Level 8
Algebra
Inequalities
Graphing Linear Inequalities in 2 Variables
2x + 3 ≥ 1 is a linear inequality in one variable, x. A linear inequality in one variable can be graphed on a
number line.
y < 2x + 3 is a linear inequality in two variables, x and y. A linear inequality in two variables can be
graphed on a set of axes.
First draw the graph of y = 2x + 3
Graphing the line for the second
1
inequality y ≤ - x + 4
5
This is a less than inequality, so shade
below the line.
The last inequality is x > 0 so shade to
the right of the line.
This last inequality needs a dashed
line because this isn't an "or equal to"
inequality, so the boundary (the line)
isn't included in the solution
The "solution" of the system is the
region where all the inequalities work,
the region where all three individual
solution regions overlap. In this case,
the solution is the shaded part in the
middle
Having already graphed the "or equal to" part
now to do the "y less than" part. In other words,
this is where you need to shade one side of the line
or the other. If I need y LESS THAN the line, do I
want ABOVE the line, or BELOW?
Naturally, I want below the line. So I shade it in
Writing Inequalities for Given Regions
Find the equations of the boundary lines. Choose a point within the given region.
Now test the co-ordinates of this point in an inequality formed from the equation of each boundary line.
Solving Quadratic Inequalities
Example
Solve the inequality x2 > 16
Answer
Finding Regions where a Number of Inequalities are True
Find the end points of the regions by solving x2 = 16
A "system" of linear inequalities is a set of linear inequalities that you deal with all at once.
Example.
Solve the following system:
2x - 3y < 12
x + 5y < 20
x>0
Answer
Rearrange the inequalities to have y on one side.
2
y≥ x-4
If x2 = 16 the x = 4 and x = -4
3
1
y≤- x+4
5
x>0
Since the inequality sign is > the symbol o is placed
on these end points.
o
o
-4
4
Now test a point in each of the three regions.
Choose a point to the left of –4 say –5. In x2 > 16, replace x with –5.
Is (-5)2 > 16? That is, is 25 > 16? Since the answer to this question is yes, solutions for the inequality are
in the region to the left of –4.
o
o
So far we have
-4
4
Then graph each individual inequality, and then finding
the overlaps of the various solutions.
Choose a point between –4 and 4, say 3. In x2 > 16 replace x with 3.
Is (3)2 = 16? That is, is 9 > 16? Since the answer to this question is no, solutions for the inequality are not
in the region between –4 and 4.
The line for the first inequality in the
2
above system, y ≥ x – 4, looks like this
Choose a point to the right of 4, say x = 6. In x2 > 16, replace x with 6.
Is 62 > 16? That is 36 > 16? Since the answer to this question is yes, solutions for the inequality are in the
region to the right of 4.
3
This inequality is a "greater than"
inequality, so shade above
the line.
We have now
o
o
-4
4
From this number line graph we see that the solutions for x2 > 16 are x < -4 and x > 4
Formulae
Using Formulae
Transforming Formulae
Example
𝑠
𝑠
The formula v = can be rearranged as s = vt or as t = .
𝑠
𝑡
𝑠
a. If r = 2.4,
𝑣
If a variable such as s is to be the subject of a formula then s must be written on its own on the left
hand side.
𝑠
When we arrange a formula such as v = , so that a variable other than v is the subject of the formula
𝑡
we are said to have transformed the formula.
We can use a flowchart to transform formulae. On the flowchart we make use of inverse operations.
Inverse operations are operations which undo each other.
This can be done using a flowchart as shown below.
r
b
𝐴
ℎ
cube
3 × 20.6
Take the cube root
4×𝜋
3
xh
bh
Since r must be positive, r = �
÷h
Begin with A
Example
Answer
ℎ
4
�
Make b the subject of the formula A = bh
4
V = x 𝜋 x 2.43
3
= 57.9 to 3 s.f.
b. If V = 20.6, 20.6 = x 𝜋 x r3.
3
To find r we need to rearrange this as r = . . .
3
Example
𝐴
Find r if V = 20.6
Answer
Written as t = , t is the subject of the formula.
The b =
Find V if r = 2.4
b.
Written as s = vt, s is the subject of the formula.
Begin with
3
a.
𝑣
Written as v = , v is said to be the subject of the formula
𝑡
4
V = πr3
Sometimes we need to rearrange formula
Multiply by 4 x 𝜋
Divide by 4 x 3.142
Divide by 3
Multiply by 3
3 × 20.6
4×𝜋
Give the answer in its simplest form
(-4)2 = (-4) x (-4)
= 16
Answer
Since x is to be expressed in terms of y and a but not in terms of t we must eliminate t. We do this as
follows
Squaring and taking the square root are inverse operations.
Rearrange y = 2at to get t =
Since both 42 = 16 and (-4)2 = 16 then both 4 and –4 are square roots of 16.
𝑦
2𝑎
Substitute t into x = at2 to get
Transforming Formulae involving Powers and Roots
Sometimes we know that the value of a variable cannot be negative. For instance if we make r (the
radius) the subject of the formula A = πr2 (The formula for the area of a circle) we know that r must
always be positive. In cases such as this when we take the square root we take just the positive
square root.
𝑦
x = a ( )2
2𝑎
x=ax
x=
Example
𝑎𝑦 2
4𝑎2
𝑦2
4𝑎2
𝑦2
V = πr h is the formula for the volume of a cylinder of radius r and height h.
x=
Make r the subject of the formula
Using function Notation
Answer
f(x) means “a function of x”. That is an expression in which the variable is x.
2
4𝑎
f(4) means “the value of the function when x is replaced by 4”
Using a flowchart
𝑉
20.6
x = at2, y = 2at
That is both 42 and (-4)2 have the answer of 16
𝜋ℎ
x 𝜋 x r3
Express x in terms of y and a.
42 = 4 x 4
= 16
±�
3
Expressing One Formula in terms of Another
Powers and Roots
Begin with r
4
Square
Multiply by πh
2
r
𝑉
Square Root
𝜋ℎ
Since r must be positive, r = �
𝑉
𝜋ℎ
πr
2
Begin with V
Example
𝑥+3
f(x) =
. find the value of f(5)
𝑥−2
Answer
𝑥+3
f(x) =
f(5) =
=
𝑥−2
5+3
5−2
8
3
=2
2
3
Expanding Brackets and Factorisation
Example
Expanding Brackets
Factorise
To expand an expression such as 2(3a – 4) we remove the brackets. We do this by multiplying everything
inside the brackets by the number outside.
a
x2 + 9x + 14
b
x2 - x – 30
That is 2(3a – 4) is expanded as 6a – 8
c
p2 + p – 30
The rectangle method of expanding is
Answers
Example
a
x2 + 9x + 14 = (x + 2)(x + 7)
Multiply out 3(x + 2a)
b
x2 - x – 30 = (x + 5)( x – 6)
since 5 x (-6) = -30 and 5 + (-6) = -1
c
p2 + p – 30 = (p – 5)(p + 6)
since –5 x 6 = -30 and –5 + 6 = 1
Answer
x
3
2a
3x
3(x + 2a) = 3x + 6a
6a
since 7 x 2 = 14 and 7 + 2 = 9
Quadratic Equations
Example
Example Find the values of n for which (n + 3)(n – 7) = 0
Expand and simplify 9x – 5 – 2(3x – 2)
Answer
Answer
Either
9x – 5 – 2(3x – 2) = 9x – 5 – 6x + 4
= 3x – 1
or
Factorising
To factorise an expression we write the expression with brackets. That is factorising is the reverse of
expanding
For instance 2x + 6 is the expanded form of 2(x + 3) while 2(x + 3) is the factorised form of 2x + 6
Notice that in 2x + 6 both 2x and 6 have a factor of 2. That is 2 is a common factor of 2x and 6
When we factorise 2x + 6 as 2(x + 3) the common factor is placed outside the bracket
Always factorise completely. If the expression in the brackets has a common factor then the factorising is not
complete
For instance 24x + 30 may be factorised as 2(12x + 15). Since 12x and 15 have a common factor of 3 the
factorising is not complete. 24x + 30 is completely factorised as 6(4x + 5)
It is a good idea to check your factorising by expanding. Suppose you have factorised 2n – 4 as 2(n – 4). By
expanding 2(n – 4) to 2n – 8 you would see that you had factorised incorrectly.
Further Expanding
b.
(a + 2b)(3a – b)
(-3 from both sides)
(+7 to both sides)
Example
Solve the quadratic equation x2 – 5x + 6 = 0
Answer
x2 – 5x + 6 = 0
(x - 2)(x – 3) = 0
Either
x-2=0
x = 0 + 2 (+ 2 to both sides)
x=2
Problems Involving Quadratic Equations
To solve problems involving quadratic equations take the following steps
Step 3 Solve the equation
Step 4 Check to see if the answers fit the question. Sometimes one of the
answers will not be possible
Answer
Using the rectangle method the expansions can be found from the following diagrams
Example
The product of two consecutive integers is 72.
a.
x
3x
3x
-2
-2x
2
-5
-b
Find the consecutive integers
-15x
a
2
3a
-ab
Answer
10
2b
6ab
-2b2
(3x – 2)(x – 5)
= 3x2 – 2x – 15x + 10
= 3x2 – 17x + 10
(a + 2b)(3a – b)
= 3a2 + 6b – ab – 2b2
= 3a2 + 5b – 2b2
or
Step 2 Rewrite the equation so 0 is on the right hand side
Expand and simplify
(3x – 2)(x – 5)
=0
= 0 -3
= -3
=0
=0+7
=7
Step 1 Write an equation
Example
a.
n+3
n
n
n–7
n
n
b.
3a
Let the first number be n and the second number be n + 1
Factorising Quadratic Expressions
n (n + 1) = 72
n2 + n = 72
n2 + n – 72 = 0
Factorising gives (n + 9)(n – 8) = 0
So n equals either 8 or -9
If n = -9 then n + 1 = -8
To expand we write without brackets. For instance (x + 5)(x – 3) is expanded as x2 + 2x – 15.
If n = 8 then n + 1 = 9.
To factorise we write with brackets. For instance x2 + 2x – 15 is factorised as (x + 5)(x – 3)
Hence the consecutive integers are
–9 and –8 or 8 and 9.
x-3=0
x = 0 + 3 (+ 3 to both sides)
x=3