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Transcript
```GCSE: Solving Linear Equations
Skipton Girlsβ High School
Objectives: (a) Solve equations, including with unknowns on both sides
and with brackets.
(b) Form equations from context (with emphasis on quality of written
communication, e.g. "Let x be...").
KEY TERMS
This is an example of a:
3
2π₯
Term
3π₯ + 2
Expression
2
5π₯
+1=2
A term is a product of numbers
and variables
?
An expression is composed
? of one or more
Equation
An equation says that the
? expressions on the
left and right hand side of the = have the same
value.
STARTER
The perimeter of this
work of art is 32.
By trial and error (or any
other method), find π.
3π + 1
9βπ
π=3
?
If we added all four sides of the painting to
get the perimeter, weβd have:
3π + 1 + 3π + 1 + 9 β π + 9 β π
= 4π + 20
And weβre told the perimeter is 32, so
ππ + ππ = ππ. Weβll see today how to
βsolveβ equations like this so we can find π.
Equations must always be βbalancedβ
We already know that the β=β symbol means each side
of the equation must have the same value.
π=4
+2
2 a
2 4
=
π + 2= 6
If we added something to one side of the equation,
what do we have to do with the other side?
+2
Equations must always be βbalancedβ
If we tripled the load on one side of the scales,
what do we have to do with the other side?
π=4
×3
a
4
a a
4 4
=
×3
3π = 12
Solving
!To solve an equation means that we find the value of
the variable(s).
3π + 1
4π + 20 = 32
9βπ
Strategy: To get π on its own on
one side of the equation, we
the things surrounding it.
Solving
Bro Note: You can probably see the answer to this in your
head because the equation is relatively simple, but this full
method is crucial when things become more complicated
4π + 20 = 32
?
-20
?
-20
?
4π =
12
?
οΈ4
Strategy: Do the opposite
operation to βget rid ofβ items
surrounding our variable.
π₯+4
οΈ4?
? 3
π=
Bro Tip: Many students find writing these
operations between each equation helpful to
remind them what theyβre doing to each side, but
youβll eventually want to wean yourself off these.
3π¦
π§
6
-4?
?
οΈ3
?
×6
π₯
π¦
π§
3π β 5 = 13
?
+5
?
+5
3π = 18
?
?
οΈ3
?
οΈ3
?
π=
6
When the solution is not a whole number
4 + 6π§ = 18
?
-4
-4?
6π§ = 14
?
?
οΈ6
οΈ6?
14 7
π§= ? =
16 3
3 = 20 + 4π₯
β17 = 4π₯
? ππ
π=β
π
Bro Note: In algebra,
we tend to give our
rather than decimals
And NEVER EVER EVER
recurring decimals.
Dealing with Fractions
π₯
3 + = 28
5
?
-3
-3?
π₯
? = 25
5
?
×5
?
×5
π₯ = ?125
Exercise 1
Solve the following equations,
showing full working.
1
2
3
4
5
6
π β 4 = 10
2π₯ + 3 = 9
5π₯ β 4 = 36
9π₯ β 2 = 61
9 = 1 + 4π¦
8π + 3 = 75
7
3π₯ = 7
8
5π₯ + 2 = 11
9
8π₯ β 2 = 3
10
3 + 10π = 7
11
5 + 3π = 4
12
7π + 23 = 11
13
14 + 9π = 3
π = ππ ?
π=π ?
π=π ?
π=π ?
π=π ?
π=π ?
π
π=
?
π
π
π=
?
π
π
π=
?
π
π
π=
?
π
π
π=β ?
π
ππ
π=β ?
π
ππ
π=β ?
π
14
15
16
17
18
19
20
21
22
23
24
N
π₯
=5
π = ππ ?
7
π
+3=8
π = ππ ?
4
π
β1=5
π = ππ ?
2
π
1+ =7
π = ππ ?
3
π₯
+3= 4
π=π ?
5
π¦
+8=5
π = βππ?
4
π
5= +9
π = βππ?
6
2π₯
3+
=7
π = ππ ?
5
5π
ππ
?
β 3 = 10
π=
6
π
3π₯
π
5+
=3
π=β ?
4
π
6π₯
π
11 =
+9
π=β ?
7
π
6
π₯+3
ππ
5
3=
+9 π=β ?
8
π
What happens if variable appears on both sides?
5π + 3 = 2π + 9
What might our strategy be?
Collect the variable terms (i.e. The terms involving a) on one side of the
equation, and the βconstantsβ (i.e. The
? individual numbers) on the other
side.
What happens if variable appears on both sides?
Strategy? Collect the variable terms on the side of the equation where
thereβs more of them (and move?constant terms to other side).
Letβs move the βπβ
terms to the left
(as 5 > 2) and the
constants to the
right.
5π + 3 = 2π + 9
This is to get rid
of the constant
term on the left.
5π = 2π + 6
-3
-3
?
?
-2a
?
-2a
3π = 6
?
?
οΈ3
?
οΈ3
π =? 2
We could
have done
these two
steps in
either order.
More Examples
11π₯ β 4 = 2π₯ β 13
3π¦ + 4 = 8π¦ β 5
9
π¦=
5
π₯ = β1?
Way weβd have
previously done itβ¦
2 = β3π₯
2?
2
π₯=
=β
β3
3
5 = 3 β 3π₯
Both methods are valid,
but I prefer the second
β itβs best to avoid
dividing by negative
numbers, and is less
?
Or where we put π₯ term on
side where itβs positive:
5 + 3π₯ = 3
3π₯ = β2
?2
π₯=β
3
3π₯ β 3 = π₯ + 5
ππ = π
?
π=π
3 β 5π₯ = 5 + 2π₯
βπ = ππ
π ?
π=β
π
Dealing with Brackets
If thereβs any brackets, simply expand them first!
2 2π₯ + 3 = 9
ππ + π = π
ππ = π
π?
π=
π
3 β 4 2π₯ β 3 = 7π₯
π β ππ + ππ = ππ
ππ β ππ = ππ
?
ππ = πππ
π=π
7 3π₯ β 1 = 21 + 14π₯
πππ β π = ππ + πππ
ππ = ππ
?
π=π
5β2 π₯+2 =4β3 2βπ₯
π β ππ β π = π β π + ππ
π β ππ = βπ + ππ
π = ππ
?
π
π=
π
Exercise 2
Solve the following.
1
2
3
4
5
6
7
8
9
10
11
12
13
3π₯ = π₯ + 4
6π¦ = 4π¦ β 4
5π₯ + 3 = 3π₯ + 7
8π¦ β 3 = 6π¦ + 7
10π₯ + 3 = 7π₯ β 3
π₯ =2βπ₯
2π§ = 9 β π§
9π‘ = 99 β 2π‘
π₯ + 6 = 2π₯ β 6
7π¦ + 1 = 9π¦ + 5
5 π₯ β 2 = 30
2 π₯+1 =9+π₯
4 π₯ β 1 = 3π₯ β 1
14
3 π₯β2 =3+π₯
15
5 2π₯ + 3 = 20
π =?π
π =?βπ
π =?π
π =?π
π =? βπ
π =?π
π =?π
π =?π
π =?ππ
π =?βπ
π =?π
π =?π
π =?π
π
π =?
π
π
π =?
π
16
17
18
19
20
21
22
N
ππ
10 β 5 π₯ β 2 = 3 + 4π₯
π =?
π
9 β 4π₯ = 4 β 9π₯
π =?βπ
π
2 3π₯ + 1 = 3 + 2 2π₯ β 1
π = ?β
π
π
3 + 2 3π₯ + 1 = 7 + π₯
π =?
π
2β π₯β2 =π₯β 2βπ₯
π =?π
6+3 π¦β2 =5 π¦+4
π =?βππ
π₯ β 2 2 β π₯ = 3 π₯ β 2 β π₯ π =?βπ
1
3
1
π
π₯ + π₯ β 2 = 2π₯ β 3
π =?
2
4
3
π
GCSE: Forming and Solving
Equations
Skipton Girlsβ High School
RECAP :: Forming/Solving Process
Worded problem
[JMC 2008 Q18] Granny swears that she is getting younger. She has
calculated that she is four times as old as I am now, but remember that 5
years ago she was five times as old as I was at that time. What is the sum of
our ages now?
Stage 1: Represent
problem algebraically
Stage 2: βSolveβ equation(s)
to find value of variables.
Let π be my age and π
be Grannyβs age.
π β 5 = 5π β 25
π = 5π β 20
5π β 20 = 4π
π = 20
β΄ π = 80
π = 4π
π β 5 = 5(π β 5)
We previously learnt how to form expressions given a worded context.
Weβll learn how to actually solve these equations formed now! Weβll first
focus on problems where the expressions have already been specified.
Example
The angles of a triangle are as
pictured. Determine π₯.
π + ππ
ππ
Step 1: Think of a sentence which would have the
word βisβ in it. Write each part of sentence as an
algebraic expression, with βisβ giving =.
π β ππ
Step 2: Solve!
Sentence with isβisβ in
it?
βSum of angles
180°β
Expr
π₯ + 10 +
2π₯1?
+ π₯ β 50 = 180 Expr 2?
4π₯ β 40 = 180
4π₯ = 220 Solve!
π₯ = 55
Another Example
π₯+4
3
π₯
10
The rectangle and triangle have the same
area. Determine the width of the rectangle.
βArea βisβ
of triangle
is area of rectβ
sentence?
Expr
Expr 2?
3 π₯ 1?
+ 4 = 5π₯
3π₯ + 12 = 5π₯
12 = 2π₯
π₯ =Solve!
6
Therefore width of
rectangle is 6 + 4 = 10
1 The following diagram shows the angles
3π₯
2π₯ + 10
2π₯ + 20
120
βTotal angle
is 360β
Expr 2?
Expr
1?
ππ + ππ + ππ + ππ + ππ + πππ = πππ
ππ + πππ = πππ
ππ = Solve!
πππ
π = ππ°
2 The area of the triangle is 1 more than the
π₯+4
5
N
π₯
12
area of the parallelogram. Determine π₯.
Area1?
of π«
Expr
is 1 more than area of par
ππ = π +
ππ + ππ + π = ππ
π = ππSolve!
Exprπ 2?
π+π
[JMO 1999 A9] Skimmed milk contains 0.1% fat and pasteurised whole milk contains
4% fat. When 6 litres of skimmed milk are mixed with π litres of pasteurised whole
milk, the fat content of the resulting mixture is 1.66%. What is the value of π?
Fat content of skimmed + Fat content of pasteurised = fat content of mixture
?
6 × 0.001
+ 0.04π
= 0.0166 π + 6
β
π=4
Exercise 3
(See printed sheet)
1 The ages of three cats are π, π + 4 and 2π β 3.
7 [JMC 1998 Q18] The three angles of a
Their total age is 21. Determine π.
ππ + π = ππ
π=π
triangle are π₯ + 10 °, 2π₯ β 40 °,
3π₯ β 90 °. Which statement about the
triangles is correct? It is:
A right-angled isosceles
B right-angled, but not isosceles
C equilateral
D obtuse-angled and isosceles
E none of A-D
Solution: C
?
2 Three angles in a triangle are π₯, 2π₯ + 20° and
7π₯ β 10°. What is π₯?
?
π = ππ°
3 Two angles on a straight line are π₯ + 30° and
2π₯ β 60°. What is π₯?
?
π = ππ°
?
4 The perimeter of this rectangle is 44. What is π₯?
Solution: 7
2π₯ + 3
?
8
The following triangle is isosceles.
Determine its perimeter (by first
determining π₯).
π₯β2
5 The area of this rectangle
is 48. Determine π₯.
?
π=π
3π₯ + 10
π₯+5
π₯+5
4
6 An equilateral triangle has lengths
3π₯ + 6, 5π₯, 5π₯. What is π₯?
π=π
?
5π₯ β 50
?
ππ + ππ = ππ β ππ β π = ππ°
Perimeter = ππ + πππ + πππ = πππ
Exercise 3
9
(See printed sheet)
[JMO 2013 A7] Calculate the value of π₯
in the diagram shown?
Solution: 36
?
10 In the following diagram,
π΄π· = π΅π·, π΄π΅ = π΄πΆ, β πΆπ΄π· = 7π₯ and
β π΄π΅π· = 9π₯. Determine π₯.
π΄
7π₯
π·
9π₯
π΅
πΆ
ππ + ππ + ππ = πππ
?
π=π
11
A very large jug of π₯ litres of orange
squash is 10% concentrate (the rest
water). When 5 litres of concentrate
is added the jug is now 12%
concentrate.
a. Form an equation in terms of
π₯ (by considering the
amount of concentrate we
have).
π. ππ + π = ?
π. ππ(π + π)
b. Hence determine π₯.
π. ππ + π = π. πππ + π. π
?
π. πππ = π. π
π = πππ ππππππ
Tougher Examples
[JMC 2013 Q7] After tennis training, Andy collects twice as
many balls as Roger and five more than Maria. They collect
35 balls in total. How many balls does Andy collect?
Let π be the number of balls Roger collects.
Then Andy collects ππ balls and Maria collects ππ β π.
Total balls collected:
π + ππ + ππ
? β π = ππ
ππ β π = ππ
π=π
So Andy collected π × π = ππ balls.
[TMC Regional 2014 Q9] In a list of seven consecutive numbers a
quarter of the smallest number is five less than a third of the largest
number. What is the value of the smallest number in the list?
Let smallest number be π. Then numbers are π, π + π, π + π, π + π, π + π, π + π, π + π
π
π
Therefore π π = π π + π β π
π
π
π = π?+ π β π
π
π
π
π = π β π = ππ
ππ
So smallest number is 36.
1 The length of the rectangle is three
times the width. The total perimeter is
56m. Determine its width.
Let π be the width of the rectangle.
Then the length is ππ.
Then the perimeter is:
π + ππ +?π + ππ = ππ
ππ = ππ
π=π
β΄ The width is 7m.
Bro Reminder: You
should usually start
with βLet β¦β
2 In 4 years time I will be 3 times as old as I was 10 years ago.
How old am I?
Let π be my age.
Then π + π = π π β ππ
π+π=
? ππ β ππ
ππ = ππ
π = ππ
Exercise 4 (Maths Challenge Questions)
1 The sum of 5 consecutive numbers
is 200. What is the smallest
number?
Solution: 38 (a quick non-algebraic
method is to realise the middle
? of the five
number is the average
numbers, i.e. 40)
2
3
In 5 years time I will be 5 times as
old as I was 11 years ago. Form a
suitable equation, and hence
determine my age.
π + π = π π β ππ
π = ππ ?
In 6 years time I will be twice as old
as I was 8 years ago. Determine my
age.
π+π=π πβπ
?
π = ππ
4
I have three times as many cats as Alice
but Bob has 7 less cats than me. In total
we have 56 cats. How many cats do I
have?
Solution: 27
?
5
Bob is twice as old as Alice at the
moment. In 4 years time their total age
will be 71. What is Aliceβs age now?
4 years time: Alice π + π Bob ππ + π
ππ + π = ππ? β π = ππ
6
[TMC Final 2012 Q1] A Triple Jump
consists of a hop, step and jump. The
length of Keithβs step was three-quarters
of the length of his hop and the length of
his jump was half the length of his step.
If the total length of Keithβs triple jump
was 17m, what was the length of his
hop, in metres?
Solution: 8 metres
?
Exercise 4
7
[JMO 2003 A6] Given a βstartingβ number, you double it
and add 1, then divide the answer by 1 less than the
starting number to get the βfinalβ number. If you start
final number is 3. What starting number gives the final
number 4?
ππ + π
=π
πβπ
ππ + π = π π β π
?
π
π=
π
Exercise 4
8
[JMO 2012 A3] In triangle π΄π΅πΆ, β πΆπ΄π΅ = 84°; π·
is a point on π΄π΅ such that β πΆπ·π΅ = 3 × β π΄πΆπ·
and π·πΆ = π·π΅. What is the size of β π΅πΆπ·?
10 [JMC 2012 Q24] After playing 500
games, my success rate in Spider
Solitaire is 49%. Assuming I win every
game from now on, how many extra
games do I need to play in order that my
success rate increases to 50%?
A 1
B 2 C 5 D 10
E 50
Let π be the number of extra games
played. Then, giving 245 games were
won before:
πππ + π = π. π πππ + π
π = ππ
Solution: Let β π¨πͺπ« = π, then β πͺπ«π© = ππ
and β π¨π«πͺ = πππ β ππ. Angles in π«π¨πͺπ«:
ππ + πππ β ππ + π = πππ
π = ππ°
πππβπππ
= ππ°
?
?
(Note: itβs easier to just exploit the fact itβs multiple
choice and try the options!)
π
9
[JMO 2005 A8] A large container holds 14 litres
of a solution which is 25% antifreeze, the
remainder being water. How many litres of
antifreeze must be added to the container to
make a solution which is 30% antifreeze?
Let π be the amount of antifreeze added in
litres. 25% of 14 is 3.5. Thus:
π. π + π = π. π ππ + π
π=π
?
N1
[JMO 2008 A9] In the diagram, πΆπ· is the
bisector of angle π΄πΆπ΅.
Also π΅πΆ = πΆπ· and π΄π΅ = π΄πΆ. What is
the size of angle πΆπ·π΄?
Let β π«πͺπ© = π. Filling in the angles
using the information we find
angles in π«π«π¨πͺ are π, πππ β ππ
and πππ β ππ. This gives π = ππ° .
β΄ β πͺπ«π¨ = πππ β π × ππ
= πππ° (Note: this is not intended to
?
be a full proof!)
Exercise 4
N2 [JMO 2010 A10] Inn the diagram, π½πΎ
and ππΏ are parallel. π½πΎ = πΎπ = ππ½ =
ππ and πΏπ = πΏπ = πΏπΎ. Find the size
of angle π½ππ.
60°
π₯
π₯ β 60°
60°
300
β 2π₯
π₯
π₯
π₯ β 60°
π₯
Since π±π² and π΄π³ are parallel, β π²π±π΄ and
β π±π΄π³ are cointerior so add to πππ°.
If we let β π²πΆπ³ = π, we can eventually
find the angles as pictured.
π β ππ + ππ + π β ππ + π = πππ
π = ππ° β΄ β π±π΄πΆ = ππ β ππ = ππ°
?
N3
[JMO 2013 B2] Pippa thinks of a number. She adds
1 to it to get a second number. She then adds 2 to
the second number to get a third number, adds 3
to the third to get a fourth, and finally adds 4 to
the fourth to get a fifth number.
Pippaβs brother Ben also thinks of a number but he
subtracts 1 to get a second. He then subtracts 2
from the second to get a third, and so on until he
has five numbers.
They discover that the sum of Pippaβs five
numbers is the same as the sum of Benβs five
numbers. What is the difference between the two
numbers of which they first thought?
Let π be Pippaβs first number. Then her numbers
are π, π + π, π + π, π + π, π + ππ. Sum is ππ +
ππ.
Let π be Benβs first number. Then his numbers are
π, π β π, π β π, π β π, π β ππ. Sum is ππ β ππ
Weβre told ππ + ππ = ππ β ππ and the
difference between their two starting numbers is
π β π. Rearranging:
ππ + ππ = ππ β ππ
ππ β ππ = ππ
β΄πβπ=π
?
Exercise 4
(See printed sheet)
N4 [JMO 2005 B4] In this figure π΄π·πΆ is a straight
line and π΄π΅ = π΅πΆ = πΆπ·. Also, π·π΄ = π·π΅.
Find the size of β π΅π΄πΆ.
(Full proof needed)
Full proof: