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GCSE: Further Simultaneous Equations

GCSE: Further Simultaneous Equations. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Last modified: 15 th December 2013. Starter. Solve the following simultaneous (linear) equations. 2x + 3y = 8 4x – y = -5. x = -0.5 y = 3 . ?. Recap: Equation of a circle. y.

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GCSE: Further Simultaneous Equations

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  1. GCSE:Further Simultaneous Equations Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 15th December 2013

  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 y The equation of this circle is: x2 + y2 = 25 ? 5 ! The equation of a circle with centre at the origin and radius r is: x2 + y2 = r2 x -5 5 -5

  4. Quickfire Circles ? 1 3 4 1 3 4 -1 -3 -4 -1 -3 -4 ? ? x2 + y2 = 1 x2 + y2 = 9 x2 + y2 = 16 ? ? 8 6 10 8 10 6 -8 -10 -6 -8 -10 -6 ? x2 + y2 = 64 x2 + y2 = 100 x2 + y2 = 36

  5. Motivation 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? y 10 ? STEP 1: Rearrange linear equation to make x or y the subject. x = y + 2 x – y = 2 x2 + y2 = 100 STEP 2: Substitute into quadratic and solve. (y + 2)2 + y2 = 100 y2 + 4y + 4 + y2 = 100 2y2 + 4y – 96 = 0 y2 + 2y – 48 = 0 (y + 8)(y – 6) = 0 y = -8 or y = 6 ? x 10 10 -2 10 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 = x2 – 3x + 4 y – x = 1 STEP 1: Rearrange linear equation to make x or y the subject. y = 1 + x STEP 3: Use either equation to find the values of the other variable. When x = 1, y = 2 When x = 3, y = 4 ? ? STEP 2: Substitute into quadratic and solve for one variable. 1 + x = x2 – 3x + 4 x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x = 1 or x = 3 STEP 4 (OPTIONAL): Check that your pairs of values work. 2 = 12 – (3 x 1) + 4 C 4 = 32 – (3 x 3) + 4 C ? ?

  7. Exercises y = x2 + 7x – 2 y = 2x – 8 x2 + y2 = 8 y = x + 4 y = x2 y = x + 2 x2 + y2 = 5 x – 2y = 5 y = x2 – x – 2 x + 2y = 11 y = x2 – 2x – 2 x = 2y + 1 ? x = -3, y = -14 x = -2, y = -12 1 x + y = 1 x2 + y2 = 1 x2 – y2 = 15 2x + 3y = 5 x2 – y2 = 15 2x + 3y = 5 y = x2 – 3x x = y – 9 x2 – 4y + 7 = 0 y2 – 6z + 14 = 0 z2 – 2x – 7 = 0 [Source: BMO] 7 ? x = 1, y = 0 x = 0, y = 1 2 ? x = -2, y = +2 x = -8, y = 7 x = 4, y = -1 ? 8 3 x = -3, y = -14 x = -2, y = -12 ? 9 x = 5, y = -3 x = 191/59, y = -255/59 ? 4 10 ? x = 1, y = -2 ? x = 2 – √13, y = 11 – √13 X = 2 + √13, y = 11 + √13 ? x = 3, y = 4 x = -5/2, y = 27/4 5 N ? 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 = 3, y = 1 x = -1/2, y = -3/4 6

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