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How do we solve Simultaneous Equations?
• We can graph the equations and look for intersection(s) if they exist.
• We can use algebraic Substitution
• We can use algebraic Elimination
• We can also use our graphics calculators to do the above methods.
In NCEA Level 1 and 2 we saw these methods for 2-Dimensional shapes (like a line and a parabola).
This year we will also investigate 3-Dimensional planes. Three equations with three unknowns!
Simultaneous Equations: Graphing
3 1
7 2y x
y x
If two lines intersect then that point is on both linesie It meets the equation of both lines
eg
(5) (2) 3 (5) (2) 7 the point satisfies both equations
We can see the lines intersect at (2,5)
3 7y x y x
Sub (2,5) into each line to check:
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
Solving simultaneous equations with Fx 9750 Calc 3 9 1
2 4 2 2
x y
x y
Select Equation Mode
Simultaneous Equations F1
2 unknowns F1 (X and Y)
Enter Coefficients
Then F1
Gives Final Answer
So X = 3 and Y = -2 is solution to simultaneous equations
Note: and and number
must line up in both equations
x y
Solving 2 linear equations: 3 cases
Case 1: Lines intersect
1 UNIQUE Solution
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
2 4
5x y
x y
Add equations
3x = 9
x = 3
Solving 2 linear equations: 3 cases
Case 2: Lines are parallel
No Solution (Equations are “Inconsistent”)
y
x
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
– 3
– 3
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
2
2
2 2
4 1
x
x y
y
Subtract 2 from 1:
0 = 2
Solving 2 linear equations: 3 cases
Case 3: Both lines are the same
Infinite number of solutions (Equations are dependent)
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
1
2 2
2
y
x y
x
Sub 2 into 1
-x + (x+2) = 2
-x + x + 2 = 2
2 = 2
Three equations with three unknowns
•3 unknowns (x,y,z) are like point (x,y) but we also give a third value z as the “height”.
•If ax + by = constant is the equation of a line
•ax + by + cz = constant is the equation of a “plane”. This is a surface (2-Dimensional) in a 3-D space.
•We can also hold a piece of paper in any direction and describe it’s surface as being on a plane of equation ax + by + cz = 0.
z
x
y
Intersection of Planes: 3 CasesCase 1: Planes intersect at ONE point 1 UNIQUE Solution
Corner of room
The solution of a 3 variablesimultaneous equation is apoint that is common to all 3Equations (planes)
Exercise
2x+y+z=3
x-y+3z=20
4x+2y+z=1
-3x+2y-4y=11
2x-z=16
-4x+y-5y=11
x+2y+z=2
x+y=3-2z
2x+y+3z=1
3a-4b+4c=0
a-b+2c=4
2a+3b-14=0
Intersection of Planes: 3 CasesCase 2: 2 or 3 Planes are parallel or lines of intersection are parallel
No Solution, Equations Inconsistent
Intersection of Planes: 3 Cases3 Planes are parallel No Solution, Equations Inconsistent
If 3 planes are parallel then the coefficients of each equation is a multiple of the others. This does not how ever apply to the constants!
From P 184
2 8 4 (1)
2 8 7 (2)
2 4 16 19 (3)
x y z
x y z
x y z
Coefficients of (1) and (2) are the same.
Coefficients of (3) are a multiple of (2).
As all equations are multiples of each other 3 planes are parallel
Constant is different.
Intersection of Planes: 3 Cases 2 Planes are parallel No Solution, Equations Inconsistent
In this case 2 of the equations representing the parallel lines are either the same or multiples of each other. The other the non parallel line is different. Again this does not apply to the constants
From P 184
5 1
5 2
2 3 6 18
x y z
x y z
x y z
5 1
2 10 2 3
2 3 6 18
x y z
x y z
x y z
orParallelParallel
Intersection of Planes: 3 CasesCase 2 cont: lines of intersection are parallel
No Solution, Equations Inconsistent
Again attempt to solve will result in a contradiction like 4 = 10
In this type, one set of coefficients is a linear combination of the other two. Again this does not apply to the constants
From P184
1
2 3 4 6
12 13 22 8
x y z
x y z
x y z
2( ) 5(2 3 4 ) 12 13 22x y z x y z x y z
2(1) 5(2) 3
Intersection of Planes: 3 CasesCase 3: Infinite number of points of intersection as all three planes
intersect on common line.
Equations are dependent
Attempt to solve will result in for example 3 = 3 or similar
The third equation is a linear combination of the other 2 equations. This applies to constants as well.
Common line to all 3 planes
192
4 2 5 (1)
3 5 4 (2)
6 5 8 13 (3)
Sigma P
x y x
x y z
x y z
(1) 2 (2) 3
for both sides of the equation
Recognising the situation
As all equations are multiples of each other 3 planes are parallel
2 equations are multiples of each so 2 planes are parallel
Contradiction eg 0=4 and one set of coefficients is a linear combination of the other two
“truth” eg 3 = 3 or similar and one set of coefficients is a linear combination of the other two. This applies to constants as well.
No solution
No solution
No solution
Multiple solutions
Inconsistent
Inconsistent
Inconsistent
Dependent
One solution? Inconsistent? Dependent?
x+2y-8z=4x+2y-8z=72x+4y-16z=19
x+2y+z=12x+4y+2z=32x-3y+6z=18
x+y+z=12x-3y+4z=63x-2y+5z=7
2x+3y+z=2x+y+z=14x+5y+3z=7
x+y+z=63x-y+2z=182x+2y-z=0
Exercise:
P192 10.02
P195 10.03
Solving 3 variable simultaneous equations
4x+2y+z=1x-y+3z=202x+y+z=3
-3x+2y=11+4z2x-z=16-4x-5z=11-y
x-2y+z=113y-z=-43+5x2x+3z-y=25
x+z=2-2yx+y+2z=3y=-2x-3z+1
3x-27=-2y2x-4y+5z=-16-6x+2y-5z=-22
7x-6y+2z+28=0-3x+2y-4z=185y-3z=31+4x
x=y+z2y-14=x+z2(y-3x)-z=-11
3x=-3y+2z+117x=5y-3z-64x=y+2z
Application of
Simultaneous Equations
Solve the system of equations:
x+2y=-z
3x-y+z=21
2z-y=19
Simultaneous Eqaution
Sarah Sue and Sally go to get snacks.
The same day Steve went to the same chain and ordered25 of each. How much does it cost him?
No. of ck wings
No. of sushi
No. of wedges
Total cost
Sarah 30 20 15 154.75
Sally 40 24 10 161.7
Sue 50 16 12 188.8
Simultaneous Equations
• A man goes shopping and buys 5 cans of catfood, 2 packets of biscuits and 2 cans of fruit. This costs him $17.74.
On another occasion he buys 4 cans of cat food, 3 packets of biscuits and one can of fruit, at a cost of $15.44.
Later again, he buys 6 cans of cat food, 4 packets of biscuits and 3 cans of fruit for $25.80.
How much would it cost for him to buy 3 cans of cat food, 5 packets of biscuits and 2 cans of fruit?
In Rugby League there are 3 ways of scoring: a try, a place kick, and a drop goal. These are each worth a points, b points and c points respectively.
Auckland scored 5 tries, 3 place kicks, and 1 dropped goal, total 27
Brisbane scored 6 tries, 2 place kicks, and 0 dropped goal, total 28
Canberra scored 2 tries, 5 place kicks, and 2 dropped goal, total 20
What are a b and c??5 3 1 27 (1)
6 2 0 28 (2)
2 5 2 20 (3)
t p g
t p g
t p g
(3) 2 (1)
2 5 2 20
(10 6 2 54)
t p g
t p g
8 34 (4)t p
(2) 2(4)
6 2 28
16 2 68
t p
t p
10 40
4
t
t
sub in (4)
8(4) 34
32 34
2
2
p
p
p
p
Sub in (1)
5(4) 3(2) 27
20 6 27
1
g
g
g
Try=4
Place kick =2
Drop goal=1
A competitor in a TV show can choose to answer a question from 3 different categories: hard, medium, and easy. Hard=10pts, Medium =5pts, Easy=2pts.
The competitor answered 26 questions, gaining 161 points, the competitor answered 20 more hard and medium questions than easy questions.
Find the number of each type of questions answered
10 5 2 161 (1)
26 (2)
20 (3)
h m e
h m e
h m e
20 (3)rearrangedh m e (2) (3)
26
20
2 2 46 (4)
h m e
h m e
h m
(1) 2 (3)
10 5 2 161
2 2 2 40
12 7 201 (5)
h m e
h m e
h m
(5) 6 (4)
12 7 201
(12 12 252)
5 75
15
h m
h m
m
m
A competitor in a TV show can choose to answer a question from 3 different categories: hard, medium, and easy. Hard=10pts, Medium =5pts, Easy=2pts.
The competitor answered 26 questions, gaining 161 points, the competitor answered 20 more hard and medium questions than easy questions.
10 5 2 161 (1)
26 (2)
20 (3)
h m e
h m e
h m e
20 (3)rearrangedh m e 2 2 46 (4)h m
12 7 201 (5)h m 15m
sub in(4)
2 2(15) 46
2 16
8
h
h
h
sub in (2)
(8) (15) 26
23 26
3
e
e
e
The competitor answered
8 hard, 15 medium and 3 easy questions
Answered in context
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
– 2
– 2
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
Determine the equation of the Parabola that passes through the Three points (0,5) (0,1) and (3,-10)
The general equation of a parabola is y=ax2 + bx + c
P199
Ex 10.04
Numerical Solution of Equations:
Some equations cannot be solved exactly.
We now need a method to attack these problems so we can get an approximate answer.
It is very important in our approximation to know how far we are away from the exact value. We need to have a range of values that contain the exact solution.
We will study 3 main Methods to solve these difficult problems:
1. Graphically
2. The Bisection Method
3. The Newton-Raphson Method
Bisection method
1x
2x
Root
1 23Find (the midpoint)
2
x xx
3x
First pick x1 and x2 as initial interval to contain root
1( )f x
2( )f x
3( )f x
Bisection method: the basics (cont)
1x
2x
Root
3x
Remember we need one point below line and one above.
So x3 now becomes our new x1
1( )f x
2( )f x
3( )f x
3x
Bisection method: the basics (cont)
2x
Root
1x1x
1( )f x
2( )f x
3( )f x1( )f x
1x
Bisection method: the basics (cont)
2x
Root
1 23Find new (the midpoint)
2
x xx
3x
2( )f x
1( )f x
3( )f x
1x
Bisection method: the basics (cont)
2x
Root
3x
Remember we need one point below line and one above.
So x3 now becomes our new x2 this time
2( )f x
1( )f x
3( )f x
1x
Bisection method: the basics (cont)
2x
Root
3x2x
We keep repeating this process until we are happy with how small our interval has become.
“We keep performing more iterations”
Bisection Method
• We make use of the “change of sign” to find out if the solution is contained within our interval.
• The Bisection is very slow but it is easy to use and always works on any continuous function.
• Sometimes we need to watch if there are more than one solution we need to start with an interval that contains the solution we want.
The Newton Raphson Method
The Newton Raphson Method:
• The Newton-Raphson Method uses tangents to the graph of a function to generate successive approximations to a root of the given function.
• The calculations are relatively easy and we approach the root very quickly compare to bisection method (as long as we choose a good starting value).
Newton-Raphson Method
y
x
1x
1( )f x
2x
2( )f x
3x
As you can see we are rapidly approaching the actual root (α) with just a few iterations
Tangent
Next approximation
Newton-Raphson Method
• Pick a starting point x1.
• Draw a tangent from the graph at x1 to the x-axis. This is now our new approximation x2.
• Repeat the process until you are within desired accuracy.
Cases where the Newton-Raphson method can fail
• When the function has more than one root, convergence may be to an unexpected value.
• When the starting point is a turning point of the function, then the tangent is horizontal. Newton-Raphson formula would be undefined.
The net profit function from a juice bar can be modelled by the equation p=-0.1t0.5 + 4t-31
Where t is the time in years and p is the net profit in thousands of dollars.
To find out after how many years the juice bar makes a net profit of $30,000 it is necessary to solve the equation
-0.1t0.5 + 4t-31 =0 Use either bisection method or Newton-Raphson method, to
solve the equation to find after how many years the net profit is $30,000.
You must state your starting values, give your answer correct to two decimal places.
An attempt is made to solve the equation
2 + 0.5x2 = 0 using the bisection method.
3x-5
The table below gives a set of values for
f(x)= 2 + 0.5x2 for x=0 and x=3
3x-5
Since there is a change of sign in the function between x=1 and x=2 it is reasonable to assume that a real root of f(x)=0 lies in the interval, but when an attempt is made to find the root by the bisection method it does not exist.
Justify why there is no real root in the inteval
x 0 1 2 3
f(x) -0.4 -0.5 4 5
There is a vertical asymtote at x=5/3. For values less than x=3/5, the expression is negative and getting smaller as x approaches 5/3. for values greater than x=5/3 the equation is positive and getting larger as x approaches x=5/3.
There is a change of sign, but because of the existence of the vertical asymptote there is no real root between 1 and 2.
Now try a test point (0,0) to see if it satisfies equation 1
0 – 0≤ 4
0 ≤ 4 True
Since it does satisfy, we shade out other side side
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
Finding the feasible areaFind the region defined by:
Now try a test point (0,0) to see if it satisfies equation 2
0 + 0 ≥ -2
0 ≥ -2 True
Since it does satisfy, we shade out other side side
x + y -2 2
x < 2
y 2x
+ 4
3
1
- ≤
x-2 2 4 6 8 10
y
-2
2
4
6
8
10
Find the feasible region and the vertices of the boundary:
First Boundary lines:
y > 0 2
0 1
y x+3
x+
3
y 7 4
x
y = 0
y
0
x+3 3
y
2
4
1
7
x
x
x+y=7
Find the region defined by:
x-2 2 4 6 8 10
y
-2
2
4
6
8
10
First Boundary lines:
y > 0 2
0 1
y x+3
x+
3
y 7 4
x
y = 0
y
0
x+3 3
y
2
4
1
7
x
x
Find the region defined by:
x-2 2 4 6 8 10
y
-2
2
4
6
8
10
First Boundary lines:
y > 0 2
0 1
y x+3
x+
3
y 7 4
x
y = 0
y
0
x+3 3
y
2
4
1
7
x
x
Equation 3:
0≤0+3, true, shade other side
Find the region defined by:
x-2 2 4 6 8 10
y
-2
2
4
6
8
10
First Boundary lines:
y > 0 2
0 1
y x+3
x+
3
y 7 4
x
y = 0
y
0
x+3 3
y
2
4
1
7
x
x
Find the region defined by:
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
2
2
4
4
6
6
8
8
10
10
– 2
– 2
y > 0
y x+3
0
x
+y
1
7 4
2
3
x
A
D
C
B
A(2,5)B(7,0)C(0,3)D(0,0)
Finding maximum profitFind the maximum of the function f(x,y) = 2x + 3y given the following constraints:
x + y -2
x
y 2x + 4
2
1
2
3
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
Identify and label Vertices:
A (-2,0)
B (2, 8)
C (2, -4)
A
B
C
Linear Programming Example 1 (cont)Find the maximum of the function f(x,y) = 2x + 3y given the following constraints:
x + y -2
x
y 2x + 4
2
1
2
3
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
A
B
C
Vertex f(x,y) = 2x + 3y Value of f(x,y)
A (-2,0)
B (2, 8)
C (2, -4)
2(-2) + 3(0)2(-2) + 3(0)
2(2) + 3(8)2(2) + 3(8)
2(2) + 3(-4)2(2) + 3(-4)
-4-4
2828
-8-8
This tells us that the maximum of f(x,y) = 28 and it occurs when x = 2 and y = 8.
Incidentally the minimum of f(x,y) = -8 when x = 2 and y = -4
2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20
y
x
2
2
4
4
6
6
8
8
10
10
12
12
14
14
16
16
18
18
20
20
2
2
4
4
6
6
8
8
10
10
12
12
14
14
16
16
18
18
20
20
Linear Programming Example 2Calculate the minimum value of f(x,y) =4x+2y
with the constraints x 2, y 3, x+3y 15,
2x+y 20
f(x,y) = 4x+2y
Remember we “Shade out” what we don’t want5
3
xy
2 10y x
Calculate the minimum value of f(x,y) =4x+3y
with the constraints x 2, y 3, x+3y 15,
2x+y 20
2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20
y
x
2
2
4
4
6
6
8
8
10
10
12
12
14
14
16
16
18
18
20
20
2
2
4
4
6
6
8
8
10
10
12
12
14
14
16
16
18
18
20
20
Linear Programming Example 2 (cont)
f(x,y) = 4x+2yIdentify and label Vertices:
A (2,6)
B (3,4)
C (6, 3)
A
B C
Calculate the minimum value of f(x,y) =4x+3y
with the constraints
x 2, y 3, x+3y 15,
2x+y 20
Linear Programming Example 2 (cont)
Vertex f(x,y) = 4x + 3y Value of f(x,y)
A (2,6)
B (3,4)
C (6,3)
4(2)+3(6)4(2)+3(6)
4(3)+3(4)4(3)+3(4)
4(6)+3(3)4(6)+3(3)
2626
2424
3535
This tells us that the minimum of f(x,y) = 24 and it occurs when x = 3 and y = 4.
2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20
y
x
2
2
4
4
6
6
8
8
10
10
12
12
14
14
16
16
18
18
20
20
2
2
4
4
6
6
8
8
10
10
12
12
14
14
16
16
18
18
20
20
AB C
A farmer has 10 hecters of land in which to plant two crops, corn and wheat and he has to plant at least 7 hectares of his land to make it profitable.
The farmer has only $1200 to spend on seed and each hectare of corn costs $200 and each hectare of wheat $100
The farmer also wants to get the planting done in 12 hours. It takes him an hour to plant a hectare of corn and 2 hours to plant a hectare of wheat.
The profit the farmer makes is $500 per hectare from corn and $300 per hectare from wheat.
Work out how many hectares of corn and wheat the farmer should plant to maximise profit.
