1. Simultaneous Equations

2. Unit 4:Mathematics Aims • Introduce Simultaneous Equations. Objectives • Recognise various ways to solve simultaneous equations.

3. Simultaneous Equations • Equations such as 2x + 3y = 8 contain 2 variables: x and y. To solve equations containing 2 variables, 2 equations are needed, e.g. 2x + 3y = 8 and 3x + 2y = 7 which need to be solved at the same time i.e. simultaneously. • There are two ways of solving simultaneous equations • by elimination • graphically

4. 1. By elimination Solve the simultaneous equations x + y = 6 (1) x – y = 2 (2) signs Different Add Same Subtract D AS S Substitute this x value into the easiest equation (usually when there are no negatives. In this case, equation 1). Hafaliad equation (1) + (2) 2x = 8 x = 4 y = 2

5. Solving Simultaneous Equations D A S S Eg 2x + y = 7 equation(1) x + y = 4 (2) Substitute into equation (2) (1) – (2) 3 + y = 4 (from both sides) x = 3 y=1 1) 3x – y = 1 -------(1) x + y = 3 -------(2) Substitute into equation (2) 1 + y = 3 (from both sides) (1) + (2) 4x = 4 y = 2 x = 1

6. Sometimes we need to eliminate the x terms instead of the y terms x + 3y = 7 -------(1) x – y = 1 -------(2) Signs are the Same so Subtract (1) – (2) 3y - - y = 3y + y = 4y and 7 + 1 = 8 4y = 8 Substitute into equation (1) y = 2 x + (3 x 2) = 7 x + 6 = 7 (-6) x = 1

7. Solve the equations 4x – y = 2 and 3x + 2y = 18 • 4x – y = 2 eqn 1 • 3x + 2y = 18 eqn 2 • In this case neither the x’s nor the y’s have the same number in front of them. This can be remedied by multiplying one equation or both equations by a scalar (number) or scalars. • 1 × 28x – 2y = 4 eqn 3 • 3x + 2y = 18 eqn 2 • 3 + 211x = 22 eqn 4

8. 3x + 4y = 18 eqn 1 • 4x – 3y = -1 eqn 2 • Eliminate y, so multiply eqn 1 by 3 and eqn 2 by 4 • 1 × 39x + 12y = 54 eqn 3 • 2 × 416x – 12y = - 4 eqn 4 • 3 + 425x = 50 eqn 5 •  x = 2 • Sub in 13(2) + 4y =18 • 4y = 18 – 6 = 12 •  y = 3 • Check 24(2) – 3(3) = 8 – 9 = -1 = RHS

9. Solving Simultaneous Equations Graphically • If the two lines represented by the two equations are plotted on a graph, then the co-ordinates of the point where the two lines cross are the solution to the equation. • Examples • 1. Solve x + 3y = 6 and 2x + y = 7 • For the line x + 3y = 6, • when x = 0, 3y = 6 hence y = 2 • when y = 0, x = 6 • Plot a line joining the points (0, 2) and (6, 0)

10. y 8 6 4 2 x 2 4 6 8 0 • For the line 2x + y = 7, • when x = 0, y = 7 and when y = 0, 2x = 7 i.e. x = 3.5 • Plot a line joining the points (0, 7) and (3.5, 0) • The point where the 2lines cross is (3, 1). • Therefore the solutionto the equations is x = 3 and y =1 • x + 3y = 3 + 3(1) • = 6 = RHS

11. Solve the equations 4x + 5y = 40 and x – y = 1 • For the equation 4x + 5y = 40, • when x = 0, 5y = 40  y = 8 • when y = 0, 4x = 40  x = 10 • Plot a line joining the points (0, 8) and (10, 0) • For the equation x – y = 1, • when x = 0, -y = 1  y = -1 • when y = 0, x = 1 • Plot a line joining the points (0, -1) and (1, 0)

12. y 8 6 4 2 x 2 8 6 4 0 10 -2 • The point where the 2 lines cross is (5, 4). • Therefore, the solution to the equations isx = 5 and y =4. • 4x + 5y = 4(5) + 5(4) • = 20 + 20 = 40

13. Solve the equations 3x – 2y = 0 and 3x + 4y = 18 • For the equation 3x – 2y = 0, • when x = 0, -2y = 0  y = 0 • when x = 2, -2y = -6  y = 3 • Plot a line joining the points (0, 0) and (2, 3) • For the equation 3x + 4y = 18, • when x = 0, 4y = 18  y = 4.5 • when y = 0, 3x = 18  x = 6 • Plot a line joining the points (0, 4.5) and (6, 0) Eryl Owen Jones

14. y 8 6 4 2 x 2 8 6 4 0 • The point where the 2 lines cross is (2, 3). • Therefore, the solution to the equations isx = 2 and y =3. • 3x – 2y = 3(2) – 2(3) • = 6 + 6 = 0