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GCSE: Further Simultaneous Equations Dr J Frost Last modified: 31 st August 2015.

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Presentation on theme: "GCSE: Further Simultaneous Equations Dr J Frost Last modified: 31 st August 2015."— Presentation transcript:

1 GCSE: Further Simultaneous Equations Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com Last modified: 31 st August 2015

2 Starter Solve the following simultaneous (linear) equations. 2x + 3y = 8 4x – y = -5 x = -0.5 y = 3 ?

3 RECAP :: Equation of a circle x y 5-5 5 The equation of this circle is: x 2 + y 2 = 25  The equation of a circle with centre at the origin and radius r is: x 2 + y 2 = r 2 ?

4 Quickfire Circles 1 1 x 2 + y 2 = 1 3 -3 3 x 2 + y 2 = 9 4 -4 4 x 2 + y 2 = 16 8 -8 8 x 2 + y 2 = 64 10 -10 10 -10 x 2 + y 2 = 100 6 -6 6 x 2 + y 2 = 36 ? ? ? ? ? ?

5 Motivation x y 10 -2 x – y = 2 x 2 + y 2 = 100 Given a circle and a line, we may wish to find the point(s) at which the circle and line intersect. How could we do this algebraically? STEP 1: Rearrange linear equation to make x or y the subject. x = y + 2 STEP 2: Substitute into quadratic and solve. (y + 2) 2 + y 2 = 100 y 2 + 4y + 4 + y 2 = 100 2y 2 + 4y – 96 = 0 y 2 + 2y – 48 = 0 (y + 8)(y – 6) = 0 y = -8 or y = 6 STEP 3: Use either equation to find the values of the other variable. When y = -8, x = -6 When y = 6, x = 8 ? ? ?

6 Your Go y = x 2 – 3x + 4 y – x = 1 STEP 1: Rearrange linear equation to make x or y the subject. y = 1 + x STEP 2: Substitute into quadratic and solve for one variable. 1 + x = x 2 – 3x + 4 x 2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x = 1 or x = 3 STEP 3: Use either equation to find the values of the other variable. When x = 1, y = 2 When x = 3, y = 4 ? STEP 4 (OPTIONAL): Check that your pairs of values work. 2 = 1 2 – (3 x 1) + 4  4 = 3 2 – (3 x 3) + 4  ? ? ?

7 Exercises y = x 2 + 7x – 2 y = 2x – 8 x 2 + y 2 = 8 y = x + 4 y = x 2 y = x + 2 x 2 + y 2 = 5 x – 2y = 5 y = x 2 – x – 2 x + 2y = 11 y = x 2 – 2x – 2 x = 2y + 1 x = -3, y = -14 x = -2, y = -12 x = -2, y = +2 x = -3, y = -14 x = -2, y = -12 x = 1, y = -2 x + y = 1 x 2 + y 2 = 1 x 2 – y 2 = 15 2x + 3y = 5 x 2 – y 2 = 15 2x + 3y = 5 y = x 2 – 3x x = y – 9 x 2 – 4y + 7 = 0 y 2 – 6z + 14 = 0 z 2 – 2x – 7 = 0 [Source: BMO] x = 2 – √ 13, y = 11 – √ 13 X = 2 + √ 13, y = 11 + √ 13 x = 3, y = 4 x = -5/2, y = 27/4 x = 1, y = 0 x = 0, y = 1 x = -8, y = 7 x = 4, y = -1 x = 1, y = 2, z = 3 (Add equations, then complete the squares – you’ll end up with a sum of squares which must each be 0) x = 5, y = -3 x = 191/59, y = -255/59 ? ? ? ? ? ? ? ? ? ? 1 2 3 4 5 7 8 9 10  6 x = 3, y = 1 x = -1/2, y = -3/4 ?


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