Transcript Algebraic Expressions

Algebraic Expressions
Writing expressions
Using letter symbols for unknowns
In algebra, we use letter symbols to stand for numbers.
These letters are called unknowns or variables.
Sometimes we can work out the value of the letters and
sometimes we can’t.
For example,
We can write an unknown number with 3 added on to it as
n+3
This is an example of an algebraic expression.
Writing an expression
Suppose Jon has a packet of biscuits and he
doesn’t know how many biscuits it contains.
He can call the number of biscuits in the full
packet, b.
If he opens the packet and eats 4 biscuits, he can write an
expression for the number of biscuits remaining in the
packet as:
b–4
Writing an equation
Jon counts the number of biscuits in the packet
after he has eaten 4 of them. There are 22.
He can write this as an equation:
b – 4 = 22
We can work out the value of the letter b.
b = 26
That means that there were 26 biscuits in the full packet.
Writing expressions
When we write expressions in algebra we don’t usually use
the multiplication symbol ×.
For example,
5 × n or n × 5 is written as 5n.
The number must be written before the letter.
When we multiply a letter symbol by 1, we don’t have to
write the 1.
For example,
1 × n or n × 1 is written as n.
Writing expressions
When we write expressions in algebra we don’t usually use
the division symbol ÷. Instead we use a dividing line as in
fraction notation.
For example,
n
n ÷ 3 is written as
3
When we multiply a letter symbol by itself, we use index
notation.
n squared
For example,
n × n is written as n2.
Writing expressions
Here are some examples of algebraic expressions:
n+7
Add 7 to a number n
5–n
Decrease n from a number 5
2n
Double the number n or 2 × n
6
n
Divide 6 by a number n
4n + 5
Add 5 to 4 lots of a number n
n3
a number n multiplied by itself 3 times
or n × n × n
3 × (n + 4)
or 3(n + 4)
a number n is increased by 4 and then the
Result is times 3.
Writing expressions
Miss Green is holding n number
of cubes in her hand:
Write an expression for the number of cubes in her hand if:
She takes 3 cubes away.
n–3
She doubles the number of
cubes she is holding.
2n
or
2×n
Translate the following into algebraic expressions:
①7 more than x
②x is decreased by two.
③Double x and add 11
④One third of a number
⑤3 less than double a number.
⑥Two numbers have sum 8. If one of them is x,
then the other is ……….
⑦Two numbers are in the ratio 1:3. If the larger one
is x then the smaller one is ……..
⑧Three consecutive integers in ascending order are
x, …… and ……..
⑨The sum of two numbers is 7. One of the numbers
is a. What is the other number ( in terms of a) ?
⑩The total value of x, 20 cent coins and (x-2), 50
cent coins
Algebraic Expressions
Forming and Solving equations
Constructing an equation
Ben and Lucy have the same number of sweets.
Ben started with 3 packets of sweets and ate 11 sweets.
Lucy started with 2 packets of sweets and ate 3 sweets.
How many sweets are there in a packet?
Let’s call the number of sweets in a packet, n.
We can solve this problem by writing the equation:
3n – 11 = 2n – 3
The number of
Ben’s sweets
is the
same as
the number of
Lucy’s sweets.
Solving the equation
Let’s solve this equation by transforming both sides of the
equation in the same way.
3n – 11 = 2n – 3
Start by writing the equation down.
-2n
Subtract 2n from both sides.
-2n
n – 11 = –3
+11
+11
n = 8
Always line up the equals signs.
Add 11 to both sides.
This is the solution.
We can check the solution by substituting it back into the
original equation:
3 × 8 – 11 = 2 × 8 – 3
Constructing an equation
I’m thinking of a number.
When I multiply the number by 4, I get the same answer
as adding 9 to the number.
What number am I thinking of?
Let’s call the unknown number n.
We can solve this problem by writing the equation:
4n
The number
multiplied by 4
=
n+9
is the
same as
the number
plus 9.
Solving the equation
Let’s solve this equation by transforming both sides of the
equation in the same way.
4n = n + 9
-n
-n
Start by writing the equation down.
Subtract n from both sides.
3n = 9
Always line up the equals signs.
÷3
Divide both sides by 3.
÷3
n = 3
This is the solution.
We can check the solution by substituting it back into the
original equation:
4 × 3 =3 + 9
Constructing an equation
Find the value of x.
(65 – 2x)o
(2x + 5)o
Remember, vertically opposite angles are equal.
We can solve this problem by writing the equation:
2x + 5 = 65 – 2x
Solving the equation
Let’s solve this equation by transforming both sides of
the equation in the same way.
2x + 5 = 65 – 2x
+2x
+2x
Add 2x to both sides.
4x + 5 = 65
-5
4x
-5
=
÷4
x
60
÷4
=
Subtract 5 from both sides.
15
Divide both sides by 4.
This is the solution.
Check:
2 × 15 + 5 =65 – 2 × 15
Rectangle problem
The area of this rectangle is 27 cm2.
Calculate the value of x and
use it to find the height of the
rectangle.
8x – 14
2x + 1
Opposite sides of a
rectangle are equal.
We can use this fact to write
an equation in terms of x.
Rectangle problem
The area of this rectangle is 27 cm2.
8x – 14 = 2x + 1
–2x
–2x
6x – 14 = 1
8x – 14
2x + 1
+14
+14
6x = 15
÷6
÷6
x = 2.5
If x = 2.5 we can find the height of the rectangle using
substitution:
8 × 2.5 – 14 = 20 – 14 = 6 cm
Rectangle problem
The area of this rectangle is 27 cm2. What is its width?
y
Let’s call the width of the
rectangle y.
8x – 14
2x + 1
The dimensions of the rectangle
are 6 cm by 4.5 cm.
If the height of the
rectangle is 6 cm and the
area is 27 cm2 then we
can find the width by
writing the equation:
6y = 27
÷6
÷6
y = 4.5
Select the correct equation
Veronica has 58 cents
and buys 4 chocolate
bars.
Thomas has $1 and
buys 7 chocolate bars.
They both receive the same amount of change.
If c is the cost of one chocolate bar, which equation could
we use to solve this problem?
A: 4c + 58 = 7c + 100
B: 58 – 4c = 1 – 7c
C: 58 – 4c = 100 – 7c
D: 4c – 58 = 7c – 1
In pairs now do:
Ex 11C ( pg 249) – 2,4,5,7 and 9
Homework :
Ex 11B ( pg 246) - 1,2 (only a’s and c’s ), 3a
Ex 11 C ( pg 249) – 1, 3, 6 and 8
Remember:
Unit 2 Test : December 7th 2014