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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
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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... ..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
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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??.. juriemagic  Oct 14, 2020
#7
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Looks good!

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

juriemagic  Oct 14, 2020