# Non-Linear Simultaneous Equation

• Jan 19th 2011, 02:07 AM
sxsniper
Non-Linear Simultaneous Equation
Hello,

I've just started on Non-linears and am Stuck on my first problem.

Heres the equations:

x^2 + y^2 =13
y = x+1

This is how far I've got, I would appreciate it if someone could show me where I'm going wrong.

• x^2 + (x+1)(x+1) = 13
• x^2 + (x^2+x+x+1) =13
• 2x^2 + 2x + 1 = 13
• 2x^2 + 2x -12 = 0
• x^2 + x -6 = 0

Now I'm stuck.
I've done this over and over and keep getting x^2=3 yet I know it should be x=2
• Jan 19th 2011, 02:15 AM
jacs
That looks good so far, now you need to factorise the quadratic to solve for x, you should get two solutions:
\$\displaystyle x^2+x-6=0\$
(x - 2)(x + 3) = 0
x = 2, x = -3
then from there substitute both in to find your respective y values
• Jan 19th 2011, 03:47 AM
sxsniper
Nice, of course it is! Thanks.

The answers should come out as x = 2 and y = -3

However if I put this in to the second equation y=x+1

Then y = 2+1 which is 3, not -3.
What's the deal with that?
I have a serious issue with +s and -s.

Are the answers I was given wrong?

If I put it in the first:
x^2 + y^2 = 13 so;
• 2^2 + -3^2 = 13
• 4 + -9 = -3 ...which isn't 13!

I know +9 makes 13, but WHY does -3^2 = 9?! as -3^3 = -27.
I think I'm missing some serious basic maths.
If I am, any suggestions on things I should practice or learn?

I was stuck straight in the top set at school, and was rarely taught the basics, just throw straight in to the difficult things.

EDIT:
OK now I understand a - * - = +

That still doesn't answer why "y = 2+1 is 3, not -3.
• Jan 19th 2011, 05:28 AM
HallsofIvy
Quote:

Originally Posted by sxsniper
Nice, of course it is! Thanks.

The answers should come out as x = 2 and y = -3

NO, x= 2 and y= -3 can't be the correct answer: it does not satisfy the second equation y= x+ 1.

I suspect you have misunderstood what is given as the two correct answers. From the quadratic equation you got, x= 2 or x= -3. Then from y= x+ 1, y= 2+1= 3 or y= -3+ 1= -2. The two answers to the problem are "x= 2 and y= 3" and "x= -3 and y= -2".

Quote:

However if I put this in to the second equation y=x+1

Then y = 2+1 which is 3, not -3.
What's the deal with that?
x= 2, y= -3 is NOT a correct answer!

[quote]I have a serious issue with +s and -s.

Are the answers I was given wrong?

If I put it in the first:
x^2 + y^2 = 13 so;
• 2^2 + -3^2 = 13
• 4 + -9 = -3 ...which isn't 13!

You have written the second equation incorrectly- all of y is squared:
\$\displaystyle 2^2+ (-3)^2= (2)(2)+ (-3)(-3)= 4+ 9= 13\$

Oh, by the say, 4- 9 is not -3, either, it is -5.

Quote:

I know +9 makes 13, but WHY does -3^2 = 9?! as -3^3 = -27
I think I'm missing some serious basic maths.
Again, it is NOT "-3^2"- the negative is as much part of y as the "3" is: (-3)^2= (-3)(-3)= +9.

Quote:

If I am, any suggestions on things I should practice or learn?

I was stuck straight in the top set at school, and was rarely taught the basics, just throw straight in to the difficult things.

EDIT:
OK now I understand a - * - = +

That still doesn't answer why "y = 2+1 is 3, not -3.
I hope you don't mean exactly what you are writing here. If you have 2 banana cream pies on the table and put one more banana cream pie on the table, you have three banana cream pies, not negative three!
• Jan 19th 2011, 07:03 AM
sxsniper
That's my point, on the answers page, it says -3
• Jan 19th 2011, 07:56 AM
sxsniper
Ahhh, you know I've been trying to solve this for about 8 hours flat now.

Bloody answers page must be wrong, I got 3 plenty of times! And was scratching my eye balls out trying to figure out WHY oh WHY am I not getting -3.

Thank yooou.
• Jan 19th 2011, 09:07 AM
Wilmer
Want some practice? Try:

a^2 + b^2 + c^2 = 142
b = a + 1
c = 2a - 1