Simultaneous Equations Unit 4 Mathematics Aims Introduce Simultaneous

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Simultaneous Equations

Simultaneous Equations

Unit 4: Mathematics Aims • Introduce Simultaneous Equations. Objectives • Recognise various ways to

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

 • Solving Simultaneous Equations Graphically – If the two lines represented by the

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

 • For the line 2 x + y = 7, • when x

• For the line 2 x + y = 7, • when x = 0, y = 7 and when y = 0, 2 x = 7 i. e. x = 3. 5 • Plot a line joining the points (0, 7) and (3. 5, 0) • The point where the 2 lines cross is (3, 1). • Therefore the solution to the equations is x = 3 and y =1 • x + 3 y = 3 + 3(1) • = 6 = RHS y 8 6 4 2 0 2 4 6 8 x

– Solve the equations 4 x + 5 y = 40 and x–y =

– Solve the equations 4 x + 5 y = 40 and x–y = 1 – – For the equation 4 x + 5 y = 40, when x = 0, 5 y = 40 y = 8 when y = 0, 4 x = 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)

– The point where the 2 lines cross is (5, 4). 8 – Therefore,

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

– Solve the equations 3 x – 2 y = 0 and 3 x

– Solve the equations 3 x – 2 y = 0 and 3 x + 4 y = 18 – – For the equation 3 x – 2 y = 0, when x = 0, -2 y = 0 when x = 2, -2 y = -6 y = 3 Plot a line joining the points (0, 0) and (2, 3) – – For the equation 3 x + 4 y = 18, when x = 0, 4 y = 18 y = 4. 5 when y = 0, 3 x = 18 x = 6 Plot a line joining the points (0, 4. 5) and (6, 0)

8 – The point where the 2 lines cross is (2, 3). – Therefore,

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

4 y 3 X - 2 y =1 2 x + 3 y =12

4 y 3 X - 2 y =1 2 x + 3 y =12 2 1 -4 -3 -2 -1 0 -1 -2 -3 -4 x 0 1 2 3 4

x+y=4 x–y=2 (1) (2) 4 y 3 2 1 -4 -3 -2 -1 0

x+y=4 x–y=2 (1) (2) 4 y 3 2 1 -4 -3 -2 -1 0 -1 -2 -3 -4 x 0 1 2 3 4

4 2 x + 3 y = 6 3 x + 2 y =

4 2 x + 3 y = 6 3 x + 2 y = 7 y 3 2 1 -4 -3 -2 -1 0 -1 -2 -3 -4 x 0 1 2 3 4