simultaneous equations

Simultaneous Equations

2

AimsAims• Introduce Simultaneous

Equations.

ObjectivesObjectives• Recognise various ways

to solve simultaneous equations.

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 equations1. by elimination2. graphically

x + y = 6 (1)x – y = 2 (2)

signs Different Add Same Subtract D A S S

2x = 8

x = 4

Solve the simultaneous equations

1. By elimination

Hafaliad equation (1) + (2)

Substitute this x value into the easiest equation (usually when there are no negatives. In this case, equation 1).

y = 2

Solving Simultaneous Equations D A S S

Eg 2x + y = 7 equation(1) x + y = 4 (2)

(1) – (2) x = 3

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

(1) + (2)

4x = 4

x = 1

Substitute into equation (2)

3 + y = 4 (from both sides)

y=1

Substitute into equation (2)

1 + y = 3 (from both sides)

y = 2

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

3y - - y = 3y + y = 4y and 7 + 1 = 8

4y = 8

y = 2

(1) – (2)

Substitute into equation (1)

x + (3 x 2) = 7

x + 6 = 7 (-6)

x = 1

– 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 × 2 8x – 2y = 4 eqn 3– 3x + 2y = 18 eqn 2– 3 + 2 11x = 22 eqn 4

3x + 4y = 18 eqn 1– 4x – 3y = -1 eqn 2– Eliminate y, so multiply eqn 1 by 3 and eqn 2 by 4– 1 × 3 9x + 12y = 54 eqn 3– 2 × 4 16x – 12y = - 4 eqn 4– 3 + 4 25x = 50 eqn 5– x = 2– Sub in 1 3(2) + 4y =18– 4y = 18 – 6 = 12– y = 3– Check 2 4(2) – 3(3) = 8 – 9 = -1 = RHS

• 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)

• 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

0 2

2

4

4

6

6

8

8

x

y

– 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)

– 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

0 2

2

4

4

6

6

8

8

10 x

y

-2

– 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

– 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

0 2

2

4

4

6

6

8

8

x

y