Linear and quadratic equations When one equation in a pair of simultaneous equations is quadratic, we often end up with two pairs of solutions. For example, look at the equations: y = x2 + 1 and y = x + 3. Substituting equation into equation leaves you with: x2 + 1 = x + 3. A B We have to collect all the terms onto the left-hand side to give a quadratic equation of the form: ax2 + bx + c = 0. x2 – x – 2 = 0 Teacher notes Point out that we would get the same quadratic equation if we subtracted equations 1 and 2 to eliminate y. If the quadratic equation had not factorized then we would have had to solve it by completing the square or by using the quadratic formula. This slide is the first of 2 slides looking at linear and quadratic equations. The working continues on to the next slide. Factorizing the quadratic allows us to find the values of x. (x + 1)(x – 2) = 0 x = –1 or x = 2

Substituting into equations We can then substitute these values of x into one of the original equations: y = x2 + 1 or y = x + 3. To find the corresponding values of y it may be easier to substitute into the linear equation. When x = –1 we have: When x = 2 we have: y = –1 + 3 y = 2 + 3 Teacher notes You can check that these solutions also satisfy equation 1 by substituting them into the equation. y = 2 y = 5 The solutions for this set of simultaneous equations are: x = –1, y = 2 and x = 2, y = 5.

Using graphs to solve equations Teacher notes It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs. The points where the two graphs intersect give the solution to the pair of simultaneous equations. Explain that graphs are a good way to demonstrate a solution but are not used to solve linear and quadratic simultaneous equations exactly. For this we should use an algebraic method.

Linear and quadratic graphs Teacher notes Use this activity to demonstrate solutions to simultaneous equations where one equation is linear and the other is quadratic in x. Establish that most solutions do not involve whole numbers. Conclude that the graph is a good way of visualizing the solutions but that an algebraic method will give more exact answers. Investigate examples where there is only one pair of solutions and where there are no solutions.

Linear and circular equations Samira finds a different pair of simultaneous equations that do not follow the given rules: y = x + 1 and x2 + y2 = 13. What shape will the graph of x2 + y2 = 13 have? The graph of x2 + y2 = 13 is a circular graph with its centre at the origin and a radius of √13. We can solve this pair of simultaneous equations algebraically using substitution. Teacher notes This slide is the first of 3 slides taking students through linear and circular equations. The following 2 slides will take students through the working they should go through before arriving at the answer. We can also plot the graphs of the equations and observe where they intersect.

Linear and circular equations By substituting y = x + 1 into x² + y² = 13, we can find a solution. x2 + (x + 1)2 = 13 x2 + x2 + 2x + 1 = 13 Expand the bracket. 2x2 + 2x + 1 = 13 Simplify. 2x2 + 2x – 12 = 0 Subtract 13 from both sides. x2 + x – 6 = 0 Divide through by 2. Teacher notes Discuss how we can quickly expand (x + 1)2. (x + 3)(x – 2) = 0 Factorize. x = –3 or x = 2

Simultaneous equations By substituting our values of x into one of the original equations, we can find the corresponding y values. It is easiest to substitute into the equation y = x + 1 because this equation is linear. When x = –3 we have, When x = 2 we have, y = –3 + 1 y = 2 + 1 Teacher notes Verify verbally that these solutions also satisfy equation 2 by substituting them into the equation. y = –2 y = 3 The solutions for this set of simultaneous equations are: x = –3, y = –2 and x = 2, y = 3.

Linear and circular graphs Teacher notes Use this activity to demonstrate solutions to simultaneous equations where one equation is linear and the other is in the form x2 + y2 = r2. Start by demonstrating the solution to the example that was given in the previous slide: y – x = 1 x2 + y2 = 13 Establish, however, that most solutions do not involve whole numbers. Conclude that the graph is a good way of visualizing the solutions but that an algebraic method will give more exact answers. Investigate examples where there is only one pair of solutions (the line will be a tangent to the circle) and where there are no solutions.

Solving problems Alfie is trying to solve a maths puzzle he has found in a book. The puzzle has two clues that will enable it to be solved. The sum of two numbers is 56. The difference between the same two numbers is 22. Find the two numbers. Teacher notes While it would be possible to solve this problem using trial and error, students should be able to see that creating a pair of simultaneous equations is a far neater and simpler way of solving this problem. Using trial and error with these clues would be an educated estimate and it wouldn’t take too long to work out the correct answer, but the algebraic method is a more efficient method. Call the unknown numbers a and b. Use the information to write a pair of simultaneous equations in terms of a and b: a + b = 56 a – b = 22 Adding these equations gives: 2a = 78 a = 39 Substituting a = 39 into the first equation gives: 39 + b = 56 b = 17 We can check these solutions by substituting them into the second equation: a – b = 22 39 – 17 = 22 This is true and so the solution is correct. Alfie thinks the best method of solving the problem is by trial and error. Do you agree? Justify your answer. What are the two numbers?

Price juggling The Smith family and the Jones family are taking a trip to the circus. In the Smith family, there are 4 adults and 3 children. The total cost of their tickets is £47.50. In the Jones family, there are 2 adults and 6 children. Teacher notes Let us call the cost of an adult’s ticket a and the cost of a child’s ticket c. We can write: 4a + 3c = £47.50 – equation 1 2a + 6c = £44 – equation 2 8a + 6c = £95 – equation 3 6a = £51 – equation 3 minus equation 2 a = £8.50. The cost of an adult’s ticket is £8.50. Substitute the cost of an adult ticket into equation 2 to work out the cost of a child’s ticket: 2 × £8.50 + 6c = £44 6c = £27 c = £4.50. The cost of a child’s ticket is £4.50. Verify these solutions by substituting them back into the original problem. The total cost of their tickets is £44. How much does each adult ticket cost? How much does a children’s ticket cost? Show your working.

Practice makes perfect Teacher notes This activity can be used to give students some more practice of the elimination method of solving simultaneous equations.

Problem-solving hints Teacher notes This activity will act as good revision for how to tackle problems involving simultaneous equations and provide an appropriate mathematical solution.

More problems Teacher notes This activity will give the students plenty of chance to practice their simultaneous equations. Encourage students to use either the elimination method or the substitution method to achieve the correct solution. Get students to verify their solutions are correct by putting their answers back into the original equations.