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
KM
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