# Solve the equations giving your answer in 2d.p (5 Marks).: x^2 + y^2 = 36 , x = 2y + 6

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x2 + y2 = 36       (1)

x = 2y + 6          (2)

You can solve simulateous equations in three ways: Substitution, addition or subtraction. If one equation contains both x2 and y2 then you should use substitution as with this example.

1.) Substitute equation (2) into equation (1):

x = 2y + 6   therefore x2 = (2y + 6)2   so

(2y + 6)2 + y2 = 36

1 mark

2.) Expand the brackets:

(2y + 6)(2y + 6) + y2 = 36

4y2 + (2y x 6) + (2y x 6) + 36 + y2 = 36

4y2 + 12y + 12y + 36 + y2 = 36

1 mark

3.) Simplify the equation:

4y2 + 24y + 36 + y2 = 36

1 mark

4.) Subtract 36 from both sides and add the 4y2 and y2:

5y2 + 24y = 0

This is now a quadratic equation. It can either be solved using the quadratic formula or through factorising. I shall do the factorising method.

5.) Take out y from both 5y2 and 24y:

y(5y + 24) = 0            y = (y + 0)

1 mark

6.) Find the values of y that would make inside the bracket become zero:

(y + 0) = 0                            (5y + 24) = 0

y = 0                                    5y = -24

y = -24/5 = -4.8

7.) Substitute the y values into the equation to find the corresponding x values:

y = 0                                     y = -4.80

x = (2 x 0) + 6                       x = (2 x (-4.8)) +6

x = 6.00                                x = (-9.6) + 6

x = -3.60

1 mark

PS - Don't forget the 2 decimal places at the end

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