maths problem

[786ARS]

I have a maths problem, which I have tried everything to solve. Can anyone here help me??
786ARS

[cmillar6]

you will need to setup a system of 2 equations in terms of x and y and then solve for both. To do this you need to express the area of the floor in terms of x and y. Your first equation should be something like

x2 + 1/2(y)(y+1) = 5.5 (area of the trianagle is 1/2(base)(Height))

setup a similar expression interms of x and y for the second equation (get creative) and solve for x and y

[786ARS]

what i end up with are simultaneous equations -
1. x2 = 2y+1
2. 2x2+xy=11m2
3. [substitute 1 into 2] 4y+2+xy=11m2

and there is a perimeter from an earlier part of the question
3x+2y=9

but the main problem lies in getting ri of the xy part of the equation

I already know the answer, but for a 9 mark question, i need to know how to get there

[Floppyman]

Unless some hint is given of the relationship between x and y you have two unknowns and would thus need two equations to solve. I agree with cmillar6's solving methodology.

HTH

Edit...can you explain your choice of equations a little more. What is the m? meters right?

[Floppyman]

BTW, are the answers

x =2 and y = 1.5 ?

I worked through this but got to a nasty cubic:

x^3+4x^2-x-22 = 0

Graphically, I could see the solution was x = 2, but I'm not sure whether you are allowed to use any aids to help you find a solution, much less use it to justify a solution.

I'm sure there must be an easier way to do this analytically, it just hasn't occured to me yet.

[786ARS]

yep, its a C1 paper, which means no calculator - AS level maths is annoying
Thanks for helping me. I hate cubics with a passion, no easy formula to work out values of x, unlike quadratics.

[rspassey]

Originally Posted by 786ARS yep, its a C1 paper, which means no calculator - AS level maths is annoying Thanks for helping me. I hate cubics with a passion, no easy formula to work out values of x, unlike quadratics.

With cubics, or any polynomial, start with Descartes rule of sign, then the rational roots theorem, then synthetic division with the possibilities of the rational roots theorem combined with the idea of upper and lower bounds, then successive isolation.

PS, could you clarify what the values of each side are. The handwriting is a little rough to understand.

[786ARS]

sorry, the 3 sides of the square are "x", the other sides are "y" and "y+1". I just hope I dont any questions like this in my exam on the 11th.

[rspassey]

Originally Posted by cmillar6 you will need to setup a system of 2 equations in terms of x and y and then solve for both. To do this you need to express the area of the floor in terms of x and y. Your first equation should be something like x2 + 1/2(y)(y+1) = 5.5 (area of the trianagle is 1/2(base)(Height)) setup a similar expression interms of x and y for the second equation (get creative) and solve for x and y

Then wouldn't it be x^2 + .5(y * x) = 5.5

You then also know that x^2 = (y + 1)^2 - y^2

EDIT: Are you sure the perimiter is 3x + 2y = 9? I derive some very irrational numbers using this data, which do not work out correctly.

x = 2 and y = 1.5. Got it.

[786ARS]

yes, but i cant get rid of the xy term
i can substitute x^2 with 2y+1 but it still annoys me

[Floppyman]

Originally Posted by 786ARS yes, but i cant get rid of the xy term i can substitute x^2 with 2y+1 but it still annoys me

Along with what Ryan said (I don't remember all that stuff on how to solve cubics -- it's been ages since I've had a course on algebra ) is it possible to take both equations and setup a matrix?

[rspassey]

no need for a matrix. Just substitute (4.5 - 1.5x) in for y in the equation 2x^2 + yx = 11.

Solve it and factor to find that x = 2 or -11, which is erroneous. Turn it around and put in 2 for x and you get 2y = 3, therefore y = 1.5.

I don't even touch cubics.

To me, it threw me at first because it was a simple problem to work, but I overlooked the perimeter part you noted; that made it a lot easier.

[Floppyman]

OH!! I never saw the perimeter was given as part of the problem....oh man, that simplifies things a heck of a lot! I was trying to solve the problem without that given bit of information the whole time, and that's pretty painful as we found out.

I agree with you Ryan, subbing in that linear equation for the perimeter turns the whole thing quadratic and then use quadratic formula to solve the problem if need be.

LOL......nohting like overlooking the obvious

Hope it all makes sense now.

[cmillar6]

Then wouldn't it be x^2 + .5(y * x) = 5.5

oops, sorry about that, thanks for pointing out my mistake rspassey, where did this expression for the perimeter come from? I never saw that in the attachment. That makes the problem pretty easy to solve.

[rspassey]

Originally Posted by cmillar6 oops, sorry about that, thanks for pointing out my mistake rspassey, where did this expression for the perimeter come from? I never saw that in the attachment. That makes the problem pretty easy to solve.

In post #3. It can be solved without it, but that would be a lot more work.