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

Solve 5x – 3 > 3x + 11

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