Download Simultaneous equations

Simultaneous •Sketching straight lines
equations •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
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
x
y=2x+1
0
1
1
2
3
5
So (0,1) (1,3) (2,5) and
(3,7) all lie on the line
with equation y=2x + 1
3
7
Now plot the points on a grid
and join them up
.
y
7
6
.
5
4
1
Now join
then up to
give a
straight line
.
3
2
Plot the points (0,1) (1,3)
(2,5) and (3,7) on the grid
.
0
x
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
Plot 0,2) and (3,0) and join them
up with a straight line
y
7
6
5
Now join
then up to
give a
straight line
4
3
2
.
1
0
1
2
.
3
x
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
x
0
1
5
y
5
4
0
(0,5) and (5,0) and (1,4)lie on
the line x + y = 5
(0,-1) and (1,0) and ( 4,3) lie on
the line
x – y =1
x
0
4
1
y
-1
3
0
Now plot the points and find
where the two lines meet
y
7
6
5
.
4
.
x + y =1
3
(3,2)
2
X
1
0
-1
.
.
1
2
3
.
x + y =5
4
.
5
The lines
intersect at
(3,2)
x
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
ADD
x= 6
How do I
find y. It is
gone.It has
been
ADD the
two
equations
together
This gives one
equation with
one unknown
which is easy to
solve
SUBSTITUTE THE VALUE
OF X INTO ONE OF YOUR
EQUATIONS
x = 6
x +y = 8
6+y = 8
y = 2
Check by putting
the values of x
and y into , the
other equation,
x-y,and see if
you get 4
Now I have
the solution
Am I
correct?
6–2=4

Solution is x =6
and y = 2
Solve the simultaneous equations
x + 3y = 7
x - 3y = -5
ADD
Substitute x =1 into
x + 3y = 7
Remember
to check
2x = 2
x = 1
1 + 3y = 7
3y =6
y = 2
Solution is x = 1 and y = 2
1– 3X2
= 1 -6
= -5 
Solve the simultaneous equations
If I add I
get
another
equation
in x and y
3x + 2=12
no use
3x + y = 8
x + y = 4
3x + y = 8
-x -y = -4
ADD
2x = 4
x = 2
SUBSTITUTE
Put x=2 into
x + y =4
x+y=4
2 +y=4
y= 2
Solution x = 2 and y = 2
Multiply
one of the
equations
by ( -1)
3X2 + 2 = 8 
check
It changes
the sign
of
everythin
g
Using simultaneous equations to solve
problems
example
Slightly more difficult
But not for you IF
you have learned the
work so far
Solve the simultaneous equations
3x + 2y = 29
2x + 3y = 26
Multiplying by a
negative does not
eliminate any of
the letters
Adding
gives 5x +
5y = 55 no
use
That ‘s good
let’s try it but
be careful
Could I
multiply
and then
add?
Remember when we add
numbers together which are
the negatives of each other
they are eliminated
Solve the following simultaneous equations
algebraically
Multiply the equations by two
suitable numbers so that the
coeficients of x or y are the
negatives of each other
3x + 2y = 29
2x + 3y = 26
3x + 2y = 29
2x + 3y = 26
We choose y to
be eliminated
X3
X -2
Giving two new
equivalent equations
9x +6y = 87
-4x + -6y = -52
Now add
9x +6y = 87
Remember
practice
makes
perfect
-4x + -6y = -52
ADD
substitute
5x = 35
x =7
x = 7 into
9x+6y = 87
check
-4 X 7 - 6X 4
9 X 7 + 6y = 87
63 +6y
= 87
6y
= 24
= -28 –24
= -52

y =4
Solution x = 7 and y = 4
Ex 7B p 139
Using simultaneous equation to
solve problems
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
2 A  5C  26 .......(1)
4 A  2C  28 .......(2)
Solve the Problem using simultaneous equations
2 A  5C  26 . X2
4 A  2C  28 X ( - 1)
4 A  10C  52 .
- 4 A  (- 2C )  28
ADD
8C  24
C 3
Substitute C  3 into (1)
2 A  5  3  26
2 A  15  26
2 A  11
A  5 5
Conclusion
Adult Ticket costs £5.50
Children's tickets cost £3
strategy
1. Read the problem
2. Introduce two letters
3. Write two equations that describe the information
4. Solve the problem using simultaneous equations
5. check
Substitution method
This an excellent method of solving simultaneous equations
when the equations are given to you in a certain form
Example 1 Solve the equations y = 2x and y = x + 10
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
Sub x = 9 in the
equation y = 2x -8
Solution x = 9 and y =10
y = 18-8
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
a) 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
x
= 580
+ y = 250
2x + 3y = 580
X 1
x
X (-2)
+ y
ADD
Sub
= 250
2x
+ 3y
= 580
-2x
+ (-2 )y
= -500
y = 80
y = 80 into x + y = 250
x + 80 =250
x = 170
170 tickets were sold to
members
check
2 x170 + 3x80
=340 + 240
=580
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
C ) Find the weight of 1 cm3 of iron
and the weight of 1cm3 of lead.
X (-5)
X 3
3x + 4 y = 65
5x + 7y = 112
Check
5x7 + 7x11
-15x + -20 y = -325
ADD
15x + 21y = 336
y
SUB
= 35 + 77= 112
= 11
y = 11 into 3x + 4y = 65
3x + 44 = 65
3x = 21
x =7
1cm3 of iron weighs
7gms and 1cm3 of lead
weighs 11 gms
3. A rectangular window has length , l cm 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
8l +8b
ADD
X (4)
X (-1)
= 1040
-5 l + (-8) b = -770
3 l = 270
l = 90
SUB
l = 90 into 2 l + 2 b = 260
180 + 2b = 260
2b = 80
b = 40
5x90 +8x40 =
450 + 320=770
Length is 90cm and
the breadth is 40 cm
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
8
3
-3
4
5
-7
-1
-6
Find the number on the shaded brick in fig 2
34
Fig 2
16
5
-2
18
11
7
7
4
-11
34
In fig 3, two of the numbers on the bricks are represented by p
and q
-3
Show that p + 3q = 10
Fig 3
Adding the numbers in the
second row to equal the top row
simplifying
p +2q -5 q-8
p+q
q -5
-3
p
q
-5
p + 2q –5 + q -8 = -3
P + 3q – 13 = -3
P + 3q
=
10
2
Use fig 4 to write down a second equation in p and q
14
fig 4
2q+1
2q-2 3
2q
Adding the numbers in the
second row to equal the top row
simplifying
-2
8-p
5-p
5
-p
2q+1+8-p =14
2q +9 – p = 14
2q – p
d) Find the values of p and q
=5
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
ADD
SUB
5q
= 15
q
=
3
q = 3 into 2q –p =5
6 –p = 5
p=1
Check 3x3 +1 = 10
p = 1 and q = 3
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)
•
•
•
•
• •
•
•
• • • •
•
•
•
•
•
Fourth number
(N=4)
•
• • •
•
Third number
(N=3)
• •
•
•
• • •
•
•
• •
•
•
•
•
•
•
a )Write the fifth number in the pattern
Form a table of values
N
1
2
3
4
5
D= no 1
of dots
5
12
22
35
When N= 1 D = 1
When N=2
D=5
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=1D=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=1D=1
N=2
D=5
Now solve
simultaneously
1=a-b
5 = 4a -2b
1=a-b
X (-2)
5 = 4a -2b
X1
-2 = -2a +
2b
5 = 4a - 2b
ADD
SUB
3 = 2a
a = 3/2
1 = 3/2 -b
b = 1/2
Check
4x3/2-2x1/2
=12/2-2/2 =10/2=5
a = 3/3 b= 1/3