Simultaneous equations

1 / 41

# Simultaneous equations - PowerPoint PPT Presentation

Sketching straight lines Solving simultaneous equations by straight line graphs Solving simultaneous equations by substitution Solving simultaneous equations by elimination Knowing two points which lie on a line find the equation of the line Using simultaneous equations to solve problems

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Simultaneous equations' - jules

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Sketching straight lines

• Solving simultaneous equations by straight line graphs
• Solving simultaneous equations by substitution
• Solving simultaneous equations by elimination
• Knowing two points which lie on a line find the equation of the line
• Using simultaneous equations to solve problems
• Simultaneous equation by substitution
• Problems from credit past papers
Simultaneous equations
Sketching straight lines

Equations of the type

y = x ; y = x + 3 ; y = 4x - 5; x + y = 2 are equations of straight lines . The general equation of a straight line is

y = ax =b

By finding the coordinates of some points which lie on the straight lines, plotting them and joining them up the graph of the straight line can be drawn.

Table of values

To draw the graph of a straight line we must find the coordinates of some points which lie on the line.We do this by forming a table of values . Give the x coordinate a value and find the corresponding y coordinate for several points

Make a table of values for the equation y = 2x + 1

1

2

3

0

3

7

1

5

So (0,1) (1,3) (2,5) and (3,7) all lie on the line with equation y=2x + 1

Now plot the points on a grid and join them up

.

y

Plot the points (0,1) (1,3) (2,5) and (3,7) on the grid

7

6

5

4

3

2

1

.

Now join then up to give a straight line

.

.

x

0 1 2 3 4

All the points on the line satisfy the equation y = 2x + 1

Sketching lines by finding where the lines cross the x axis and the y axis A quicker method

Straight lines cross the x axis when the value of y = o

Straight lines cross the y axis when the value of x =0

Sketch the line 2x + 3y = 6

Line crosses x axis when y = 0

Line crosses y axis when x = 0

2x + 0 =6

0 + 3y = 6

3y =6

2x =6

y = 2

x =3

at ( 3,0)

at ( 0,2)

Ex 2 page 124

y

7

6

5

4

3

2

1

Now join then up to give a straight line

.

.

x

0 1 2 3 4

All the points on the line satisfy the equation 2x + 3y = 6

Solving simultaneous equations by straight line graphs

Two lines either meet at a point intersect or they are parallel they never meet

To find where two lines meet draw the graphs of both lines on the same grid and read off the point of intersection

Find where the lines x + y = 5 and x – y =1 intersect

get two points that lie on each line or better still three

x + y = 5

0

1

5

x

(0,5) and (5,0) and (1,4)lie on the line x + y = 5

y

5

4

0

x – y =1

0

4

1

x

(0,-1) and (1,0) and ( 4,3) lie on the line

y

-1

3

0

Now plot the points and find where the two lines meet

y

7

6

5

4

3

2

1

0

-1

.

.

x + y =1

.

(3,2)

X

x + y =5

.

.

x

.

1 2 3 4 5

The lines intersect at (3,2)

Solving simultaneous equations algebraically

Whoopee no more drawing graphs

But you will have to learn a strategy to solve the equations

Yikes!

Elimination method

Solve the simultaneous equations algebraically

HOW DO I SOLVE TWO EQUATIONS WITH TWO UNKNOWNS ?

x + y = 8

x - y = 4

2x = 12

This gives one equation with one unknown which is easy to solve

x = 6

How do I find y. It is gone.It has been eliminated ?

SUBSTITUTE THE VALUE OF X INTO ONE OF YOUR EQUATIONS

x = 6

x + y = 8

Now I have the solution

6 + y = 8

y = 2

Am I correct?

Check by putting the values of x and y into , the other equation, x-y,and see if you get 4

6 – 2 = 4 

Solution is x =6 and y = 2

Solve the simultaneous equations

x + 3y = 7

x - 3y = -5

Remember to check

2x = 2

x = 1

Substitute x =1 into x + 3y = 7

1 + 3y = 7

3y =6

1 – 3 X 2

y = 2

= 1 -6

Solution is x = 1 and y = 2

= -5 

Solve the simultaneous equations

If I add I get another equation in x and y

3x + y = 8

x + y = 4

3x + 2=12

no use

3x + y = 8

-x -y = -4

Multiply one of the equations by ( -1)

2x = 4

x = 2

It changes the sign of everything

SUBSTITUTE Put x=2 into x + y =4

x + y = 4 2 + y = 4

3X2 + 2 = 8 

y = 2

check

Solution x = 2 and y = 2

Slightly more difficult

But not for you IF you have learned the work so far

Solve the simultaneous equations

Adding gives 5x + 5y = 55 no use

3x + 2y = 29

2x + 3y = 26

Multiplying by a negative does not eliminate any of the letters

Could I multiply and then add?

That ‘s good let’s try it but be careful

Remember when we add numbers together which are the negatives of each other they are eliminated

Solve the following simultaneous equations algebraically

3x + 2y = 292x + 3y = 26

Multiply the equations by two suitable numbers so that the coeficients of x or y are the negatives of each other

We choose y to be eliminated

X 3

3x + 2y = 292x + 3y = 26

X -2

Giving two new equivalent equations

9x +6y = 87

-4x + -6y = -52

9x +6y = 87

-4x + -6y = -52

Remember practice makes perfect

5x = 35

x = 7

check

substitute

x = 7 into 9x+6y = 87

-4 X 7 - 6X 4

= -28 –24

= -52 

9 X 7 + 6y = 87

63 +6y = 87

6y = 24

y = 4

Ex 7B p 139

Solution x = 7 and y = 4

Problem Solving With Simultaneous Equations

Example : The problem

For the cinema

2 adults’ tickets and 5 children’s tickets cost £26

4 adults’ tickets and 2 children tickets cost £28

Find the cost of each kind of ticket.

Introduce Letters

Let the cost of adult ticket = £A

Let the cost of children’s ticket = £C

Write the equations

Solve the Problem using simultaneous equations

Conclusion

Children's tickets cost £3

strategy
• Introduce two letters
• Write two equations that describe the information
• Solve the problem using simultaneous equations
• check
Substitution method

This an excellent method of solving simultaneous equations when the equations are given to you in a certain form

Solve the equations y = 2x and y = x + 10

Example 1

We take the equation y = x + 10

And substitute 2x for y

2x = x + 10

Take out y and

replace it with 2x

x = 5

Sub x=5 into y=2x

y = 10

Now solve the equation

Solution x = 5 y = 10

Example 2

Solve the equations y = 2x –8 and y – x = 1

Sub y =2x-8 in the equation

y - x = 1

2x –8 -x = 1

x – 8 = 1

x = 9

y = 18-8

Sub x = 9 in the equation y = 2x -8

y = 10

Solution x = 9 and y =10

Check 10 –9 =1

Problems with simultaneous equations from past papers
• 1. The tickets for a sports club cost £2 for members and £3 for non- members
• The total ticket money collected was £580
• x tickets were sold to members and y tickets were sold to non-members.
• Use this information to write down an equation involving x and y

2x + 3y =580

b) 250 people bought raffle tickets for the disco. Write down another equation involving x and y.

x + y = 250

c ) How many tickets were sold to members ?

We now have two equations in x and y so let’s solve them simultaneously

2x + 3y = 580

x + y = 250

2x + 3y = 580

x + y = 250

X 1

X (-2)

2x + 3y = 580

-2x + (-2 )y = -500

y = 80

Sub y = 80 into x + y = 250

check

x + 80 =250

2 x170 + 3x80 =340 + 240 =580

x = 170

170 tickets were sold to members

2. Alloys are made by mixing metals.Two different alloys are made using iron and lead. To make the first alloy, 3 cubic cms of iron and 4 cubic cms of lead are used. This alloy weighs 65 grams

a ) Let x grams be the weight of 1 cm3 f iron and y grams be the weight of 1 cm3 of lead Write down an equation in x and y which satisfies the above equation.

3x + 4 y = 65

To make the second alloy, 5 cm3 of iron and 7cm3 of lead are used . This alloy weighs 112 grams.

b ) Write down a second equation in x and y which satisfies this condition

5x + 7y = 112

X (-5)

3x + 4 y = 65 5x + 7y = 112

X 3

Check 5x7 + 7x11

= 35 + 77= 112

-15x + -20 y = -325

15x + 21y = 336

y = 11

y = 11 into 3x + 4y = 65

SUB

3x + 44 = 65

1cm3 of iron weighs 7gms and 1cm3 of lead weighs 11 gms

3x = 21

x = 7

3. A rectangular window has length , lcm and breadth, b cm .

A security grid is made to fit this window. The grid as 5 horizontal wires and 8 vertical wires

a ) The perimeter of the window is 260 cm. Use this information to write down an equation involving l and b .

2 l + 2 b = 260

b) In total, 770 cm of wire is used. Write down another equation involving l and b

5 l + 8 b = 770

Find the length and breadth of the window

2 l + 2 b = 260

5 l + 8 b = 770

X (4)

X (-1)

8 l + 8 b = 1040

-5 l + (-8) b = -770

3 l = 270 l = 90

5x90 +8x40 = 450 + 320=770

l = 90into 2 l + 2 b = 260

SUB

180 + 2b = 260

Length is 90cm and the breadth is 40 cm

2b = 80

b = 40

4. A number tower is built from bricks as shown in fig 1.The number on the brick above is always equal to the sum of the two numbers below.

fig1

9

12

-3

8

4

-7

3

5

-1

-6

Show that p + 3q = 10

-3

p +2q -5

q-8

p+q

q -5

-3

Fig 3

p

q

-5

2

Adding the numbers in the second row to equal the top row

p + 2q –5 + q -8 = -3

P + 3q – 13 = -3

simplifying

P + 3q = 10

Use fig 4 to write down a second equation in p and q

14

2q+1

8-p

fig 4

2q-2

3

5-p

2q

-2

5

-p

Adding the numbers in the second row to equal the top row

2q+1+8-p =14

2q +9 – p = 14

simplifying

2q – p = 5

d) Find the values of p and q

3q + P = 10

2q – p = 5

Change the terms around q under q and p under p

Remember to check

3q + P = 10

2q - p = 5

Check 3x3 +1 = 10

5q = 15

q = 3

SUB

q = 3 into 2q –p =5

p = 1 and q = 3

6 –p = 5

p=1

5. A sequence of numbers is 1 ,5 ,12 ,22 ,………Numbers from this sequence can be illustrated in the following way using dots

First number ( N = 1)

• •

• •

Second number (N=2)

• • •

• •

• • • •

• •

• •

Third number (N=3)

• •

• •

• • •

• •

• •

• •

• •

• •

• •

Fourth number (N=4)

a )Write the fifth number in the pattern

Form a table of values

When N=2 D = 5

When N= 1 D = 1

b) The number of dots needed to illustrate the nth number in this sequence is given by the formula

D =aN² - bN

Find the values of a and b

N = 1 D = 1 When N=2 D = 5

D =aN² - bN

Form two equations by substituting the values of n and D into the equation D =aN² - bN

N = 1 D = 1

1 = a - b

5 = 4a -2b

N=2 D = 5

Now solve simultaneously

X (-2)

1 = a - b

X1

5 = 4a -2b

-2 = -2a + 2b

5 = 4a - 2b

Check

4x3/2-2x1/2

=12/2-2/2 =10/2=5