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
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 $\displaystyle (5x-3)(x+3) = 0$
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