STEP I need a little help please.

this is from the 2010 step 1 paper, question 1.
I have pretty much done it all, but I am having some difficulty with this last step. I am having some trouble understand their solution.

Without going through the whole question,

5x^2 + 2y^2 - 6xy + 4x - 4y = 9; (1)
6x^2 + 3y^2 - 8xy + 8x - 8y = 14: (2)

I get 2 simultaneous questions:
(x-y+2)^2=9
and
(y-2x)^2=4

I square root both sides to get

x-y+2=+/- 9 (equation 4)
and
y-2x=+/-4 (equation 5)

They then say:
"Thus we now have four possibilities (two from equation (4), and for each of these, two from
equation (5)), and we solve each one, checking our results back in the original equations."
On a table (which I can't format onto this thread properly) they solve the 4 possibilities and get values for x and y. This is the part I am stuck on, they don't show how to solve these 4 possibilities.

"Thus the solutions are 1,0 7,12 -3,-4 3,8."
They then check the four solutions by putting into the original equations 1 and 2.

Can somebody please explain to me the jump from
x-y+2=+/- 9 (equation 4)
and
y-2x=+/-4 (equation 5)
to where they get those x and y values.

Edit: here is the link to the solution: http://www.geogebra.org/static/STEP/STEP%202010%20Solutions.pdf
the part I am stuck on is the table they make at the top of page 6, where they get the x and y values form the 4 possible equations.

Thnx

(edited 12 years ago)

Reply 1

Coulson72

Here you are:

Spoiler

Reply 2

MWM

OP

Hmm, OK I get the solutions for X. Many thanks for that.

But with Y, why can't I put each x value into just one equation and get 2 y values. e.g for x=-3, y=-1 +/- 3 so y = -4 or 2. Why do I have to put each x value into both equations and then see which value of y satisfies both?
Is it because, I will end up with 8 solutions rather than 4, then I have to put each one of them into the original to find which ones are actually correct ?
So, in a way, do you put each x value into both equations as a check to see which y value is the correct and so you can omit the other 2 incorrect ones, 2 and -8 as in the first example with x=-3.
Or is is because they are simultaneous equations, so you MUST find a y value which satisfies both equations and so you must put x back into both equations ?

(edited 12 years ago)

Reply 3

DFranklin

18

What you've basically done is shown "if x, y are solutions of 5x^2 + 2y^2 - 6xy + 4x - 4y = 9, 6x^2 + 3y^2 - 8xy + 8x - 8y = 14, then x and y must satisfy some other equations that are easier to solve". You have not shown that "if x and y must satisfy these easier equations, then x, y are solutions of 5x^2 + 2y^2 - 6xy + 4x - 4y = 9, 6x^2 + 3y^2 - 8xy + 8x - 8y = 14".

The simplest way to fix this is to check that the values you get really are valid solutions.

Reply 4

ian.slater

12

Original post by MWM

this is from the 2010 step 1 paper, question 1.

I square root both sides to get

x-y+2=+/- 9 (equation 4)
and
y-2x=+/-4 (equation 5)

Thnx

It's not good practice to take the square root of both sides of an equation.

Faced with x^2 = 1 I think it's better to

x^2 - 1 = 0

(x+1)(x-1) = 0

Either x=-1 or x = 1

Otherwise when you meet x^2 = y^2 you may be tempted to go wrong.

Reply 5

MWM

OP

Hmm you have a good point, so How should I have proceeded without rooting both sides ?

Reply 6

Coulson72

Original post by ian.slater

It's not good practice to take the square root of both sides of an equation.

Faced with x^2 = 1 I think it's better to

x^2 - 1 = 0

(x+1)(x-1) = 0

Either x=-1 or x = 1

Otherwise when you meet x^2 = y^2 you may be tempted to go wrong.

so by your method you get the identical value of

x=\pm1

and i'd like to see your method work on this question, if you look at my method above you'd see it is perfectly possible to take account of the situation without making things more complicated than they already are. After factorising you still get:

(x-y+2)^2=9 \Rightarrow (x-y-1)(x-y+5)=0 \Rightarrow x-y+2=\pm3

(y-2x)^2=4 \Rightarrow (y-2x+2)(y-2x-2)=0 \Rightarrow y-2x=\pm2

Hence all you're doing is adding in a futile step

Reply 7

ian.slater

12

Original post by MWM

Hmm you have a good point, so How should I have proceeded without rooting both sides ?

(x-y+2)^2 = 9
(x-y+2+3)(x-y+2-3) = 0

Either x = y-5 or x = y+1

You then have two cases to be considered separately, each of which may lead to possible solutions.

Reply 8

ian.slater

12

Original post by Coulson72

so by your method you get the identical value of

x=\pm1

and i'd like to see your method work on this question, if you look at my method above you'd see it is perfectly possible to take account of the situation without making things more complicated than they already are. After factorising you still get: