1. ## 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

2. 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

3. 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.

4. 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".

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.

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.

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!

5. That's my point, on the answers page, it says -3

6. 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.

7. Want some practice? Try:

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