the integers, so $$ac=50\tag1$$ $$ad=30\tag2$$ $$bc=15\tag 3$$ $$bd =9 \tag 4$$

with $$(an+b)(ck+d)=5009$$

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  • $\begingroup$ Thanks for the quick reply! I have fixed my error. If we didn't know the final equation, how do we know it needs to be multiplied by the value 5 and 9 needs to be added? Essentially how do we know to use those values? $\endgroup$ – user2924127 Jul 9 '20 at 17:20
  • $\begingroup$ Edited to add clarification $\endgroup$ – Rhys Hughes Jul 9 '20 at 17:41
  • $\begingroup$ Thanks so much! That explains so much for me! $\endgroup$ – user2924127 Jul 9 '20 at 18:04

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