Solving an equation with
one unknown
From arithmetic to algebra
Modifying equations in algebra
Solving an equation with one unknown

Today, we will use the rules of arithmetic
seen last week to solve an equation with one
unknown


As for last week, the exercises for next week
are intended to get you used to solving
equations
Again, the slides provide list of properties
you can go back to for reference
Solving an equation with one unknown
From arithmetic to algebra
Solving equations
From arithmetic to algebra

Last week we saw the rules for using
addition, subtraction, multiplication and
division to perform calculations


Like the examples we’ve seen…
Today we examine how you can use the
same rules and notation to solve equations

What’s the difference ?
From arithmetic to algebra

An equation is a line of calculation that
contains an “=” sign:
24  6

An equation can contain unknown “x, y, z” the
value of which is not given
2 x  6

The numbers that are known are called
parameters
From arithmetic to algebra

With the calculations, the aim was to find the
following result (for example):
24
6
52
24
 6  ??
52

In fact, a calculation is simply an equation
where the unknown is isolated on one side of
the equation !!
From arithmetic to algebra

When an equation contains an unknown “x”
the aim is to find the value(s) of “x” for which
the equation is true

In other words, “solving an equation” means
finding the value of the unknown for which the
equality holds
2 x
60
52

Example : is the equation true if x=10? What if
x=-20?
From arithmetic to algebra

In the general case, the unknown is not
isolated, like in a calculation:
4   x  6 5  2  2

Therefore, the aim is to modify the equation
until you can isolate the unknown:
24
6 x
52

At that point, what you have is a calculation
you can solve
From arithmetic to algebra

The rules in the following section also apply
to the case where the parameters of the
equation (the numbers) are also unknown
4   x  a  b  2  c

However, you can use the same method to
isolate the unknown


It is simply the final calculation that you can’t
perform.
You have some examples in the exercises
Solving an equation with one unknown
From arithmetic to algebra
Solving equations
Solving equations

Lets try and solve the following equation

In other words find the value of x such that the
equation is true.
4   x  6 5  2  2

The goal is to isolate “x” on one side. How do
you do it?




Use the rules from last week.
Identify the grouped operations
Move the groups until you have isolated x
Important: take all the steps you need to do so !
Solving equations


When modifying the equation take advantage
of the fact that carrying out an operation on
both sides of an equation does not change
the equation
Step 1: move “2” to the other side
4   x  6 5  2  2
4  2   x  65  2  2  2
4  2   x  65  2  2  2
4  2   x  65  2
Solving equations

Step 2: if you want, you can carry out some
calculations at the same time
4  2   x  65  2
6   x  6 3

Step 3: Move ×3 to the other side by dividing
both sides by 3:
6
3
  x  6
3
3
2  x6
Solving equations

Step 4: Move -6 to the other side by adding 6
on both sides
26  x 66
26  x

Step 5: Perform the final calculation:
8 x

Step 6: ALWAYS check by replacing x in the
original equation to see if it is true!!