Simultaneous equation question

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The discussion revolves around solving a system of simultaneous equations, specifically y = 2 - x and x² + 2xy = 3. The original poster initially found the roots to be x = -1 and x = 3, leading to coordinates (-1, 3) and (3, -1). However, a participant pointed out an error in the factorization, clarifying that the correct solution for x is 1, not -1. After correcting the mistake, the final coordinates are confirmed as (-1, 3) and (1, 1), aligning with the answer in the reference book. The conversation emphasizes the importance of careful factorization and sign management in solving equations.
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Hey just need my answer to be checked on this problem
just to clarify x2 means x squared!

Solve the following

y = 2 - x
x2 + 2xy = 3

Substitute equations

x2 + 2x(2 - x) = 3
x2 + 4x - 2 x2 = 3
-x2 + 4x - 3 = 0

Factorising

(-x + 1) (x - 3) = 0

so x = -1 and 3
_____________________________________________

as y = 2 - x

y = 2 - 3
= -1

y = 2 + 1
= 3

so the co ordinates for the two roots of this curve will be

(-1, 3) and (3, -1)

is this the right answer? i believe it to be, but my answer books says differently!

Please check my answer.
Thanx in advance!
 
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james_rich said:
Hey just need my answer to be checked on this problem
just to clarify x2 means x squared!
Solve the following
y = 2 - x
x2 + 2xy = 3
Substitute equations
x2 + 2x(2 - x) = 3
x2 + 4x - 2 x2 = 3
-x2 + 4x - 3 = 0
Factorising
(-x + 1) (x - 3) = 0
so x = -1 and 3

_____________________________________________
as y = 2 - x
y = 2 - 3
= -1
y = 2 + 1
= 3
so the co ordinates for the two roots of this curve will be
(-1, 3) and (3, -1)
is this the right answer? i believe it to be, but my answer books says differently!
Please check my answer.
Thanx in advance!

You made a small mistake where I highlighted.
 
sorry, i still don't understand, i checked what i did, i can't see what's wrong with the bit highlighted!

Can anyone elaborate?
 
james_rich said:
sorry, i still don't understand, i checked what i did, i can't see what's wrong with the bit highlighted!
Can anyone elaborate?
I didn't read it all but if your factorization was correct, then x = 1 is a solution and not x = -1.
 
i think i see what i have done wrong

the x = 3 is right

but it is -x = -1!

so x = 1

so to start again...

as y = 2 - x
y = 2 - 3
= -1
y = 2 - 1
= 1
so the co ordinates for the two roots of this curve is
(-1, 3) and (1, 1)

Thanx a lot, this is the answer in the book, easy mistake i made i think! bah humbug with the positive and negative signs!

Cheers! I can sleep now!
 
Good :smile:
 
A word of advice, when you factorise like that it usually helps to have the x^2 coefficient be positive.
 

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