Document 6529622

Statistics and
Modelling
Achievement Standard 90644
Solve equations
3.4
Externally assessed
(4 credits)
Solve Systems of Equations Involving Three Variables
This is an extension of simultaneous equations solving techniques with two variables which have been
studied in earlier years.
The method to be shown is called elimination. (There are alternative methods such as substitution,
matrices and others which may also be used.)
The method of elimination involves eliminating a variable from one pair of equations and eliminating the
same variable from another pair of equations. This leads to two equations in two unknowns (or variables)
which can be solved using techniques from earlier courses.
Example
Q. Solve the system of equations 2x + y + z = 4, 4x + y – z = 7 and 8x + 2y + 2z = 10.
A. In this example, the variable which will be eliminated is z (x or y could just as easily be selected).
The equations are numbered for ease of reference:
(1) + (2) gives
2 x (2) + (3) gives
Doubling (4) gives
(5) – (6) gives
2x + y + z = 4 4x + y – z = 7
8x + 2y + 2z = 10
... (1)
... (2)
... (3)
6x + 2y = 11
16x + 4y = 24
12x + 4y = 22
... (4)
... (5)
... (6)
4x = 2
This value is then substituted in (4), (5) or (6) to get y.
Substituting in (6) gives
1
2
1
2
[solving]
6 + 4y = 22
∴ 4y = 16
∴x=
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∴y=4
The values x =
, y = 4 can now be substituted into any of (1), (2) or (3) to find z.
Substituting into (1) gives
∴ the solutions to the system of simultaneous equations is x =
Note: This solution can easily be found using the equation solving mode of most graphical calculators, such as the Casio fx-9750G.
1 + 4 + z = 4
∴ z = –1
1
2
, y = 4, z = –1
Some problems in various practical contexts need to be expressed in the form of simultaneous equations in
order to be solved.
46 Achievement Standard 90644 (Statistics and Modelling 3.4)
Example
Q.Find the values for A, B and C for the quadratic function f (x) = Ax2 + Bx + C whose graph passes through (1, 4), (2, 7) and (–1, 16).
A. Substituting the coordinates for the three given points in this formula gives
A + B + C = 4
4A + 2B + C = 7
A – B + C = 16
[substituting x = 1, y = 4 in y = Ax2 + Bx + C]
[substituting x = 2, y = 7]
[substituting x = –1, y = 16]
Solving these equations using any method gives A = 3, B = –6, C = 7.
∴ the function is f (x) = 3x2 – 6x + 7
Geometric Interpretation of Systems of Linear Equations
Linear equations in two variables can be represented as lines, and the three possible outcomes are shown
below.
y
non-parallel lines
unique solution
x
y
example:
y=x
x+y=3
y
parallel lines
no solutions
x
coincident lines
infinitely many solutions
example:
x – y = –1
x–y=3
x
reduces to 0 = 4
(a contradiction)
Solution:
(1.5, 1.5)
example:
y=x+1
2y – 2x = 2
reduces to 2 = 2
(or 0 = 0)
Similarly, linear equations in three variables can be represented as planes which can intersect in various
ways.
Consistent equations — Solution(s) exist
planes have 1 common point
unique solution
example
x + y + z = 6 ... (1)
2x — y + z = 3... (2)
x + 2y + z = 8 ... (3)
by elimination,
solution is (1, 2, 3)
This type of system can be
solved automatically using a
graphics calculator
planes meet on a line
(they are not independent)
infinitely many solutions
(simultaneous solution
results in 0 = 0 type situation)
example
x + y + z = 10
... (1)
2x + 3y + 4z = 20 ... (2)
3x + 4y + 5z = 30 ... (3)
since (1) + (2) = (3)
subtracting gives 0 = 0
infinitely many solutions
Inconsistent equations — No solution(s) exist
parallel planes
no solution
(simultaneous solution
produces a contradiction)
example
2x + y + 3z = 12 ... (1)
2x + y + 3z = 15 ... (2)
2x + y + 3z = 18 ... (3)
(2) — (1) gives 0 = 3
so no solutions
Similarly a system with two
parallel planes (and a third
plane not parallel).
no point(s) common
to all three planes
no solutions
(simultaneous solution
produces a contradiction)
example
x + 2y + z = 4
... (1)
x — 2y + 2z = 6
... (2)
3x — 2y + 5z = 8 ... (3)
(1) + (2) and (1) + (3)
and (3) — (2)
give parallel lines
so no solutions
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Solve equations 47
Example
Q. Consider the following system of equations in x, y and z:
4x + 3y + 2z = 11
3x + 2y + z = 8
7x + 5y + az = b
Give values for a and b in the third equation which make this system:
1. Inconsistent. 2. Consistent, but with an infinite number of solutions.
A. 1. Adding the first two equations gives 7x + 5y + 3z = 19
which is similar to the third equation 7x + 5y + az = b
By selecting a = 3 and b ≠ 19 (eg b = 10), a contradiction arises (since 7x + 5y + 3z = 19 and
7x + 5y + 3z = 10 cannot happen simultaneously) and the system of equations will be inconsistent.
2.By choosing a = 3 and b = 19, the third equation becomes the same as the sum of the first two (the
equations are dependent), and thus the three planes meet on a line, generating an infinite set of solutions.
Questions
Solving Systems of Equations Involving Three Variables
1. Find the solutions to the system of equations
2. Solve the system of equations
4x + y – 2z = –18
x + y + z = 0
–2x + 3z = 17
2a + 2b
= c + 14
2a + 2c = –5
4a – 4b – 3c = –9
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48 Achievement Standard 90644 (Statistics and Modelling 3.4)
3. Solve the system of equations:
4. Solve the system of equations:
3x – 3y + 2z + 6 = 12 6x + 3y + 7z = 77
3x + y = z y + 21= 3x + z + 10
2x + y + z = 13 + x
2x + 3y = z + 5
5. Only three companies sell licensed Olympic merchandise in Athenville.
In a particular week a total of $100 000 worth of the merchandise is sold. Company A sells $1 600 more
than Company B and Company C combined. Company B sells twice as much as Company C.
Set up and solve a system of equations to find the value of the merchandise each of the three companies
has sold in this particular week.
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Solve equations 49
Achievement Standard 90644 (Statistics and Modelling 3.4)
Solve equations
3.4 Solving Systems of Equations Involving Three Variables (page 47)
1. a. Rearrange the equations to standard form ax + by + cz = d
4x + y – 2z = –18 ... (1)
x + y + z =
0 ... (2)
–2x + 3z = 17 ... (3)
Eliminate z (1) + 2 × (2)
gives 6x + 3y = –18 ... (4)
3 × (2) – (3)
gives5x + 3y = –17 ... (5)
Eliminate y (4) – (5)
gives x = –1
Substitute x = –1 in (4) gives
–6 + 3y= –18
∴ 3y= –12
∴ y= –4
Substitute for x and y in 2 gives
–1 + –4 + z = 0
adding 6
dividing by 3
substituting in (1) or (3) would be equally satisfactory
∴ z=5
Solutions are x = –1, y = –4, z = 5. (A)
2. Rearrange the equations to standard form ax + by + cz = d
2a + 2b – c = 14
...(1)
2a + 2c = –5
...(2)
4a – 4b – 3c = –9
...(3)
Eliminate a (1) – (2)
gives 2b – 3c = 19
2 × (2) – (3)
gives 4b + 7c = –1
Eliminate b (5) – 2 × (4)
gives
13c = –39
∴
c = –3
Substitute c = –3 in (4) gives 2b + 9= 19
∴
b = 5
Substitute c = –3 in (2) gives 2a – 6= –5
1
13
1
∴ a×= ×  ×
1
13
1
3 7
1
15
2
8
∴ a =×
, ×b = 5, c× = –3+ (A)× 
15
2
5 8
2
8
3. Rearrange the equations to standard form ax + by + cz = d
3x – 3y + 2z = 6
...(1)
3x + y – z = 0
...(2)
2x + 3y – z = 5
...(3)
Eliminate z (1) + 2 × (2) gives
9x – y = 6
(2) – (3)
gives
x – 2y = –5
Eliminate y 2 × (4) – (5) gives
17x = 17
∴
x = 1
Substitution in (5)
gives 1 – 2y = –5
∴
y = 3
Substitute for x and y in (2) gives 3 + 3 – z = 0
∴
z = 6
hence x = 1, y = 3, z = 6
(A)
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...(4)
...(5)
solving
solving
3 7
1
solving
+
× 
5 8
2
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solving
...(4)
...(5)
50 Achievement Standard 90644 (Statistics and Modelling 3.4)
4. Rearrange the equations to standard form ax + by + cz = d
6x + 3y + 7z = 77 ...(1)
y +11 = 3x + z ...(2)
x + y + z = 13 ...(3)
(2) may be rearranged to give
–3x + y – z = –11 ...(2)
Eliminate z (1) + 7 × (2) gives –15x + 10y = 0
...(4)
(2) + (3)
gives –2x + 2y = 2
...(5)
Eliminate y (4) – 5 × (5) gives –5x = –10
∴ x = 2
Substituting in (5)
gives –4 + 2y = 2
∴ y = 3
solving
Substituting in (3)
gives 2 + 3 + z = 13
∴ z = 8
hence x = 2, y = 3, z = 8 (A)
5. Let a be the amount A sells, b be the amount B sells, c be the amount C sells
From the information given
a + b + c= 100 000
...(1)
a – b – c = 1 600
...(2)
b = 2c
...(3)
Solving (1), (2) and (3) simultaneously, gives
a = 50 800,
b = 32 800,
c = 16 400
∴ A sells $50 800 worth ∴ B sells $32 800 worth and C sells $16 400 worth. (M)
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