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Simultaneous equations

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Applying algebraic skills to linear equations

I can…

Simultaneous equations

…solve simultaneous equations graphically

…solve simultaneous equations algebraically by substitution

…solve simultaneous equations algebraically by elimination

…use context to create simultaneous equations

An introduction

Reminders

The equation can be represented by a straight line

which has a of m and passes through the point

y = mx + c

(0, c)

gradient

Systems of Equations

A System of Equations consists of two (or more) equations with at least two variables.

These are also referred to as as their solution holds true

for both equations.

Simultaneous Equations

When the System consist of two equations, with two variables, there are three

methods of finding the solution: -

- * by drawing graphs* by substitution* by elimination

…solve simultaneous equations graphically

If the lines representing the equations are drawn then the solution is the coordinates of the point where the lines intersect (meet).

Drawing straight lines

Either:

- set x = 0, find the y-coordinate from the formula, then set y = 0 and find x

- pick 2 values for x and find the corresponding values of y

- use the y-intercept, gradient and y = mx + c

Example

Solve these equations simultaneously

y= ½x + 1 & y = 7 – x

Line two - y = 7 – x

intercept =

m =

(0, 7)

Line one - y = ½x + 1

When x = 0, y = → (0, 1)

When x = 2, y = → (2, 2)

-1

½× 0 + 1 = 1

½× 2 + 1 = 2

Draw the two lines with the given information

Lines intersect at so solution is x =

y =

4

(4, 3)

3

…solve simultaneous equations graphically (continued)

Simultaneous equations are often used to solve

problems and use letters other than x & y.

Solving simultaneous equations graphically will

often only give approximate solutions and relies on

accurate drawing of graphs.

In order to get precise solutions it is better to

use one of the other methods – substitution or

elimination.

…solve simultaneous equations algebraically by substitution

the same

At the point where the lines meet, the values of x and y are in both equations.

This allows the first equation to be substituted into the second.

Examples

1) y= x + 1

y= 4x – 5

2) y= 4x + 1

2y– 5x + 4 = 0

Replace the yin the second equation with the first

= 4x – 5

(4x + 1)

x + 1

2 – 5x + 4 = 0

Make x the subject of the formula

1 = 3x – 5

6 = 3x

x = 2

8x + 2 – 5x + 4 = 0

3x + 6 = 0

3x = –6x = –2

Substitute x into the first equation to find y

x = -2,

y =

x = 2, y =

2 + 1 = 3

4 × (–2) + 1 = – 7

the solution is

(2, 3)

Check by substituting into the second equation, if it is true the solution is correct

the solution is

(–2, –7)

3 = 4 × 2 – 5

2 x (-7) - 5 x (-2) + 4 = 0

…solve simultaneous equations algebraically by substitution

Sometimes it is necessary to rearrange one of the equations first.

Example

3) y– 2x = 3

3y– 2x = 17

Rearrange the first equation

y=

2x + 3

3 – 2x = 17

(2x + 3)

Replace the yin the second equation with the first

6x+ 9 – 2x = 17

4x+ 9 = 17

4x= 8

x= 2

Make x the subject of the formula

x = 2,

y – 2 × 2 = 3

y – 4 = 3

y = 7

Substitute x into the first equation to find y

the solution is

(2, 7)

Check by substituting into the second equation, if it is true the solution is correct

3 x 7 – 2 x 2 = 17

…solve simultaneous equations algebraically by elimination

In this method the equations are added or subtracted so that one of the variables will be eliminated.

Examples

1) 2x + 3y = 35 7x– 3y = 1

Place the letters in the same order

2) 3x + 2y = 75x+ 2y = 13

3x + 2y – 5x – 2y = –2x

7 - 13 = – 6

So -2x = -6

Add/subtract to remove a letter

2x + 3y + 7x + (-3y)

= 9x

35 + 1 = 36

So 9x = 36

Solve for the remaining letter

x = 4

x = 3

× 3

Substitute into the first equation to find y

2 + 3y = 35

x 4

3 + 2y = 7

2y = –2y = –1

3y = 27y = 9

(4, 9)

the solution is

the solution is

(3, –1)

…solve simultaneous equations algebraically by elimination

Sometimes it is necessary to multiply one or both of the equations before adding or subtracting.

Examples

3) 3x+ 4y = 266x – y = 7

4) 6x+ 2y = 382x – 3y = 20

Place the letters in the same order

3(2x– 3y)= 3 x 20

4(6x – y) = 4 x 7

Multiply as required

6x– 9y= 60

24x – 4y = 28

Add/subtract to remove a letter

6x + 2y - 6x – (-9y)

= 2y + 9y = 11y

38 – 60 = -22

So 11y = -22

3x + 4y+ 24x + (-4y)

= 27x

26 + 28 = 54

So 27x = 54

Solve for the remaining letter

x= 2

y = -2

x 2

6x + 2 = 38

× (–2)

3 + 4y = 26

Substitute into the first equation to find y

4y = 20y= 5

6x = 42x= 7

the solution is

(7, –2)

the solution is

(2, 5)

…solve simultaneous equations algebraically by elimination

Example

5) 3x+ 2y = 74x + 3y = 9

Place the letters in the same order

Multiply as required

3(3x+ 2y) = 3 x 7

2(4x+ 3y) = 2 x 9

9x + 6y = 21

8x+ 6y = 18

Add/subtract to remove a letter

9x + 6y- 8x – 6y = x

21 – 18 = 3

So x = 3

Substitute into the first equation to find y

3 + 2y = 7

× 3

2y = –2y= –1

the solution is

(3, –1)

…use context to create simultaneous equations

Examples

1) A jug and two glasses hold 1·6 litres altogether.

Two jugs and three glasses hold 2·9 litres altogether.

How much does each hold?

Some problems can be solved using simultaneous equations by turning the problem into a set of equations.

If answering a question set in a particular context you must write your final answer in context.

j + 2g = 1·6

2j+ 3g = 2·9

Choose appropriate letters

2(j+ 2g)= 2 x 1.6

Multiply as required

2j + 4g = 3·2

Add/subtract to remove a letter

2j + 4g – 2j – 3g = g

3.2 – 2.9 = 0·3

So g = 0.3

Substitute into the first equation to find y

0·3

j+ 2 × = 1·6

j+ 0.6 = 1.6

j = 1

So a Jug holds 1 litre and a glass holds 0·3 litres

Applying algebraic skills to linear equations

I can…

Simultaneous equations

…solve simultaneous equations graphically

…solve simultaneous equations algebraically by substitution

…solve simultaneous equations algebraically by elimination

…use context to create simultaneous equations

An introduction

Reminders

The equation can be represented by a straight line

which has a of m and passes through the point

Systems of Equations

A System of Equations consists of two (or more) equations with at least two variables.

These are also referred to as as their solution holds true

for both equations.

When the System consist of two equations, with two variables, there are three

methods of finding the solution: -

- * by drawing graphs* by substitution* by elimination

…solve simultaneous equations graphically

If the lines representing the equations are drawn then the solution is the coordinates of the point where the lines intersect (meet).

Drawing straight lines

Either:

- set x = 0, find the y-coordinate from the formula, then set y = 0 and find x

- pick 2 values for x and find the corresponding values of y

- use the y-intercept, gradient and y = mx + c

Example

Solve these equations simultaneously

y= ½x + 1 & y = 7 – x

Line two - y = 7 – x

intercept =

m =

Line one - y = ½x + 1

When x = 0, y = → (0, 1)

When x = 2, y = → (2, 2)

Draw the two lines with the given information

Lines intersect at so solution is x =

y =

…solve simultaneous equations graphically (continued)

Simultaneous equations are often used to solve

problems and use letters other than x & y.

Solving simultaneous equations graphically will

often only give approximate solutions and relies on

accurate drawing of graphs.

In order to get precise solutions it is better to

use one of the other methods – substitution or

elimination.

…solve simultaneous equations algebraically by substitution

At the point where the lines meet, the values of x and y are in both equations.

This allows the first equation to be substituted into the second.

Examples

1) y= x + 1

y= 4x – 5

2) y= 4x + 1

2y– 5x + 4 = 0

Replace the yin the second equation with the first

= 4x – 5

2 – 5x + 4 = 0

Make x the subject of the formula

Substitute x into the first equation to find y

x = -2,

y =

x = 2, y =

the solution is

Check by substituting into the second equation, if it is true the solution is correct

the solution is

…solve simultaneous equations algebraically by substitution

Sometimes it is necessary to rearrange one of the equations first.

Example

3) y– 2x = 3

3y– 2x = 17

Rearrange the first equation

y=

3 – 2x = 17

Replace the yin the second equation with the first

Make x the subject of the formula

x = 2,

Substitute x into the first equation to find y

the solution is

Check by substituting into the second equation, if it is true the solution is correct

…solve simultaneous equations algebraically by elimination

In this method the equations are added or subtracted so that one of the variables will be eliminated.

Examples

1) 2x + 3y = 35 7x– 3y = 1

Place the letters in the same order

2) 3x + 2y = 75x+ 2y = 13

Add/subtract to remove a letter

Solve for the remaining letter

Substitute into the first equation to find y

2 + 3y = 35

3 + 2y = 7

the solution is

the solution is

…solve simultaneous equations algebraically by elimination

Sometimes it is necessary to multiply one or both of the equations before adding or subtracting.

Examples

3) 3x+ 4y = 266x – y = 7

4) 6x+ 2y = 382x – 3y = 20

Place the letters in the same order

3(2x– 3y)= 3 x 20

4(6x – y) = 4 x 7

Multiply as required

Add/subtract to remove a letter

Solve for the remaining letter

6x + 2 = 38

3 + 4y = 26

Substitute into the first equation to find y

the solution is

the solution is

…solve simultaneous equations algebraically by elimination

Example

5) 3x+ 2y = 74x + 3y = 9

Place the letters in the same order

Multiply as required

3(3x+ 2y) = 3 x 7

2(4x+ 3y) = 2 x 9

Add/subtract to remove a letter

Substitute into the first equation to find y

3 + 2y = 7

the solution is

…use context to create simultaneous equations

Examples

1) A jug and two glasses hold 1·6 litres altogether.

Two jugs and three glasses hold 2·9 litres altogether.

How much does each hold?

Some problems can be solved using simultaneous equations by turning the problem into a set of equations.

If answering a question set in a particular context you must write your final answer in context.

Choose appropriate letters

2(j+ 2g)= 2 x 1.6

Multiply as required

Add/subtract to remove a letter

Substitute into the first equation to find y

j+ 2 × = 1·6

So a Jug holds 1 litre and a glass holds 0·3 litres