Using methods of substitution solve the following simultaneous equations: y - 2x - 1 = 0 and 4x^2 + y^2 - 25 = 0

(1) 4x² + y² - 25 = 0 (2) y - 2x - 1 = 0
Rearranging (2) into an equation of y in terms of x we have:y = 2x +1
now substituting into equation (1) 4x² + (2x +1)² - 25 = 0 and now expanding this equation we have,
4x² + (4x² + 4x + 1) - 25 = 0 Expanding out the brackets8x² +4x + 1 - 25 = 0 here we simplify the equation8x² + 4x - 24 = 0 dividing through by 4 gives us,2x² + x - 6 = 0
Now we factorise to find out our x values:(2x -3)(x+2)= 0Our x values are therefore x = -2 and x = 3/2
Now we substitute back into equation (2)our y values are then y = 4 and y = -3