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.

juriemagic Oct 13, 2020

#2

#3**0 **

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

#4**0 **

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**+2 **

So, the solution x/y = -1 is not compatible with part 2.

What about the solution x/y = 3/2?

Alan Oct 13, 2020

#6**+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??..

juriemagic
Oct 14, 2020

#8**0 **

Hey hey!!!..what do you know!!!!...whoop whoop!..I am really excited!!..Thank you sir!

juriemagic
Oct 14, 2020