A5 Simultaneous equations

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A5 Simultaneous equations
KS4 Mathematics A5 Simultaneous equations

A5.1 Solving simultaneous equations graphically
Contents A5 Simultaneous equations A A5.1 Solving simultaneous equations graphically A A5.2 The elimination method A A5.3 The substitution method A A5.4 Simultaneous linear and quadratic equations A A5.5 Problems leading to simultaneous equations

Simultaneous equations
Equations in two unknowns have an infinite number of solution pairs. For example, x + y = 3 is true when x = and y = 2 x = and y = 0 x = –2 and y = and so on … 3 x y x + y = 3 We can represent the set of solutions on a graph: Ask pupils to suggest other pair of values that would solve this equation. Praise correct answers involving negative or fractional solutions. Verify that all of the suggested solutions lie on the line x + y = 3.

Simultaneous equations
Another equation in two unknowns will also have an infinite number of solution pairs. For example, y – x = 1 is true when x = 1 and y = 2 x = 3 and y = 4 x = –2 and y = –1 and so on … x y 3 y – x = 1 This set of solutions can also be represented in a graph: Ask pupils to suggest other pair of values that would solve this equation. Praise correct answers involving negative or fractional solutions.

Simultaneous equations
There is one pair of values that solves both these equations: x + y = 3 y – x = 1 We can find the pair of values by drawing the lines x + y = 3 and y – x = 1 on the same graph. x y 3 y – x = 1 The point where the two lines intersect gives us the solution to both equations. x + y = 3 This is the point (1, 2). At this point x = 1 and y = 2.

Simultaneous equations
x + y = 3 y – x = 1 are called a pair of simultaneous equations. The values of x and y that solve both equations are x = 1 and y = 2, as we found by drawing graphs. We can check this solution by substituting these values into the original equations. Tell pupils that simultaneous means ‘happening at the same time’. When two equations are described as simultaneous it means that we have to find values that solve both equations at the same time. 1 + 2 = 3 2 – 1 = 1 Both the equations are satisfied and so the solution is correct.

Solving simultaneous equations graphically
Use this activity to demonstrate the solutions to linear simultaneous equations graphically. Investigate examples where there are no solutions, that is, when the lines are parallel, and where there are infinite solutions, that is, when the lines coincide. Point out that when the solution are not whole numbers, it can be difficult to find the solution exactly from a graph. For these problems an algebraic method is required.

Simultaneous equations with no solutions
Sometimes pairs of simultaneous equations produce graphs that are parallel. Parallel lines never meet, and so there is no point of intersection. When two simultaneous equations produce graphs which are parallel there are no solutions. How can we tell whether the graphs of two lines are parallel without drawing them? Two lines are parallel if they have the same gradient.

Simultaneous equations with no solutions
We can find the gradient of the line given by a linear equation by rewriting it in the form y = mx + c. The value of the gradient is given by the value of m. Show that the simultaneous equations y – 2x = 3 2y = 4x + 1 have no solutions. Rearranging these equations in the form y = mx + c gives, y = 2x + 3 y = 2x + ½ The gradient m is 2 for both equations and so there are no solutions.

Simultaneous equations with infinite solutions
Sometimes pairs of simultaneous equations are represented by the same graph. For example, 2x + y = 3 6x + 3y = 9 Notice that each term in the second equation is 3 times the value of the corresponding term in the first equation. Both equations can be rearranged to give It is not always as obvious as it is here that two lines coincide. Stress that the only way to make sure is to rearrange both equations so that they are in the form y = mx + c. y = –2x + 3 When two simultaneous equations can be rearranged to give the same equation they have an infinite number of solutions.

Special solutions

A5.2 The elimination method
Contents A5 Simultaneous equations A A5.1 Solving simultaneous equations graphically A A5.2 The elimination method A A5.3 The substitution method A A5.4 Simultaneous linear and quadratic equations A A5.5 Problems leading to simultaneous equations

The elimination method
If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example, suppose 3x + y = 9 5x – y = 7 Adding these equations: The y terms have been eliminated. 3x + y = 9 Point out that the y terms have different signs in front of them. If we add the equations together they will be ‘eliminated’. 5x – y = 7 + 8x = 16 x = 2 divide both sides by 8:

The elimination method
Adding the two equations eliminated the y terms and gave us a single equation in x. Solving this equation gave us the solution x = 2. 3x + y = 9 5x – y = 7 To find the value of y when x = 2 substitute this value into one of the equations. Substituting x = 2 into the first equation gives us: 3 × 2 + y = 9 6 + y = 9 y = 3 subtract 6 from both sides:

The elimination method
We can check whether x = 2 and y = 3 solves both: 3x + y = 9 5x – y = 7 by substituting them into the second equation. 5 × 2 – 3 = 7 10 – 3 = 7 This is true, so we have confirmed that x = 2 y = 3 solves both equations.

The elimination method
Solve these equations: 3x + 7y = 22 3x + 4y = 10 Subtracting gives: The x terms have been eliminated. 3x + 7y = 22 3x + 4y = 10 3y = 12 divide both sides by 3: y = 4 Substituting y = 4 into the first equation gives us, The x terms have the same number in front of them, but this time the signs are the same. Subtracting the equations will eliminate the x terms. 3x + 7 × 4 = 22 3x + 28 = 22 subtract 28 from both sides: 3x = –6 x = –2 divide both sides by 3:

The elimination method
We can check whether x = –2 and y = 4 solves both, 3x + 7y = 22 3x + 4y = 10 by substituting them into the second equation. 3 × –2 + 7 × 4 = 22 – = 22 This is true and so, x = –2 y = 4 solves both equations.

The elimination method 1

The elimination method
Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example, 4x – y = 29 1 3x + 2y = 19 2 We need to have the same number in front of either the x or the y before adding or subtracting the equations. Call these equations 1 and 2 . 2 × 1 : 8x – 2y = 58 3 3x + 2y = 19 + : 11x = 77 x = 7 divide both sides by 11:

The elimination method
To find the value of y when x = 7 substitute this value into one of the equations, 4x – y = 29 1 3x + 2y = 19 2 Substituting x = 7 into 1 gives, 4 × 7 – y = 29 28 – y = 29 subtract 28 from both sides: –y = 1 y = –1 multiply both sides by –1: Check by substituting x = 7 and y = –1 into 2 , 3 × × –1 = 9 21 – 2 = 19

The elimination method
Solve: 2x – 5y = 25 3x + 4y = 3 1 2 Call these equations 1 and 2 . 3 × 1 6x – 15y = 75 3 2 × 2 6x + 8y = 6 4 3 – 4 , – 23y = 69 y = –3 divide both sides by –23: This slide shows a more difficult example where both equations have to be multiplied to make the coefficients of x the same. Discuss the alternative of multiplying equation 1 by 4 to make the coefficient of y –20. Multiplying the second equation by 5 would make the coefficient of y 20. These equations could then be added together to eliminate the terms containing y. Substitute y = –3 in 1 , 2x – 5 × –3 = 25 2x + 15 = 25 2x = 10 subtract 15 from both sides: x = 5 divide both sides by 2:

The elimination method 2

A5.3 The substitution method
Contents A5 Simultaneous equations A A5.1 Solving simultaneous equations graphically A A5.2 The elimination method A A5.3 The substitution method A A5.4 Simultaneous linear and quadratic equations A A5.5 Problems leading to simultaneous equations

The substitution method
Two simultaneous equations can also be solved by substituting one equation into the other. For example, y = 2x – 3 y = 2x – 3 1 2x + 3y = 23 2 Call these equations 1 and 2 . Substitute equation 1 into equation 2 . 2x + 3(2x – 3) = 23 expand the brackets: 2x + 6x – 9 = 23 simplify: 8x – 9 = 23 add 9 to both sides: 8x = 32 x = 4 divide both sides by 8:

The substitution method
To find the value of y when x = 4 substitute this value into one of the equations, y = 2x – 3 1 2x + 3y = 23 2 Substituting x = 4 into 1 gives y = 2 × 4 – 3 y = 5 Check by substituting x = 4 and y = 5 into 2 , 2 × × 5 = 23 = 23 This is true and so the solutions are correct.

The substitution method
How could the following pair of simultaneous equations be solved using substitution? 3x – y = 9 1 8x + 5y = 1 2 One of the equations needs to be arranged in the form x = … or y = … before it can be substituted into the other equation. Call these equations 1 and 2 . Rearrange equation 1 . 3x – y = 9 add y to both sides: 3x = 9 + y subtract 9 from both sides: 3x – 9 = y y = 3x – 9

The substitution method
3x – y = 9 1 8x + 5y = 1 2 Now substitute y = 3x – 9 into equation 2 . 8x + 5(3x – 9) = 1 expand the brackets: 8x + 15x – 45 = 1 simplify: 23x – 45 = 1 add 45 to both sides: 23x = 46 divide both sides by 23: x = 2 Substitute x = 2 into equation 1 to find the value of y. 3 × 2 – y = 9 6 – y = 9 –y = 3 y = –3

The substitution method
3x – y = 9 1 8x + 5y = 1 2 Check the solutions x = 2 and y = –3 by substituting them into equation 2 . 8 × × –3 = 1 16 – 15 = 1 This is true and so the solutions are correct. Solve these equations using the elimination method to see if you get the same solutions for x and y.

A5.4 Simultaneous linear and quadratic equations
Contents A5 Simultaneous equations A A5.1 Solving simultaneous equations graphically A A5.2 The elimination method A A5.3 The substitution method A A5.4 Simultaneous linear and quadratic equations A A5.5 Problems leading to simultaneous equations

When one of the equations in a pair of simultaneous equations is quadratic, we often end up with two pairs of solutions. For example, y = x2 + 1 1 y = x + 3 2 Substituting equation 1 into equation 2 , x2 + 1 = x + 3 We have to collect all the terms onto the left-hand side to give a quadratic equation of the form ax2 + bx + c = 0. 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. x2 – x – 2 = 0 factorize: (x + 1)(x – 2) = 0 x = – or x = 2

We can substitute these values of x into one of the equations y = x2 + 1 1 y = x + 3 2 to find the corresponding values of y. It is easiest to substitute into equation 2 because it is linear. When x = –1 we have, When x = 2 we have, You can check that these solutions also satisfy equation 1 by substituting them into the equation. y = –1 + 3 y = 2 + 3 y = 2 y = 5 The solutions are x = –1, y = 2 and x = 2, y = 5.

Using graphs to solve equations
We can also show the solutions to using a graph. y = x2 + 1 y = x + 3 –1 –2 –3 –4 1 2 3 4 6 8 10 y = x2 + 1 y = x + 3 (2, 5) The points where the two graphs intersect give the solution to the pair of simultaneous equations. (–1,2) 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. It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs, particularly when the solutions are not exact.

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.

Look at this pair of simultaneous equations: y = x + 1 1 x2 + y2 = 13 2 What shape is the graph given by x2 + y2 = 13? 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. We can also plot the graphs of the equations and observe where they intersect.

y = x + 1 1 x2 + y2 = 13 2 Substituting equation 1 into equation 2 , x2 + (x + 1)2 = 13 x2 + x2 + 2x + 1 = 13 expand the bracket: simplify: 2x2 + 2x + 1 = 13 2x2 + 2x – 12 = 0 subtract 13 from both sides: Discuss how we can quickly expand (x + 1)2. Use A1.2 Multiplying out brackets slides 35 and 36 if needed. x2 + x – 6 = 0 divide through by 2: factorize: (x + 3)(x – 2) = 0 x = – or x = 2

We can substitute these values of x into one of the equations y = x + 1 1 x2 + y2 = 13 2 to find the corresponding values of y. It is easiest to substitute into equation 1 because it is linear. When x = –3 we have, When x = 2 we have, Verify verbally that these solutions also satisfy equation 2 by substituting them into the equation. y = –3 + 1 y = 2 + 1 y = –2 y = 3 The solutions are x = –3, y = –2 and x = 2, y = 3.

Linear and circular graphs
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.

A5.5 Problems leading to simultaneous equations
Contents A5 Simultaneous equations A A5.1 Solving simultaneous equations graphically A A5.2 The elimination method A A5.3 The substitution method A A5.4 Simultaneous linear and quadratic equations A A5.5 Problems leading to simultaneous equations

Solving problems The sum of two numbers is 56 and the difference between the two numbers is 22. Find the two numbers. Let’s call the unknown numbers a and b. We can use the given 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

Solving problems Substituting a = 39 into the first equation gives,
subtract 39 from both sides: So the two numbers are 39 and 17. We can check these solutions by substituting them into the second equation, a – b = 22: 39 – 17 = 22 This is true and so our solution is correct.

Solving problems The cost of theatre tickets for 4 adults and 3 children is £ The cost for 2 adults and 6 children is £44. How much does each adult and child ticket cost? Let’s 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 1 2a + 6c = 44 2 Dividing equation 2 by 2 gives, a + 3c = 22 3 We can now subtract equation 3 from equation 1 to eliminate the terms containing c.

1 – 3 , Solving problems 4a + 3c = 47.50 – a + 3c = 22 3a = 25.50
1 – 3 , 3a = 25.50 a = 8.50 divide both sides by 3: Substitute a = 8.50 in 3 : c = 22 3c = 13.50 subtract 8.50 from both sides: Verify these solutions by substituting them back into the original problem. c = 4.50 divide both sides by 3: The cost of an adult’s ticket is £8.50 and the cost of a child’s ticket is £4.50.

Solving problems Remember, when using simultaneous equations to solve problems: 1) Decide what letters to use to represent each of the unknown values. 2) Use the information given in the problem to write down two equations in terms of the two unknown values. 3) Solve the simultaneous equations using the most appropriate method. 4) Check the values by substituting them back into the original problem.