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Hi dear friends,

I am wanting to solve a question for a pupil I have, but cannot seem to achieve it. There are 2 questions to this sum. please help...

 

the equation is:

\(2x^2-xy=3y^2\)

 

Q1) calculate the value of the ratio \(x \over y\)

Q2) Hence, calculate the values of x and y if \(x+y=2\)

 

Thank you all for helping out, it is as always very much appreciated.

 Oct 13, 2020
 #1
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2x^2-xy=3y^2  

2(1)^2-1*1=3(1)^2

 

2-1=3

 

45

 
 Oct 13, 2020
 #2
avatar+31093 
+1

Part 1 is as follows:

Let's see you attempt part 2.

 
 Oct 13, 2020
 #3
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Hi Alan,

 

gosh, I don't even have a slightest clue where to start...how to start....I'm looking at this..but just do not know at all what to do..I have just googled..found something..maybe...frown..just give me a minute or so, I'll come back to you..

 
juriemagic  Oct 13, 2020
edited by juriemagic  Oct 13, 2020
 #4
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Alan,

I really do not know....I'm thinking something like this:

 

\({x \over y}=-1\)

 

so then \(x=-y\)

so then \(x+y=0\)

However, it is given that \(x+y=2\)

 

uuhh, Alan...I don't know...this makes me feel so utterly stupid!

 
juriemagic  Oct 13, 2020
 #5
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So, the solution x/y = -1 is not compatible with part 2.

What about the solution x/y = 3/2?
 

 
 Oct 13, 2020
 #6
avatar+748 
+1

Alan, Thanx for walking me through this...honestly, you will never realise how much this means!!..okay, x:y = 3:2

 

that gives \(2x=3y\)

which can go to \(2x-3y=0\)

 

Nothing can be factored..soooo, mabe then I should look at simultaneous equations???, meaning...

\(2x-3y=0.....(1)\)

\(x+y=2\)

\(x=2-y.....(2)\)

 

(2) in (1)

\(2(2-y)-3y=0\)

\(4-2y-3y=0\)

\(-5y=-4\)

\(y= {4 \over 5}...(3)\)

 

(3) in (2)

\(x=2- {4 \over5}\)

\(x={6 \over5}\)

 

So x then is 6\5 and y is 4\5

 

Is this correct maybe??..sad

 
juriemagic  Oct 14, 2020
 #7
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Looks good!

 
Alan  Oct 14, 2020
 #8
avatar+748 
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Hey hey!!!..what do you know!!!!...whoop whoop!..laughlaughI am really excited!!..Thank you sir!

 
juriemagic  Oct 14, 2020

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