Download Chapter 6 – Solving equations Defining an equation 6.1 Up to now

Chapter 6 – Solving equations
6.1
Defining an equation
Up to now we have looked mainly at expressions. An expression is
an incomplete statement and has no equal sign. Now we want to
look at equations. An equation has an = sign and a letter
(variable) in it. It is a complete statement.
2x + 3 is an expression or sometimes we call it a function.
But 2x + 3 = 7 is an equation. It has an = sign in it and the
unknown variable, x. This statement is only true for some specific
value of x. In this case x=2 because 2(2) + 3 = 7. In other words
the LHS will only equal the RHS when x is 2.
Now consider 2x + 3x = 5x. In this case although the statement
has an equal side the LHS = RHS for all values of x. We call this an
identity.
With the RHS we can see it is not an identity. So it is also an
equation but this has two unknown variables, x and y in it.
E6.1
A6.1
6.2
2x + 4y = 5 – 2x is an equation with two unknowns, x and y
although the x appears twice in the equation.
(i)
Is 2x + 3 = x + 4 an equation? How many variables?
(ii)
Is 5x + 2x = 7x an equation? Explain.
(iii)
Is 5x + 3y + 2z = 10 an equation? How many variables?
(iv)
Is 7x2 - 3 = 5x an equation? How many variables?
(v)
Is 7x2 – 3 an equation? Explain.
(i)
2x + 3 = x + 4:Yes, an equation with 1 variable.
(ii)
5x + 2x = 7x: No, it is an identity.
(iii)
5x + 3y + 2z = 10:Yes, an equation with 3 variables.
(iv)
7x2 - 3 = 5x:Yes, an equation with 1 variable.
(v)
7x2 – 3: No, it is an expression.
Compare x + 3 =5 with x+x=2x.
x + 3 = 5 is an equation because it is only true for x = 2.
Now consider
x+ x = 2x.
This looks like an equation but is strictly not an equation.
Consider the LHS: x + x is 2x. This the same as the RHS. So
this equation is true for any or all values of x. It is an identity
rather than an equation.
So when the LHS can be shown to equal the RHS, we call it an
68
E6.2
A6.2
6.3
identity.
Is x2 + 2x + 1 = (x+1)2 an equation? Consider the LHS and
RHS.
It is an identity as the RHS = LHS (expand the RHS as a perfect
square)
We will be concerned mainly with equations, rather than identities.
We want to learn how to solve an equation. This means finding the
one (or 2) value(s) for the unknown variable that makes the
statement true i.e. LHS = RHS.
We have seen that we can test for a solution by verification.
For example: Test if x=2 is a solution of the equation
3x + 4 = 5x.
We substitute x = 2 in the LHS viz:
LHS = 3(2) + 4 = 6 + 4 = 10
E6.3
A6.3
6.4
Now substitute x = 2 in the RHS viz:
5(2) = 10 = LHS.
So LHS = RHS so x = 2 is a solution.
Test if x = 3 is a solution of the equation of the equation:
2x2 – 3 = 7x – 6.
2x2 – 3 = 7x – 6.
RHS= 2(32) – 3 = 15
LHS = 7(3) – 6 = 15
Yes x=3 is a solution.
But how do we solve an equation when we do not have a value to
substitute? We are going to learn some techniques to do this.
Consider the very simple equation: x + 7 = 10.
E6.4
We can ‘guess’ the result x = 3. So one way is to obtain a solution
is by inspection. This applies to easy, convenient cases only.
Solve by inspection : x – 12 = 8.
A6.4
6.5
x = 20
Let’s be more introspective:
Solve x + 15 = 24. By inspection we know x = 9 is the solution.
So we can write:
x + 15 = 24 Note: LHS has 2 terms and the RHS has 1 term.
69
x = 24 – 15 Note: LHS has 1 term and the RHS has 2 terms.
So x = 9.
Observe that the term 15 has moved over the = sign and has
changed sign.
E6.5
A6.5
This shows us that when we move a term over the = sign
it changes sign.
Solve: x + 23 = 70 by moving the constant term from the LHS to
the RHS.
6.6
x + 23 = 70
x = 70 – 23
So x = 47
Similarly with x – 12 = -2 we proceed:
E6.6
x = -2 + 12 = 10, moving the -12 term.
Solve x – 2.3 = 4.4
A6.6
6.7
E6.7
A6.7
6.8
Solve x – 2.3 = 4.4
x = 4.4 + 2.3 = 6.9
Now consider solving the equation:
3x = 15
We know the solution by inspection is x = 5.
Formally we write:
3x  15
15
x   5
3
Note the factor of 3 has moved from the numerator of the LHS to
the denominator of the RHS i.e. it has been ‘cross multiplied’.
Solve 5x = 100 by moving the factor 5 from the LHS to the RHS.
5x  100
100
x 
 20
5
x
Now solve  2 .
7
We get: x = 2(7) = 14
Another example:
Solve
70
3x
2
7
x 
2(7)
3
So when we move a factor over the = sign it is cross multiplied i.e.
moves from numerator(denominator) of one side of the = sign to
the denominator(numerator) of the other side.
E6.8
Solve (i) 3x = 12
x
(ii)  12
5
3x
(iii)
 12
5
A6.8
(i) 3x  12
x 
12
4
3
x
 12
5
 x  12(5)  60
3x
(iii)
 12
5
12(5) 4(5)
x 

 20
3
1
Layout of a solution to an equation
When we were dealing with expressions we ‘joined or connected’
each variant of the expression with an = sign.
e.g. 2x + 1 + x + 5 = 3x + 6.
(ii )
6.9
With equations we must not use the = sign or rather use it with
caution i.e. do not introduce a new = sign in addition to the one
that is already there.
For example in the equation of box 6.6 writing the following is not
correct:
3x  15
15
x  5
3
The first = sign of the second line is not appropriate because the
suggestion is that 15 = 5. Instead of = sign we use the therefore
symbol,  or the ‘implication arrow’  or i.e.
71
E6.9
Use any additionl, new = signs only between trivial
numerical values.
Replace any incorrect ‘=’ signs in the following:
2x + 7 – x = 27
= x = 27 – 7 = 20
A6.9
6.10
2x + 7 – x = 27
 x = 27 – 7 = 20
Further tools for solving equations
We will develop a technique that involves ‘Unpacking an
expression until we isolate the unknown variable’. What does this
mean?
Lets take a non mathematical example:
E6.10
A6.10
6.11
Suppose I (make) ‘pack’ a present as follows. I take a cricket ball,
wrap it in foil then place it in bubble wrap then place it in a box
and then seal the box with tape. To isolate the ball the present will
have to unpacked in the right order:
Remove the tape, then remove from the box then remove bubble
wrap and then finally remove the foil.
How do you undo: put my socks on then put more shoes on!
Remove shoes then socks.
Here is an equation with mixed mode expressions (addition with
multiplication) 7x + 11 = -10. To solve this equation we will need
to isolate (UNPACK) the x by moving both a term and a factor to
the RHS. But in the right order. What is the order?
The left hand side shows that the x has been multiplied by 7 first
and then has 7 added to it. So to unpack we first move the + 11
term from the LHS to the RHS and then move the factor 7 to the
LHS. This is shown below:
7x  11  - 10
 7x  - 10 - 11  - 21
(move term from LHS to RHS)
- 21
 -3
(move factor from LHS to RHS)
7
So x = 3 is the solution.
Check: LHS = 7(-3) + 11 = -21 + 11 = -10 = RHS.
x
E6.11
Are the = signs ok? Additional = signs in a line can be used to do
trivial simplifications.
Solve: 12x – 3 = 1
72
A6.11
6.12
12x  3  1
 12x  1  3  4
4
 x   13
12
(move term from LHS to RHS)
(move factor from LHS to RHS)
Now consider this case where the unknown appears more than
once in the equation: 3x + 7 = 5 – 2x.
Move terms so that the terms in the unknown, x are on one side of
the equal sign and the constant terms are on the other side:
3x  2x  5 - 7
 5x  - 2
x 
-2
5
-
(simplify both sides)
2
5
E6.12
Solve 7x + 9 = 3x – 7
A6.12
7x  9  3x - 7
 7x - 3x  - 7 - 9
 4 x  - 16
- 16
 -4
4
Track this example:
9 - 7x  3x - 1
 9  1  3x  7x
10  10x
x 1
x 
6.13
E6.13
It is Ok to have the unknown on the RHS if it is convenient.
Solve 4x – 8 = 7 + 15x by having the unknown terms on the RHS.
A6.13
4x - 8  7  15x
 -8 - 7  15x  4x
 - 15  11x
11
 x  - 15
6.14
Now consider 8 – 3x = 12 – 8x.
Note the coefficients of the x terms are both negative.
73
We can change the sign of each term of both sides of the
equation:
Change the sign of each term of 8 – 3x = 12 – 8x to get
- 8  3x  - 12  8x
12 - 8  8x - 3x
 4  5x
 x  54
E6.14
Solve 10 – 3x = 12 – 2x by changing the sign of each term first.
A6.14
10  3x  12  2x
 10  3x  - 12  2x
 -10  12  2x - 3x
 2  -x
6.15
x  2
Now consider the equation:
3(x – 5) = 20
We have a choice of unpacking as follows:
3(x - 5)  20
x - 5 
x 
20
3
20
20 5 20  15 35
2
5
 

 11
3
3 1
3
3
3
Or first removing the brackets viz:
3(x - 5)  20
 3x - 15  20
 3x  20  15  35
x 
E6.15
35
2
 11
3
3
Solve 5(x – 11) = 10 by unpacking directly and then by removing
brackets first.
74
A6.15
5(x - 11)  10
10
2
5
 x  2  11  13
 x - 11 
First removing the brackets:
5(x - 11)  10
 5x - 55  10
 5x  10  55  65
65
x 
 13
5
6.16
Now consider the algebraic examples:
i. ax + b = c ii. a(x – b) = c
ax  b  c
i.  ax  c - b
c-b
c b
x 
OR x  
a
a a
a(x  b)  c
ii.  x  b 
x 
c
a
c
c - ab
- b OR x 
a
a
a(x  b)  c
 ax  ab  c
repeat of ii.  ax  c - ab
E6.16
c - ab
c ab c
OR x   -b
a
a a a
Solve : d + a(x – b) = e by both methods.
A6.16
d  a(x  b)  e
x 
 a(x  b)  e - d
e-d
a
e-d
e d
x 
 b ORx    b
a
a a
Alternative method:
d  a(x  b)  e
x  b 
 d  ax  ab  e
 ax  e - ab - d
x 
e - ab - d
a
OR x 
e ab d
e d
- -  b  a a a
a a
75
6.17
4
35
x
Note the unknown, x is in the denominator.
Consider the equation:
Track the following solution:
4
35
x
4
 53 2
x
Now take reciprocals of both sides to get
x 1
 .
4 2
x is now in the numerator.

x 
E6.17
A6.17
6.18
4
2
2
Solve 5 
4
3
x
4
3
x
4
   5  3  2
x
4
 2
x
Now take reciprocals of both sides to get
x 1
  .
4 2
x is now in the numerator.
4
x   2
2
Now consider the algebraic case:
a
bc
x
a
 cb
x
x
1
x 1 1
 
Note :  
is incorrect. See * below.
a c-b
a c b
a
x 
c-b
5
*Remember the reciprocal of a difference is not the difference of
the reciprocals.
76
E6.18
Solve
a
c
xb
A6.18
a
c
xb
xb 1


a
c
a
x  b 
c
a
x   b
c
6.19
Consider the equation 3x2 + 5 = 17.
We can proceed by the unpacking method as follows:
3x 2  5  17
 3x 2  17 - 5  12
12
4
3
 x  2 and x  - 2
 x2 
Note the unpacking of the square is by square rooting. The proper
language is ‘the inverse of squaring is square rooting’.
E6.19
Note also the square root provides two roots, +ve & -ve.
Solve 2x2 – 11 = 7
A6.19
2x 2  11  7
 2x 2  7  11  18
18
 x2   9
2
x   3
 x  - 3 and x  3.
6.20
Now we will use ‘the inverse of rooting is squaring’ to unpack this
equation: 2(√x – 3) = -14
77
2( x - 3)  - 14
- 14
 x - 3
 7
2
 x  7  3  4
 x   4  16
2
E6.20
Solve 5 – 2√x = 23
A6.20
5 - 2 x  23
 2 x  23 - 5  18
 x
18
2
9
 x   9  81
Zero Product Equations (ZPE)
What can you say about the ? in the following:
2
6.21
(?)(?) = 0. We can say that at least one of the ? is a zero.
If a product is zero then at least one factor of a product
must be zero.
Here is an example of a ZPE: xy = 0.The LHS is a product and the
RHS is zero.
So in the ZPE equation, xy = 0 we have two solutions:
x = 0; y =0
E6.21
A6.21
6.22
Now x(x+2) = 0 is also a ZPE.
By setting each factor to zero we get the two solutions:
x = 0; And x+2=0 i.e. x = -2.
Solve x(x – 3) = 0
x(x – 3) = 0
Setting each factor = 0 we get.
x = 0 ; x – 3 = 0  x = 3.
Consider the example: (x+2)(2x-3) = 0
This is a ZPE.
So x+2 = 0 i.e. x = -2 And 2x-3 = 0 i.e. 2x = 3 i.e. x =
E6.22
Solve (x-3)(2-x) = 0
78
2
3
A6.22
6.23
In the ZPE, (x-3)(2-x) = 0 set each factor to zero to get:
x – 3 = 0  x = 3. And 2 – x = 0  x = 2.
Quadratic equations
Here is a quadratic equation: x2 + 4x = 0
The x2 is the quadratic term. In its present form it is not a ZPE.
To solve it, make it into a ZPE by factorising the LHS. Note the
RHS is already zero.
E6.23
A6.23
6.24
x2 + 4x = 0
i.e. x(x+4) = 0 So x = 0 and x = -4.
Solve: 4x - x2 = 0
4x - x2 = 0
x(4 – x) = 0 It is now a ZPE.
 x = 0 and x = 4.
Now consider the quadratic equation:
x2 + 4x = -3.
Note the RHS is not zero. To make this into a ZPE we first make
the RHS = 0 by moving the term, -3 from the LHS to the RHS viz:
x2 + 4x + 3 = 0.
Now make LHS into a product. So let’s factorise the LHS using our
quadratic pair rule.
(x + 3)(x + 1) = 0
Note: Now we have a ZPE.
E6.24
So solve to get x = -3 and x = -1.
Solve the quadratic equation, x2 + 4 = -4x
A6.24
x2 + 4 = -4x.
Make the RHS = 0:
x2 + 4x + 4 = 0.
Factorise the LHS using our quadratic pair rule:
(x + 2)(x + 2) = 0
79
 x = -2. Note only 1 (repeated) root
You may spotted that the LHS is a perfect square.
(x+2)2 = 0 So x = -2.
6.25
Here is another quadratic equation in which the coefficient of the
quadratic term is not 1.
Solve 3x2 + 2x – 5 = 0
To make this into a ZPE we need to factorise the LHS.
We will find the quadratic pair and then use grouping.
3x 2  2x - 5  0
 3x 2  5x  3x - 5  0
 x(3x - 5)  1(3x - 5)  0
 (x  1)(3x - 5)  0
 x  1  0  x  -1 And
3x - 5  0  3x  5  x 
E6.25
5
3
Solve 2x2 – 9x – 5 = 0
Hint: (?)(?) = -10
? + ? = -9
A6.25
2x2 – 9x – 5 = 0
Now: (-10)(1) = -10
-10 + 1 = -9
2x 2  10x  x - 5  0
 2x(x - 5)  1(x - 5)  0
 (2x  1)(x - 5)  0
 2x  1  0  x  -0.5 And
x -5  0  x  5
80
81
Exercise 6
1. Solve by unpacking
i. 3(4x + 2) = 15
2.
(i)
ii. 3(5 – x) = -4 iii. 2(√x – 3) =10 iv. (x – 3)2 = 1
Solve the equations by converting to ZPE:
– 4x + 7x2 = 0 (ii) x2 – 2x – 8 = 0 (iii) X2 + 5 = 9
3. Solve the equation 5x2 – 8x – 4 = 0
82