Examples
:
1)
(
x
+ 1)(
x
+ 2)
=
x
(
x
+ 2) + 1(
x
+ 2)
or
(
x
+1)(
x
+ 2)
=
x
2
+ 2 + 2
x
+
x
=
x
2
+ 3
x
+2
or
x
1
x
x
2
x
2
2
x
2
2)
(
x
 2)(2
x
+ 3)
=
x
(2
x
+ 3)  2(2
x
+3)
= 2
x
2
+ 3
x
– 4
x
 6
= 2
x
2
–
x
– 6
or
(
x
 2)(2
x
+ 3) = 2
x
2
– 6 + 3
x
– 4
x
= 2
x
2
–
x
– 6
or
x
2
2
x
2
x
2
4
x
4
(
x
+1)(
x
+ 2)
=
x
2
+ 2
x
+
x
+ 2
=
x
2
+ 3
x
+2
(2
x
+3)(
x
 2)
= 2
x
2
+ 3
x
 4
x
 6
= 2
x
2

x
 6
3
3
x
6
5
EXERCISE A
Multiply out the following brackets and simplify.
1
7(4
x
+ 5)
2
3(5
x
 7)
3
5
a
– 4(3
a
 1)
4
4
y
+
y
(2 + 3
y
)
5
3
x
– (
x
+ 4)
6
5(2
x
 1) – (3
x
 4)
7
(
x
+ 2)(
x
+ 3)
8
(
t
 5)(
t
 2)
9
(2
x
+ 3
y
)(3
x
– 4
y
)
10
4(
x
 2)(
x
+ 3)
11
(2
y
 1)(2
y
+ 1)
12
(3 + 5
x
)(4 –
x
)
Two Special Cases
A Perfect Square:
A Difference of two squares:
(
x
+
a
)
2
= (
x + a
)(
x + a
)
=
x
2
+ 2
ax
+
a
2
(
x  a
)(
x + a
)
=
x
2
–
a
2
(2
x
 3)
2
= (2
x
– 3)(2
x
– 3) = 4
x
2
– 12
x
+ 9
(
x
 3)(
x
+ 3)
=
x
2
– 3
2
=
x
2
– 9
EXERCISE B
Multiply out
1.
(
x
 1)
2
2.
(3
x
+ 5)
2
3.
(7
x
 2)
2
4.
(
x
+ 2)(
x
 2)
5.
(3
x
+ 1)(3
x
 1)
6.
(5
y
 3)(5
y
+ 3)
Harder expansions
Use the same methods but with more multiplications and more terms in the answer
EXERCISE C
Multiply out
1.
(
x
−
2
)(
x
2
+
x
+
1
)
2.
(
x
+
1
)(
x
+
2
)(
x
+
3
)
3.
(
x
−
2
)(
x
2
−
x
−
1
)
4.
(
x
−
4
)(
x
−
1
)(
x
+
1
)
5.
(
2
x
−
1
)(
2
x
2
−
3
x
+
5
)
6.
(
x
−
2
)(
x
−
3
)(
x
+
1
)
6
Chapter 2:
LINEAR EQUATIONS
When solving an equation, you must remember that whatever you do to one side must also be
done to the other.
You are therefore allowed to
add the same amount to both side
subtract the same amount from each side
multiply the whole of each side by the same amount
divide the whole of each side by the same amount.
If the equation has unknowns on both sides, you should collect all the letters onto the same side
of the equation.
If the equation contains brackets, you should start by expanding the brackets.
A linear equation is an equation that contains numbers and terms in
x
.
A linear equation does not contain any
2
3
or
x
x
terms.
Example 1
:
Solve the equation
64 – 3
x
= 25
Solution
:
There are various ways to solve this equation.
One approach is as follows:
Step 1
:
Add 3
x
to both sides (so that the
x
term is positive):
64 = 3
x
+ 25
Step 2
:
Subtract 25 from both sides:
39 = 3
x
Step 3
:
Divide both sides by 3:
13 =
x
So the solution is
x
= 13
Example 2
:
Solve the equation 6
x
+ 7 = 5 – 2
x
.
Solution:
Step 1
: Begin by adding 2
x
to both sides
8
x
+ 7 = 5
(to ensure that the
x
terms are together on the same side)
Step 2
:
Subtract 7 from each side:
8
x
= 2
Step 3
:
Divide each side by 8:
x
= ¼
Exercise A
:
Solve the following equations, showing each step in your working:
1)
2
x
+ 5 = 19
2)
5
x
– 2 = 13
3)
11 – 4
x
= 5
7
4)
5 – 7
x
= 9
5)
11 + 3
x
= 8 – 2
x
6)
7
x
+ 2 = 4
x
– 5
Example 3
:
Solve the equation
2(3
x
– 2) = 20 – 3(
x
+ 2)
Step 1
:
Multiply out the brackets:
6
x
– 4 = 20 – 3
x
– 6
(taking care of the negative signs)
Step 2
:
Simplify the right hand side:
6
x
– 4 = 14 – 3
x
Step 3
:
Add 3x to each side:
9
x
– 4 = 14
Step 4
:
Add 4:
9
x
= 18
Step 5
:
Divide by 9:
x
= 2
Exercise B:
Solve the following equations.
1)
5(2
x
– 4) = 4
2)
4(2 –
x
) = 3(
x
– 9)
3)
8 – (
x
+ 3) = 4
4)
14 – 3(2
x
+ 3) = 2
5)
5(2
x
– 4) = –8(3 –
x
)
6)
what is wrong here?
3
x
– 2(
x +
3) =
x
 11
7)
3(6
x
– 2) = 2(4x – 7) + 3
8)
2(3 –7
x
) = 3(10  2
x
)
8
Equations containing Fractions
When an equation contains a fraction, the first step is usually to multiply through by the
denominator of the fraction.
This ensures that there are no fractions in the equation.
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 Fall '13
 Quadratic equation, Elementary algebra