First, rearrange the first equation:
3y = x - 27
y = x/3 - 9
Square this equation to make y^2 the subject:
y^2 = (x/3 - 9)^2 = (x/3 - 9)(x/3 - 9) = (x^2)/9 - 3x - 3x + 81 = (x^2)/9 - 6x + 81
Now you can substitute this for y^2 in the second equation, then rearrange into the form ax^2 + bx + c = 0:
(x^2)/9 - 6x + 81 = 4x
(x^2)/9 - 10x + 81 = 0
x^2 - 90x + 729 = 0
Factorise the equation, then equate to zero and zolve:
(x - 9)(x - 81) = 0
x - 9 = 0 --> x = 9
x - 81 = 0 --> x = 81
Using these x values, find the corresponding y values:
y^2 = 4x ∴ y = √4x
When x = 9, y = √(4*9) = √36 = ±6
When x = 81, y = √(4*81) = √324 = ±18
Now we need to test whether each y co-ordinate is positive or negative:
When x = 9 and y = 6: x - 3y - 27 = 9 - 18 - 27 ≠ 0
When x = 9 and y = -6: x - 3y - 27 = 9 + 18 - 27 = 0
When x = 81 and y = 18: x - 3y - 27 = 81 - 54 - 27 = 0
When x = 81 and y = -18: x - 3y - 27 = 81 + 54 - 27 ≠ 0
Therefore, the co-ordinates of the points of intersection are (9, -6) and (81, 18)
