+1124 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.
+1124 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..
+1124 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!
+33661 So, the solution x/y = -1 is not compatible with part 2.
What about the solution x/y = 3/2?
+1124 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??..
+1124 Hey hey!!!..what do you know!!!!...whoop whoop!..
I am really excited!!..Thank you sir!
