Download Linear Simultaneous Equations

2013
Year 10 Core Mathematics
Module 4
Linear Simultaneous Equations
Name:
Home Group:
Contents:
Set 1 Graphical Solution of Simultaneous Linear Equations.
Set 2 Solving Simultaneous Linear Equations using Substitution.
Set 3 Solving Simultaneous Linear Equations using Elimination.
Set 4 Writing equations from worded statements.
Set 5 Problem Solving using Simultaneous Linear Equations.
Graphical solution of simultaneous
linear equations
Simultaneous linear equations
•
•
•
•
Any two linear graphs will meet at a point, unless they are parallel.
At this point, the two equations simultaneously share the same x- and y-coordinates.
This point is referred to as the solution to the two simultaneous linear equations.
Simultaneous equations can be solved graphically or algebraically.
Graphical solution
•
•
•
•
This method involves drawing the graph of each equation on the same set of axes.
The intersection point is the simultaneous solution to the two equations.
An accurate solution depends on drawing an accurate graph.
Graph paper or graphing software can be used.
Use the graph of the given simultaneous equations below to
determine the point of intersection and, hence, the solution
of the simultaneous equations.
x + 2y =4
y = 2x — 3
Point of intersection (2, 1)
Solution: x = 2 and y = I
For the following simultaneous equations, use substitution to check if the given pair of coordinates,
(5, —2), is a solution.
[1]
3x — 2y = 19
4y + x = —3
[2]
3x 2y = 19
4y + x = —3
[1]
[2]
—
Check equation [1]:
LHS = 3x 2y
= 3(5) 2(-2)
= 15+4
=19
LHS = RHS
—
—
RHS = 19
Check equation [2]:
LHS = 4y + x
= 4(-2) +5
= —8+ 5
=-3
LHS = RHS
RHS = —3
In both cases, LHS = RHS. Therefore, the
solution set (5, —2) is correct.
Set 1
Use the graphs below of the given simultaneous equations to write the point of
intersection and, hence, the solution of the simultaneous equations.
a x+y= 3
bx+y=2
x—y=1
3x — y = 2
C y—x=4
3x + 2y = 8
e y — 3x = 2
x—y=2
2
d y + 2x = 3
2y + x = 0
f 2y — 4x = 5
4y+ 2x=5
For the following simultaneous equations, use substitution to check if the given pair of
coordinates is a solution.
3x + 2y = 31
y— x =4
a (7,5)
b (3,7)
2y+x= 17
2x+3y= 28
C (9,1)
x + 3y = 12
d (2,5)
x+y= 7
2x+ 3y= 18
5x — 2y = 43
e (4, —3)
y = 3x — 15
f (6, —2)
x — 2y = 2
4x + 7y = —5
3x+y= 16
g (4, —2)
2x + y = 6
h (5,1)
y — 5x = —24
3y + 4x = 23
x — 3y = 8
i (-2, —5) 3x — 2y = —4
j (-3, —1) y —x = 2
2y — 3x = 7
2x — 3y = 11
3
Solve each of the following pairs of simultaneous equations using a graphical method.
a x +y=5
b x+2y=10
2x + y = 8
3x+y=15
c 2x+3y=6
d x-3y=-8
2x — y = —10
2x + y = —2
f y+2x=6
• 6x+5y=12
2y + 3x = 9
5x + 3y = 10
Answers — Graphical solution of simultaneous
linear equations
1 a (2,1)
d (2, —1)
b (1,1)
e (-2, —4)
C Yes
2 a No
b Yes
e Yes
i No
3 a (3, 2)
e (2,0)
f No
j Yes
g No
b (4, 3)
f (3,0)
c (-3, 4)
c (0,4)
f (-0.5, 1.5)
d No
h Yes
d (-2, 2)
Solving simultaneous linear
equations using substitution
• There are two algebraic methods which can be used to solve simultaneous equations.
• They are the substitution method and the elimination method.
Substitution method
• This method is particularly useful when one (or both) of the equations is in a form where
one of the two variables is the subject.
• This variable is then substituted into the other equation, producing a third equation with only
one variable.
• This third equation can then be used to determine the value of the variable.
Solve the following simultaneous equations using the substitution method.
y = 2x — 1 and 3x + 4y = 29
V = 2x — 1
3x + 4y = 29
Substituting (2x — 1) into [2]:
3x + 4(2x — 1) = 29
3x + 8x — 4 = 29
1 lx — 4 = 29
1 lx = 33
x=3
Substituting x = 3 into [1]:
y = 2(3) — 1
= 6—1
=5
Solution: x = 3, y = 5 or (3, 5)
[3]
Set 2
Solve the following simultaneous equations using the substitution method. Check your
solutions using technology.
a x=-10 + 4y
b 3x + 4y = 2
x=7+5y
3x + 5y = 21
C 3x+y=7
d 3x + 2y = 33
y = 41 — 5x
x = —3 — 3y
f 4x+y= 9
e y = 3x — 3
y = 11 — 5x
—5x + 3y = 3
x=-5-2y
h x = —4 — 3y
5y+x=-11
—3x — 4y = 12
x = 7 + 4y
j x=14 +4y
2x+y=-4
—
2x + 3y = —18
3x + 2y = 12
I y= 2x+ 1
x=9 —4y
—5x — 4y = 35
2
Solve the following pairs of simultaneous equations using the substitution method.
Check your solutions using technology.
Answers — Solving simultaneous linear
a y = 2x — 11 and y = 4x + 1
equations using substitution
1 a (2, 3)
b (2, —1)
c (3, —2)
b y = 3x + 8 and y = 7x — 12
d (7,6)
e (3,6)
f (2,1)
c y=2x— 10 andy=-3x
i (-1, —2)
g (-1, —2)
h (-4, 0)
d y = x — 9 and y = —5x
j (6, —2)
k (3, 1 .)
I (-3, —5)
e y = —4x — 3 and y = x — 8
f y = 0.5x and y = 0.8x + 0.9
2 a (-6, —23)
c (2, —6)
b (5, 23)
1
d
e (1, —7)
f
(-3 —1 5)
Solving simultaneous linear
equations using elimination
• Elimination is best used when the two equations are given in the form ax + by = k.
• The method involves combining the two equations so that one of the variables is eliminated.
• Addition or subtraction can be used to reduce the two equations with two variables into
one equation with only one variable.
WORKED EXAMPLE
Solve the following pair of simultaneous equations using the elimination method.
2x + y = 7
—2x — 3y = —9
—2x — 3y = —9
2x+y=7
[1] + [2]:
—2x — 3y + (2x + y) = —9 +7
—2x — 3y + 2x + y = —2
—2y = —2
Y=1
Substituting y = 1 into 121:
2x + 1 = 7
2x = 6
x=3
Solution: x = 3, y = 1 or (3, 1)
111
[2]
Set 3
1
Solve the following pairs of simultaneous equations by adding equations or by subtracting equations
to eliminate either x or y.
a 3x+y= 7
2x — y = 3
b x+y=12
x—y=2
C
8x+3y = 8
5x — 3y = 5
d 11x+3y= 5
h 10x+y= 34
7x — 3y = 13
e 8x+ 6y = 2
f 5x+ 2y = 9
g 4x + 3y = 11
2x — 6y = 3
2x+ 2y = 6
4x+5y =13
2x + y = 10
r+3y=10
x—y=-2
5x-3y = —2
—5x+y= 4
4x — 3y = 0
—4x+y=-8
I 8x+ 7y = 7
1
k
3 Solve each of the following equations using the elimination method.
a x+2y= 12
3x — 2y = 12
c 6x+ 5y = —13
—2x + 5y = —29
e x — 4y = 27
3x — 4y = 17
g —5x + 3y = 3
—5x+y=-4
4
a
C
e
g
5
a
c
e
g
b 3x + 2y = —23
5x + 2y = —29
d 6x — 5y = —43
6x—y=-23
f —4x+y=-10
4x — 3y = 14
h 5x — 5y = 1
2x — 5y = —5
Solve the following pairs of simultaneous equations.
x+2y=4
b 3x+2y= 19
3x — 4y = 2
6x — 5y = —7
d 6x + y = 9
—2x + 3y = 3
5x — 6y = —3
—3x + 2y = 3
f 5x+y=27
x+ 3y= 14
4x + 3y = 26
3x+y=10
h 2x+ 5y= 14
—6x + 5y = —14
3x + y = —5
—2x + y = —6
Solve the following pairs of simultaneous equations.
2x+3y= 16
b 5x — 3y = 6
3x + 2y = 19
3x — 2y = 3
3x + 2y = 6
d 2x + 7y = 3
4x+3y= 10
3x + 2y = 13
f —3x + 7y = —2
2x — 3y = 14
4x + 2y = 14
3x — 5y = 21
—4x + 5y = —9
h 2x+5y=-6
2x + 3y = 21
3x + 2y = 2
6 Solve the following simultaneous equations using an appropriate method.
a 7x + 3y = 16
y = 4x — 1
b 2x+y= 8
4x+ 3y= 16
10x — 7y = —7
Answers - Solving simultaneous linear
equations using elimination
1
a (2,1)
b (7,5)
c (1,0)
d (1, -2)
e (0.5, 1/3)
f (1,2)
g (2,1)
h (3,4)
1(1,3)
j (-1, -1)
k (3,4)
1(0,1)
3 a (6, 3)
d (-3, 5)
g 43
4 a (2, 1)
d (1,3)
g (4, 2)
5 a (5,2)
b (-3, -7)
f (2, -2)
h (2,1)
b (3,5)
C
e (2,4)
(3,3)
f (5,2)
h (-3, 4)
b (3,3)
d (5, -1)
e (7,0)
g (6, 3)
h (2, -2)
6 a (1,3)
(2, -5)
C
e (-5, -8)
C
(-2, 6)
f (3.1)
b (4,0)
Writing equations from worded statements
1. The multiplication sign between the number and the pronumeral is omitted (not written). For example, 2 x x is written as 2x.
2. Division of algebraic terms is usually written as a fraction with the number that is being divided (the dividend) in the
numerator and the number by which we are dividing (the divisor) in the denominator. For example, x ÷ 2 is written as Lc .
2
The table below shows some common words and expressions and their mathematical meaning.
Operation
Word expression
'the sum of', 'is added to', 'plus'
'the difference between', 'is subtracted from', 'minus'
'the product of', 'is multiplied by', 'times'
'the quotient of', 'is divided by'
'is equal to', 'the result is', 'gives'
Write an equation for each of the following statements, using x to represent the unknown number.
a Five times a certain number gives 15.
b When 12 is subtracted from a certain number, the result is 6.
a 'Five times' means '5 x'.
'A certain number' means x.
'Gives 15' means '= 15'.
5 xx= 15
5x= 15
b '12 is subtracted from' means `- 12'.
'A certain number' means x.
'The result is 6' means '= 6'.
x - 12 = 6
Set 4
Try these
Write an equation for each of the following statements, using x to represent the unknown number.
1 When 3 is added to a certain number, the result is 100.
2 Nine times a certain number is 72.
'3 is added to' means
'Nine times' means
'A certain number' means
'A certain number' means
'The result is 100' means
'Is 72' means
Equation: x +
Equation:
=
= 72
4 Dividing a certain number by 5 gives 12.
3 When 4 is subtracted from a certain number, the result is 19
'4 is subtracted from' means
'Dividing by' means
'A certain number' means
'A certain number' means
'The result is 19' means
'Gives 12' means
Equation:
Equation:
5 The difference between a certain number and 7 is 3
6 The product of a certain number and 12 is 88.
'The difference between' means
'The product of' means
'A certain number' means
'A certain number' means
'Is 3' means
'Is 88' means
Equation:
Equation
7 The sum of 23 and a certain number is 50.
8 When a certain number is divided by 10, the result is 3.
'The sum of' means
'A certain number' means
'A certain number' means
Is divided by' means
'Is 50' means
'The result is 3' means
Equation
Equation
9 Seven times a certain number gives 56
10 Five more than a certain number is 9
'Seven times' means
'Five more than' means
'A certain number' means
'A certain number' means
'Gives 56' means
Is 9' means
Equation
Equation
Answers
x + 3 = 10
n — 4 = 19
-
7=3
n + 23 = 50
7n = 56
9x = 72
5
12n = 88
n =3
_
10
n+5=9
Problem solving using simultaneous
linear equations
• Many word problems can be solved using simultaneous linear equations.
• Follow these steps.
• Define the unknown quantities using appropriate pronumerals.
• Use the information given in the problem to form two equations in terms of these
pronumerals.
• Solve these equations using an appropriate method.
• Write the solution in words.
• Check the solution.
Ashley received better results for his Maths test than for his English test. If the sum of the two
marks is 164 and the difference is 22, calculate the mark he received for each subject.
Let x= the maths mark.
Let y= the English mark.
x+y= 164
x—y= 22
[1] + [2]:
[1]
[2]
2x=186
x = 93
Substituting x = 93 into [1]:
x+y= 164
93 + y = 164
y= 71
Solution:
Maths mark (x) = 93
English mark (y)= 71
Set 5
1
Rick received better results for his Maths test than for his English test. If the sum of his
two marks is 163 and the difference is 31, find the mark for each subject.
3 Find two numbers whose difference is 5 and whose sum is 11.
4 The difference between two numbers is 2. If three times the larger number minus double the
smaller number is 13, find the two numbers.
5 One number is 9 less than three times a second number. If the first number plus twice the
second number is 16, find the two numbers.
6 A rectangular house has a perimeter of 40 metres and the length is 4 metres more than the
width. What are the dimensions of the house?
7 Mike has 5 lemons and 3 oranges in his shopping basket. The cost of the fruit is $3.50. Voula,
with 2 lemons and 4 oranges, pays $2.10 for her fruit. How much does each type of fruit cost?
8 A surveyor measuring the dimensions of a block of land finds that the length of the block is
three times the width. If the perimeter is 160 metres, what are the dimensions of the block?
11 If three Magnums and two Paddlepops cost $8.70 and the difference in price between a
Magnum and a Paddlepop is 90 cents, how much does each type of ice-cream cost?
12 If one Redskin and 4 Golden roughs cost $1.65, whereas 2 Redskins and 3 Golden roughs cost
$1.55, how much does each type of sweet cost?
14 The difference between Sally's PE mark and Science mark is
12, and the sum of the marks is 154. If the PE mark is the higher
mark, what did Sally get for each subject?
15 Mozza's cheese supplies sells six Mozzarella cheeses and eight Swiss cheeses to Munga's deli
for $83.60, and four Mozzarella cheeses and four Swiss cheeses to Mina's deli for $48. How
much does each type of cheese cost?
16 If the perimeter of the triangle in the diagram is 12 cm and the length of the rectangle is 1 cm
more than the width, find the value of x and y.
Answers — Problem solving using
simultaneous linear equations
1 Maths mark = 97, English mark =66
3 8 and 3
4 9 and 7
5 6 and 5
6 Length = 12 m and width = 8 m
7 Lemons cost 55 cents and oranges cost 25 cents.
8 Length 60 m and width 20 m
11 Paddlepops cost $1.20 and a Magnum costs $2.10.
12 Cost of the Golden rough = 35 cents and cost of the
Redskin = 25 cents
14 PE mark is 83 and Science -mark is 71.
15 Mozzarella costs $6.20, Swiss cheese costs $5.80.
16 x=3 andy=4
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