Solving Simultaneous
When you solve simultaneous
equations you are finding where
two lines intersect
Q. a) Find the equation of each line.
b) Write down the coordinates of the point of intersection.
4
y=x+2
3
(1,3)
y = 2x + 1
2
1
2
1
0
1
2
3
4
1
2
3
4
Q. a) Find the equation of each line.
b) Write down the coordinates where they meet.
y=x
2
1.5
y=-x-1
1
0.5
2
1.5
1
0.5
0
0.5
(-0.5,-0.5)
1
1.5
2
0.5
1
1.5
2
Q. a) Plot the lines:
y=x
y = 2x - 1
b) Write down the coordinates where they meet.
2
1.5
1
(1,1)
0.5
2
1.5
1
0.5
0
0.5
1
1.5
2
0.5
1
1.5
2
Draw each pair of lines on the same diagram and solve the
equations by finding the point of intersection
1
3
5
7
9
x + y = 10
x-y=4
y=x
x+y=9
y = 2x - 5
y=x–2
y=8-x
y=x+2
y = 2x + 3
y=x–2
2
4
6
8
10
y = 2x
y=x+4
y=x-3
y=5-x
x+y=2
x - y = -10
y = 3x
y=x+6
2x + y = 7
y=x+1
We can use straight line theory to work out real-life problems
especially useful when trying to work out hire charges.
Q.
I need to hire a car for a number of days.
Below are the hire charges for two companies.
Complete tables and plot values on the same graph.
Anrnold Palmer Car Hire
Per Day
0
1
2
3
4
5
Total Cost £
£100
120
140
160
180
200
4
5
240
300
Swinton Direct Car Hire
Per day
0
1
2
Total Cost £
0
60
120
3
180
Total Cost £
Summarise data !
Who should I hire
the car from?
Up to 2 days
Swinton
Over 2 days Arnold
Days
Key steps
1. Fill in tables
2. Plot points on the same graph ( pick scale carefully)
3. Identify intersection point ( where 2 lines meet)
4. Interpret graph information.
13-Mar-16
Created by Mr. Lafferty Maths Department
Algebraic Method : Solving by substitution
“Solve simultaneously”
y = 2x + 1 (1)
y = - x + 7 (2) The graphs of these lines cross where y = y
They are both equal to y, so:
2x + 1 = - x + 7
+x
+x
3x + 1 = 7
-1 -1
Put x = 2 into
equation (1)
y = 2x + 1
y=22+1
y=5
3x = 6
x=2
These lines would cross at (2 , 5)
Algebraic Method
: Solving by substitution extended
y = 2x + 1
3x + 2y = 9
Put y = 2x + 1
into
3x + 2y = 9
“Solve simultaneously by
substitution method”
Put x = 1 into
y = 2x + 1
y=2x1+1
3x + 2(2x + 1) = 9
y=2+1
3x + 4x + 2 = 9
7x + 2 = 9
7x = 7
x=1
y=3
Simultaneous Equations - Substitution
Solve each of the sets of equations below
using the method of substitution.
1
2x + 3y = 9
y = 2x – 5
2
5x + y = 22
y=x+4
3
4x – 3y = 7
y=x+1
4
3x + 2y = 8
y = 2x – 3
5
8x + 5y = 16
y = 2x - 4
6
y=x+8
x + y = 12
2x + 3y = 9
y = 2x – 5
Put y = 2x – 5
into 2x + 3y = 9
2x + 3(2x – 5) = 9
2x + 6x - 15 = 9
8x = 24
x=3
Put x = 3 into
y = 2x – 5
y=6-5
y=1
5x + y = 22
y=x+4
Put y = x + 4
into 5x + y = 22
5x + x + 4 = 22
6x + 4 = 22
6x = 18
x=3
Put x = 3 into
y=x+4
y=3+4
y=7
7
10x + 2y = 6
y = 3x – 5
8
5x + y = 15
y=3–x
9
y=x+8
x + 3y = 44
10
4x + 3y = 10
y=2–x
11
5x – 2y = 16
y = 2x – 7
12
y = 10 + x
3x + 2y = 25
13
4x + 7y = 37
y=x-1
14
y = 2x
3x + 2y = 42
7
y = 2x - 5
y = 4x – 3
8
y=x+4
y = 3x - 7
9
y=x+2
4x + 3y = 27
10
5x - 2y = 10
y=2–x
11
y=x+9
2x + 3y = 37
12
y = 4x
x + y = 12
13
y = 6x
2x + y = 4
14
y = 1.5x
3x + 2y = 24
Set 2 Solve each of the sets of equations below
using the method of substitution
1
2x + y = 13
y=x+1
2
x + y = 13
y=x-2
3
2x + y = 15
y = 3x
4
y = 2x +7
y=x+4
5
y=7-x
y = 3x - 5
6
y=1-x
3x + 2y = 8
Simultaneous Equations
Elimination
Solving Simultaneous Equations: Elimination
3x + 4y = 26 (1)
(2)
7x - y = 9
(3) = (2) x 4
(1)
(1)+(3)
“Solve simultaneously”
Make either coefficient
of x or y ‘same size’
28x - 4y = 36
3x + 4y = 26
31x
= 62
x= 2
Put x = 2 into
equation (1)
3x + 4y = 26
32 + 4y = 26
6 + 4y = 26
4y = 20
y=5
Simultaneous Equations - Elimination 1
Solve each of the sets of equations below
using the method of elimination.
1
x + y = 11
x–y=7
2 x+y=9
x–y=2
3
x–y=4
x + y = 24
4 2x + y = 7
x–y=2
5
3x + y = 7
x–y=5
6 2x – y = 7
x+y=5
7
5x + 2y = 6
x – 2y = 6
8
4x + y = 14
3x – y = 7
9
x + 3y = 1
x – 3y = 13
10
x + 2y = 6
-x + y = 0
11
4x + 3y = 14
2x – 3y = 16
12
x+y=7
2x – y = 8
13
3x – 2y = 10
x + 2y = 6
14
5x + 2y = 19
x – 2y = -1
Elimination (1b)
1
x + y = 13
x–y=5
2
x - y = 14
x+y=6
3
x–y=1
x+y=9
4
3x + y = 13
x–y=3
5
5x + y = 12
x–y=0
6
2x – y = 11
4x + y = 25
7
3x + 2y = 20
x – 2y = -4
8
7x + 2y = 4
3x – 2y = -4
9
2x + 3y = 15
2x – 3y = -3
10
6x + 2y = 2
-6x + y = -8
11
9x + 2y = 15
x – 2y = -5
12
x + 5y = 8
2x – 5y = -14
13
3x – 7y = 5
x + 7y = 11
14
11x + 2y = 5
x – 2y = 7
Elimination (2)
1
3x + 2y = 8
x–y=1
2
5x + 3y = 11
2x – y = 0
3
7x + 2y = 13
x-y=7
4
4x - 5y = 3
x+y=3
5
3x - 2y = 5
x+y=0
6
6x + 5y = 17
-3x + 2y = -4
7
5x + 9y = -8
x – 3y = 8
8
2x + 3y = 9
-x + y = -2
9
4x - 7y = 7
x + y = 10
10
5x + 4y = 1
x-y=2
11
5x + 3y = 16
3x – 2y = 2
12
2x + 5y = 12
13x – 2y = 9
13
7x – 3y = 20
3x + 4y = -2
14
x + 2y = 1
3x – 7y = 29
Elimination (2b)
1
5x + 3y = 4
2x – y = 6
2
3x + y = 9
7x – 2y = 8
3
4x + 3y = 19
x-y=3
4
6x - 5y = 1
x+y=2
5
2x - 3y = -2
-x + 5y = 8
6
5x - 8y = 10
3x + 4y = 6
7
2x + y = 9
5x – 3y = 39
9
5x + 4y = 14
3x - 2y = 4
11
3x +7y = 14
-x +3y = 6
8
2x + 3y = 11
3x - y = 11
10
-x + 6y = 1
2x - 5y = 5
12
6x - 4y = -2
5x + 2y = 9
Elimination (2c)
1
3
5
7
9
11
3x + 2y = 10
5x – 3y = 4
7x + 2y = 11
2x - 3y = -4
3x - 4y = -6
2x + 5y = 19
8x - 3y = 2
5x +2y = 9
3x - 7y = 7
2x + 3y = -3
10x - 3y = -13
4x +5y = 1
2
4
6
8
10
12
2x - 5y = 7
3x + 4y = -1
6x - 5y = 12
4x + 3y = 8
9x - 5y = 14
2x + 3y = -1
4x - 5y = 18
5x + 6y = -2
5x - 2y = 11
4x + 3y = -5
9x + 5y = -1
4x - 3y = 10
Elimination (2d)
1
5x + 3y = 11
2x + y = 4
3
8x + 5y = -2
3x + 4y = -5
5
9x + 4y = 1
3x + 2y = -1
7
10x + 3y = 1
3x +2y = -3
9
3x + 7y = -1
2x + 3y = 1
11
10x + 7y = 14
3x +5y = 10
2
7x + 2y = 17
3x + y = 8
4
4x + 5y = 18
x+y=4
6
5x + 6y = 12
3x + 5y = 10
8
4x + 7y = 1
x + 3y = -1
10
5x + 2y = 16
4x + 5y = 6
12
6x + 5y = 8
x + 3y = -3
5 pens and 3 rubbers cost £0·99 while 1 pen
and 2 rubbers cost £0·31. Make two
equations and solve them to find the cost of
each.
Let p = pen, r = rubber
5p + 3r = 99
p + 2r = 31
p = 15, r = 8
3 plum trees and 2 cherry trees cost
£161 while 2 plum trees and 3 cherry
trees cost £154. Make two equations
and solve them to find the cost of each.
Let p = plum tree, c = cherry tree
3p + 2c = 161
2p + 3c = 154
p = 35, c = 28
4 bottles of red wine and 5 bottles of
white wine cost £35·50 while one
bottle of each cost £8. Find the cost of
each type of wine.
Let r = red wine, w = white wine
4r + 5w = 35∙5
r+w=8
r = £4.50, w = £3.50
Tickets on sale for a concert are £25 each for the
front section and £20 each for the back section of
the theatre.
If 212 people attended the concert and the total
receipts were £4980, how many of each price of
ticket were sold ?
Let f = front seat, b = back seat
f + b = 212
25f + 20b = 4980
f = 148, b = 64
Meals in a restaurant are available at £28
per person for the fish courses menu and
£30 per person for the meat courses
menu. There are 95 guests attending the
function in the restaurant and the total
bill came to £2 760.
How many guests chose each menu ?
Let f = fish course, m = meat course
f + m = 95
28f + 30m = 2760
f = 45, m = 50
Calendars cost £9 and £5 each and are on sale in a
card shop.
If 760 calendars are sold and the total takings for
them were £5 240, how many of each price of
calendar were sold ?
Let x = £9 calendars, y = £5 calendars
x + y = 760
9x + 5y = 5240
x = 360, y = 400