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I'm aware this is pretty basic, I'm having trouble with directly and inversely proportional variations.

I have this equation

$$ A = \frac{BC^2\sqrt{D}}{E.F^3} $$

So, as I understand it, if $B$ gets bigger, $A$ should also get bigger as they are directly proportional because they follow the $y=kx$ pattern.

Also if $E$ gets bigger $A$ should get smaller as they are inversely proportional because they follow the $y=\frac{k}{x}$ pattern.

So far so good, however now I'm asked to know how much A will vary depending of what I do to the other unknowns.

For example:

What would happen to $A$ if $B$ was doubled?

I understand $A$ would grow porportionally, as stated above, but It also asks for the specific amount it will grow, for example, if $B$ was doubled then $A$ will also double.

And I cannot comprehend how to do this, or what if $E$ were halved, I know $A$ would grow, but how much?!

kjetil b halvorsen
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undefined
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  • If $B$ is doubled $B \to 2B$ then $A$ is also doubled $A \to 2A$ as you can see from the linear dependence. You might want to brush up on the laws of arithmetic (associativity of multiplication). $$ $$ If $E$ is halved $E \to E/2$ then $A \to 2A$. Again, it follows from the laws of arithmetic – Yuriy S Jan 24 '18 at 00:20
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    Is it because multiplying $B$ by two is the same as multiplying the whole right side of the equation by two? (associativity of multiplication). And then if I multiply the whole right side by two I would have to multiply the left side by the same ammount to preserve the equality? – undefined Jan 24 '18 at 00:25
  • @ undefined, pretty much yes – Yuriy S Jan 24 '18 at 00:28
  • Thanks, could you ellaborate on the more "complicated" variables? I'm having trouble with exponents (or radicands). I appreciate your help, but maybe an answer would be more appropiate than comments. – undefined Jan 24 '18 at 00:31

2 Answers2

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@Nilknarf has dealt with the example for $B$ in the other answer. I thought I could add to what happens with $F$.

Suppose $F$ is doubled. Then the step to take is to replace the place where $F$ was with $2F$. So you end up with $$\text{something}=\frac{BC^2\sqrt D}{E(2F)^3}=\frac{BC^2\sqrt D}{8EF^3}=\frac18\frac{BC^2\sqrt D}{EF^3}=\frac18 A$$So $A$ would decrease by a factor of $8$.

This method is better in my opinion, since there is no "guesswork" involved (you said you would have divided both sides by $2$ instead).

B doubled:

$$\frac{(2B)C^2\sqrt D}{EF^3}=2\times\frac{BC^2\sqrt D}{EF^3}=2A$$So $A$ doubles.

C doubled: $$\frac{B(2C)^2\sqrt D}{EF^3}=\frac{4BC^2\sqrt D}{EF^3}=4A$$ So $A$ quadruples.

D doubled: $$\frac{BC^2\sqrt {2D}}{EF^3}=\sqrt2\times\frac{BC^2\sqrt D}{EF^3}=\sqrt2\times A$$So $A$ increases by factor of $\sqrt 2$.

E doubled: $$\frac{BC^2\sqrt D}{(2E)F^3}=\frac12\frac{BC^2\sqrt D}{EF^3}=\frac12 A$$So $A$ halves.

John Doe
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  • I find the whole process explained (using arithmetic properties) really helpful, would It bother you to do it for each variable? – undefined Jan 24 '18 at 00:41
  • @undefined I have quickly sketched how to do it for each variable when it comes to doubling. Using this same method, you can see what happens when some quantity halves too - but instead of replacing the variables with twice themselves, you can replace them with half of themselves. Example: $E$ halves: $$\frac{BC^2\sqrt D}{(E/2)F^3}=2\times \frac{BC^2\sqrt D}{EF^3}=2A$$So $A$ would double. – John Doe Jan 24 '18 at 00:46
  • Thank you so much, I will mark your answer as the accepted one not because @Nilknarf's was wrong, but because it's friendlier to people of my "math level". People who would have this kind of question on the first place. – undefined Jan 24 '18 at 00:49
  • Thanks, glad to help :) – John Doe Jan 24 '18 at 00:50
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It is given that $$A=\frac{BC^2\sqrt D}{EF^3}$$ By multiplying both sides by $2$, it follows that $$2A=\frac{2BC^2\sqrt D}{EF^3}$$ or $$(2A)=\frac{(2B)C^2\sqrt D}{EF^3}$$ You can see now that replacing $A$ by $2A$ and $B$ by $2B$ results in the equation remaining solved. Thus, when $B$ is doubled, or replaced by $2B$, $A$ must be replaced by $2A$, or doubled.

Here's another example. Suppose that $F$ is doubled. By dividing both sides by $8$, it follows that $$A/8=\frac{BC^2\sqrt D}{8EF^3}$$ or $$(A/8)=\frac{BC^2\sqrt D}{E(2F)^3}$$ Thus, since $A\to A/8$ and $F\to 2F$ satisfy the equation, it follows that if $F$ is doubled, then $A$ must by reduced by a factor of $8$.

Here's yet another example. Suppose that $D$ is doubled. Then, by multiplying both sides by $\sqrt 2$, we have $$A\sqrt 2=\frac{\sqrt{2}BC^2\sqrt D}{EF^3}$$ or $$(A\sqrt 2)=\frac{BC^2\sqrt{2D}}{EF^3}$$ So when $D$ is doubled, $A$ is increased by a factor of $\sqrt 2$.

Franklin Pezzuti Dyer
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  • I understand the first one, however for the second one, in my head, I would've multiplied $F$ by two and divided $A$ by two. I understand it's because of the power of three (2^3=8, I get it) but I just can't follow the reasoning. Maybe another example would help, $C$ or $D$ seem similar. – undefined Jan 24 '18 at 00:29
  • @undefined If you found my answer helpful, don't forget to $\color{green}{\checkmark}$! :D – Franklin Pezzuti Dyer Jan 24 '18 at 00:34