Mr Kidman's Classroom
Revise the key ideas
Linear simultaneous equations
- Simultaneous equations must both be true for the same pair of values. Their solution is the intersection of their graphs.
Intersection of x plus y equals seven and x minus y equals one - Elimination adds or subtracts equations to remove a variable. If its coefficients are equal, subtract; if they are opposites, add.
- Worked example: x + y = 7 and x − y = 1. Add to get 2x = 8, so x = 4. Substitute into x + y = 7 to obtain y = 3.
- Worked example: 2x + 3y = 13 and 3x + 2y = 12. Multiply the first by 3 and the second by 2 to obtain 6x + 9y = 39 and 6x + 4y = 24. Subtract: 5y = 15, so y = 3 and x = 2.
- When multiplying an equation, multiply every term on both sides. Keep negative signs and the equals sign aligned when eliminating.
- Substitution is useful if one variable is already isolated. From y = 2x + 1 and x + y = 10, substitute: x + 2x + 1 = 10, giving x = 3 and y = 7.
- Check a candidate pair in both original equations. Satisfying one equation alone is not enough.
Modelling and graphical solutions
- Worked example: Two adult and three child tickets cost £31; three adult and two child tickets cost £34. Let a and c be the adult and child ticket prices in pounds. Set 2a + 3c = 31 and 3a + 2c = 34. Eliminating gives c = £5 and a = £8.
Multiply the first equation by 3 and the second by 2: 6a + 9c = 93 and 6a + 4c = 68. Subtracting gives 5c = 25, so c = 5. Then 2a + 15 = 31 gives a = 8.
- Graphical intersections give approximate solutions limited by scale. Parallel distinct lines have no solution; equations describing the same line have infinitely many pairs.
- Define the variables with units before modelling, and interpret the pair in the context rather than reporting bare x and y values.
Higher — linear and quadratic pairs
- Substitute the linear expression into the quadratic equation to obtain an equation in one variable. Solve it, then recover the other coordinate for each root.
- Worked example: y = x + 2 and y = x2. Set x2 = x + 2, giving (x − 2)(x + 1) = 0. The solutions are (2, 4) and (−1, 1).
Intersections of y equals x plus two and y equals x squared - For x2 + y2 = 25 and y = x + 1, substitute the whole bracket: x2 + (x + 1)2 = 25. Expand before solving; (x + 1)2 is not x2 + 1.
- A line can meet a parabola or circle twice, once or not at all. Keep corresponding x and y values paired; do not mix coordinates from different solutions.
