Year 10 Algebra & Quadratics: A paving tile has the shape and dimensions shown....

JoshRulz

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Hi there,

I'm currently studying Algebra & Quadratics in my Year 10 class. I came across this question which I found incredibly difficult to understand. My teacher was kind enough to help me out with it but my teacher, parents and uncle all don't understand how they got their answers. Could somebody please help me? It'd be much appreciated.

Question:

Algebra & Quadratics, Part 1 of Chapter 3
Factorising Using Common Factors 3.2
Open-Ended
Q13.
A paving tile has the shape and dimensions shown (x x x x y). Find two sets of values for x and y so that the volume (in cm3) has numerically the same value as the surface area (in cm2). The values do not have to be integers.
I've been told by my teacher that the answers are x = 12, y = 3; x = 8, y = 4

If somebody could help me understand how they got that answer, it'd be awesome, it's been baffling my whole family and I for the past week. I tried asking people on Quora but I have since had no luck.

Thanks.

Kind Regards,
Josh.
 
Write an equation that says volume = surface area.

Then solve that equation for y. (Or you could solve for x if you prefer.)

Pick any x such that both x and y will be positive, and calculate y. So there are infinitely many solutions; several involve integers.

Let's see your equation and any other work you can do, and we can discuss further.
 
Hi Dr. Peterson,

Sorry I took so long to reply, I've been trying to finally answer the question. I've done what you recommended, and have worked out that the volume (432) and the surface area (36) are the answers to part of it. Basically saying that the volume = surface area has been converted to 432 = 36. I'm still not quite getting it though. I've tried solving the equation for both x and y and still cannot find the answer.
Would it be possible if I could send through a photo maybe of the question and then you could help me work it out, step by step if possible?
Thanks,

Kind Regards, Josh.
 
It sounds like you misunderstood what I said. The volume and surface area are not known fixed numbers; where did you get them from?

Please show me the expressions you wrote for volume and surface area, and the resulting equation you had. Then I can show you what I mean by solving for y (in terms of x).
 
Hi Dr. Peterson,

I got the numbers from my teacher, she kindly checked the answers as we tried to work out the answer together.
Basically, I got the numbers 432 and 36 by going 12 x 12 x 3 = volume; 8 x 4 = surface area.
I got to about here and then got confused by how the numbers could mean the same, though I would suggest I misunderstood.
Thanks.

Kind Regards, Josh.
 
You appear to be working backward from the answer, which is inappropriate, since it is clearly stated that they asked for, and gave, only two of many answers. You also seem to be totally misreading the problem, as in any solution there will be only one pair of x and y. The two pairs are two different solutions.

Clearly you know that the volume is [MATH]x\cdot x\cdot y = x^2 y[/MATH].

The surface area of that same tile will be [MATH]2x^2 + 4xy[/MATH], since the top and bottom are each x by x, and each of the four sides is x by y.

So the equation you have to solve is [MATH]x^2y = 2x^2 + 4xy[/MATH]. That says that the numerical values of the volume and surface area are equal.

Now, I said to solve this for y as a function of x. Can you do that?
 
Hi Dr. Peterson,

Sorry for misunderstanding the question, I believe I understand now. ?
I think I can solve it, I can understand how you got x2y an. And I also think I understand how you've calculated the surface area. That being said, the equation I have to solve is still seemingly trick and hard for me to understand. This is all certainly new to me (I don't remember the schoolbook ever teaching me this at this degree)
I need to make 2x2 + 4xy = x2y. Which would mean that I need to add 2x2 + 4xy to get x2y correct?
And if that's correct, then 2x2 + 4xy would have to equal 8x2y?
 
I need to make 2x2 + 4xy = x2y. Which would mean that I need to add 2x2 + 4xy to get x2y correct?
And if that's correct, then 2x2 + 4xy would have to equal 8x2y?
No. The equation just says that 2x2 + 4xy and x2y are equal. Multiplying the right side by 8 would make them unequal. Maybe you didn't mean what you wrote. In any case, the goal is to solve for y. (You wouldn't want to solve for x.)

It's quite possible that you haven't seen equations that look this complicated; but I would expect you to have solved equations in x and y for y. For instance, you might have solved something like 4x + 3y = 6 for y, by getting y alone on one side, treating x as if it were just some number. You have to do the same sort of thing here.

Look at where y is in the equation. It's in the second term on the left, and in the term on the right. You want to get them together on one side. Give it a try.

If you're not ready for that, try this: I'll replace x with some arbitrary number, say 5. Then the equation becomes 50 + 20y = 25y. How would you solve that? Do it, and then try the actual equation.
 
Hi Dr. Peterson,

I was feeling a bit overwhelmed by it all, so I did the equation where you replaced x for 5. I would solve that by working out the HCF?
I'm having a quick flick through some of the previous questions I've answered correctly in my book and one of them is quite similar (Factorise 8x2 + 12x). I needed to find the HCF of the particular terms. In your equation, the HCF of 50 and 20 would be 10. Meaning I would then need to divide each term to find the quotient? That would mean the answer to that question would be 10y(5y + 2)?
So if that was correct, that'd mean I would have my question (x x x x y) = (x x x). But then that wouldn't give me numerical numbers which is what they are looking for...
I did at one stage think it was going to be a guess and check question, and at one stage I did wonder if it was decimal numbers, but it seems it's not, according to the answers my teacher gave me. The book is also making out like they're the only answers which is confusing me more.
I don't see how the question is even possible in a way, considering that they're pretty much asking me to make (12 x 12 x 3) = (12 x 12), which is theoretically impossible. Or at least I would have thought.
 
Maybe I should just show you what I did, in case this is not something you're meant to do at all.

First, to solve me equation 50 + 20y = 25y, you just want to get the y terms together, so you subtract 20y from both sides and get 50 = 5y. Then divide by 5, and you have y = 10.

Going back to the actual equation, [MATH]2x^2 + 4xy = x^2y[/MATH], we do exactly the same thing. We see y in two terms, so we subtract [MATH]4xy[/MATH] from both sides and have [MATH]2x^2 = x^2y - 4xy[/MATH]. We want to solve for y, so we factor the HCF xy out of both terms (equivalent to combining like terms in the simpler example): [MATH]2x^2 = (x - 4)xy[/MATH]. Finally, we can divide both sides by [MATH](x - 4)x[/MATH], and we end up with [MATH]y = 2x/(x - 4)[/MATH].

What this tells us is that for any x greater than 4 (to make sure y will be positive), we can find a value of y so the original equation will be true.

The book's two answers come from choosing x=12 and x=8, respectively:

If [MATH]x = 12[/MATH], then [MATH]y = 2x/(x - 4) = 2(12)/(12 - 4) = 3[/MATH]: x=12, y=3
[MATH]V = x^2y = 432[/MATH], [MATH]SA = 2x^2 + 4xy = 432[/MATH]​
If [MATH]x = 8[/MATH], then [MATH]y = 2x/(x - 4) = 2(8)/(8 - 4) = 4[/MATH]: x=8, y=4
[MATH]V = x^2y = 256[/MATH], [MATH]SA = 2x^2 + 4xy = 256[/MATH]​
But there are lots of other answers, some with integers, some not. For example,

If [MATH]x = 5[/MATH], then [MATH]y = 2x/(x - 4) = 2(5)/(5 - 4) = 10[/MATH]: x=5, y=10
[MATH]V = x^2y = 250[/MATH], [MATH]SA = 2x^2 + 4xy = 250[/MATH]​
If [MATH]x = 6[/MATH], then [MATH]y = 2x/(x - 4) = 2(6)/(6 - 4) = 6[/MATH]: x=6, y=6
[MATH]V = x^2y = 216[/MATH], [MATH]SA = 2x^2 + 4xy = 216[/MATH]​
If [MATH]x = 7[/MATH], then [MATH]y = 2x/(x - 4) = 2(7)/(7 - 4) = \frac{14}{3}[/MATH]: x=7, y=4 2/3
[MATH]V = x^2y = 228 \frac{2}{3}[/MATH], [MATH]SA = 2x^2 + 4xy = 228 \frac{2}{3}[/MATH]​
 
Hi Dr. Peterson,

So basically, I needed to subtract from both sides and then divide what's remaining which would give me the answer.
So in the event that I had a question that was 180 + 30y = 60y; I would need to subtract 30 from 60, which would again give me 30, and then divide it by 180, which would equal 6?
 
Yes, that's how you can solve a simple linear equation in one variable. Your problem was "just" a bigger version of that ... well disguised.
 
Hi Dr. Peterson,

Thank you so much for helping. I'm sure my teacher will be incredibly pleased that you helped me. I must give most of the credit to you though. I really appreciate your help and I honestly couldn't have done it without you.
 
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