1. ## Simultaneous equations

Am I correct in thinking that the solutions of x and y in these simultaneous equations

y - 3x = 4
x^2 + y^2 = 34

are x = -1.5 and y = -0.5

2. Originally Posted by Natasha1
Am I correct in thinking that the solutions of x and -0.5+y in these simultaneous equations

y - 3x = 4
x^2 + y^2 = 34

are x = -1.5 and y = -0.5
No if you put them into the 2nd equation you get

$\displaystyle \displaystyle \left( -\frac{3}{2}\right)^2+\left( -\frac{1}{2}\right)^2=\frac{10}{4} \ne 34$

How did you get this solution?

3. No. Your solutions do not satisfy the second equation.

4. By sustitution

5. No, for if you put those values into the second equation you won't get an answer of 34. What was your working?

From eq1: $\displaystyle y = 4+3x$

You can sub that into eq2 to give a quadratic in x: $\displaystyle x^2+(4+3x)^2 = 34$

Eventually you can simplify that down to $\displaystyle (5x-3)(x+3) = 0$

6. I did this call equ1 -3x + y = 4 and equ2 x^2 + y^2 = 34

By squaring all of equ1 you get 9x^2 + y^2 = 16 call that equ3

Equ3 - equ2 is = to 9x^2 - x^2 + y^2 - Y^2 = 16 - 34 which simplifies to 8x^2 = -18 which brakes down to x = -3/2

when I sustitute x = -3/2 back into say equ1 i get y = -1/2

7. I get this

x^2 + (4 + 3x)^2 -34 = 0
x^2 + 9x^2 + 24x + 16 -34 = 0
10x^2 + 24x -18 = =0

8. Originally Posted by Natasha1
I did this call equ1 -3x + y = 4 and equ2 x^2 + y^2 = 34

By squaring all of equ1 you get 9x^2 + y^2 = 16 call that equ3. e^(i*pi): You don't get that you would get (-3x+y)(-3x+y) = 9x^2-6xy+y^2

Equ3 - equ2 is = to 9x^2 - x^2 + y^2 - Y^2 = 16 - 34 which simplifies to 8x^2 = -18 which brakes down to x = -3/2

when I sustitute x = -3/2 back into say equ1 i get y = -1/2
10x^2 + 24x -18 =0
I got that. If you divide through by 2 you'll find it factors nicely (although no reason why you can't use the formula)

9. I see, I just made it far too complicated. Thanks for the help, you are seriously clever guys!