MTH5P1: Equations

This unit explains equations

Solving Equations

An equation states that two quantities are equal. This means that the left hand and right hand sides of the equals sign are equivalent, they balance. The equation may contain an unknown quantity that we wish to find. In the equation, 5π‘₯π‘₯ + 10 = 20, the unknown quantity is π‘₯π‘₯. This means that 5 multiplied by something that is then added to 10, will equal 20.

To solve an equation means to find all values of the unknown quantity so that they can be substituted to make the left side equal to the right side and vice versa.

Each such value is called a solution, or alternatively a root of the equation. In the example above,

the solution is π‘₯π‘₯ = 2 because when 2 is substituted, both the left side and the right side equal 20 ∴ the sides balance. The value π‘₯π‘₯ = 2 is said to satisfy the equation.
Sometimes we are required to rearrange or transpose the equation to solve it: essentially what we do to one side we do to the other.

Hence, as above 5π‘₯π‘₯ + 10 = 20, we first rearrange by subtracting 10 from the LHS and the RHS.
5π‘₯π‘₯ + 10 (βˆ’10)= 20(βˆ’10);5π‘₯π‘₯=10
Then we rearrange further to get the 𝒙𝒙 on its own; so we divide both LHS and RHS by 5.

To check if this is correct we substitute π‘₯π‘₯ 𝑓𝑓𝑓𝑓𝑓𝑓 2 ;(5Γ—2)+10=20∴10+10=20

This was a two-step equation – which will be covered later.

Four principles to apply when solving an equation:
Work towards the variable: Our aim is to get the variable by itself on one side of the equation. (So for the above we would aim to get the π‘₯π‘₯ by itself on one side)

(𝒙= )

Use the opposite mathematical operation: thus to remove a constant or coefficient, we do the opposite on both sides:

Maintain balance: β€œWhat we do to one side, we must do to the other side of the equation.”

Check: Substitute the value back into the equation to see if the solution is correct.

One-step Equations

ADDITION EXAMPLE: 𝒙𝒙+(βˆ’πŸ“πŸ“)=πŸ–πŸ–
Step 1: The constant, (βˆ’5), is the first target. So we need to do the opposite of plus (βˆ’5) which is to subtract (βˆ’5 ) from the LHS and the RHS:
π‘₯+(βˆ’5)βˆ’(βˆ’πŸ“πŸ“)=8βˆ’(βˆ’5)
π‘₯=8+5 (Here we apply our knowledge of negative and positive integers from Module 1.)
∴π‘₯=13
Check 13+(βˆ’5)=8
13βˆ’5=8
SUBTRACTION EXAMPLE: π‘₯βˆ’6=(βˆ’4)

Step 1: The constant, (βˆ’6) is the target. The opposite of subtract 6 is to add 6.
π‘₯βˆ’6+6=(βˆ’4)+6
So π‘₯=(βˆ’4)+6 ∴π‘₯π‘₯=2
Check 2βˆ’6=(βˆ’4)

Two-step Equations

The following equations require two steps to single out the variable.
ADDITION EXAMPLE: 𝟐𝟐𝟐 +πŸ”πŸ”=𝟏𝟏𝟏𝟏
In words: What number can we double then add six so we have a total of fourteen?
Step 1: The constant, 6, is our first target. If we subtract 6 from both sides, we create the following equation:
2π‘₯+6βˆ’6=14βˆ’6 (The opposite of +6 is βˆ’6)
2π‘₯=8 (+6βˆ’6 = 0)
Step 2: The only number left on the same side as the variable is the coefficient, 2. It is our second target. If we divide both sides by two, we create the following equation: (Note: between the 2 and the π‘₯ is an

Check. If we substitute 4 into the equation we have: 2Γ—4 +6=14
8+6=14
14=14

SUBTRACTION EXAMPLE:
Solve for j: 3π‘—βˆ’5=16
Step 1: 3π‘—βˆ’5=16 (The first target is 5)
3π‘—βˆ’5+5=16+5 (Opposite of βˆ’5 𝑖s +5)
Thus, 3𝑗=21

MULTI-STEP EXAMPLE:

Set your work out in logical clear to follow steps

Solving Equations with Fractions

So far we have looked at solving one and two step equations. The last example had fractions too, which we will explore more deeply in this section. First, let us go back and revise terms. In Module 5 we covered terms, but it is important to remember that in algebra, terms are separated by a plus (+) or minus (βˆ’) sign or by an equals (=) sign. Whereas variables that are multiplied or divided are considered one term.
For example, there are four terms in this equation: πŸ–ab+πŸ‘ab +πŸ’ =πŸ•0
and also: πŸ‘(𝒙+𝟐)+πŸ–βˆ’πŸ”=𝟐0
When working with fractions, it is generally easier to eliminate the fractions first, as follows:

Step 1: Eliminate the fraction, to do this we work with the denominator first, rather than multiply by the reciprocal as this can get messy. So both sides of the equation will be multiplied by 5, which includes distributing 5 through the brackets :

Solving Equations with Variables on Both Sides

Solving equations with variables on both sides can be difficult and requires some methodical mathematical thinking. Remembering that both sides are equivalent, the goal is to get all of the constants on one side of the equation and the variables on the other side of the equation. As we have explored previously, it is helpful to deal with the fractions first.
We can begin by solving the equation on the front cover of this module:

 

 

 

 

 

 

EXAMPLE ONE:

Solve for : 3π‘₯+5+2π‘₯=12+4π‘₯

Work towards rearranging the equation to get all of the constants on the RHS and the variables on the LHS:
3π‘₯+5+2π‘₯βˆ’4π‘₯=12+4π‘₯βˆ’4π‘₯ Let’s start by subtracting 4π‘₯π‘₯ from both sides.
The result: 3π‘₯+5+2π‘₯βˆ’4π‘₯=12
3π‘₯+5βˆ’5+2π‘₯βˆ’4π‘₯=12βˆ’5 Now subtract 5 from both sides.
3π‘₯+2π‘₯βˆ’4π‘₯=7 Now simplify by collecting like terms. (3+2βˆ’4=1)
1π‘₯π‘₯=7 ∴π‘₯=7
Check: 3π‘₯+5+2π‘₯=12+4π‘₯

21+5+14=12+28 (Substitute the variable for 7)
40=40

EXAMPLE TWO:

Solve for π‘₯: 6π‘₯+3(π‘₯+2)=6(π‘₯+3)
Again, work towards moving all the constants on one side and the variables on the other. First we could distribute through the brackets which will enable us to collect like terms:
6π‘₯+3π‘₯+6=6π‘₯+18 (Distribute through the brackets)
6π‘₯+3π‘₯+6βˆ’6=6π‘₯+18βˆ’6 (Subtract 6 from both sides first, to get constants on RHS)
The result: 6π‘₯+3π‘₯=6π‘₯+18βˆ’6
9π‘₯βˆ’6π‘₯=6π‘₯βˆ’6π‘₯+18βˆ’6 (Subtract 6π‘₯ from both sides to get variables on LHS) 3π‘₯=12 ∴ π‘₯=4
Check: 6π‘₯+3(π‘₯+2)=6(π‘₯+3);24+18=6Γ—7 ∴ 42=42
Multiples and Factors

This section will provide you with a refresh on factors and multiples. These concepts relate to understanding fractions, ratios and percentages, and they are integral to factoring when solving more difficult problems.
Multiples, common multiples and the LCM
A multiple of a given a number, into which that number can be divided exactly.
For example: multiples of 5 are 5, 10, 15, 20, 25, 30…
A common multiple is a multiple in which two or more numbers have in common.
For example: 3 and 5 have multiples in common 15, 30, 45…
The lowest common multiple LCM is the lowest multiple that two numbers have in common.
For example: 15 is the LCM of 3 and 5

Factors, common factors and the HCF

Factor – a whole number that can be multiplied a certain number of times to reach a given number.
3 is factor of 15 because it can be multiplied by 5 to get 15
β€’ A common factor is a factor that two or more numbers have in common.
3 is a common factor of 12 and 15
β€’ The highest common factor – HCF
What is the HCF of 20 and 18?
We list all of the factors first.
The factors of 20: (1, 2, 4, 5, 10, 20)
The factors of 18: (1, 2, 3, 6, 9, 18)
The common factors are 1 and 2 and the highest common factor is 2
β€’ Proper factors: All the factors apart from the number itself
The factors of 18 (1, 2, 3, 6, 9, 18)
Thus the proper factors of 18 are 1, 2, 3, 6, 9
β€’ Prime number: Any whole number greater than zero that has exactly two factors – itself and one
2, 3, 5, 7, 11, 13, 17, 19…
β€’ Composite number: Any whole number that has more than two factors

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20…

As the factor trees show, some numbers have many factors. This is a helpful way to find the factors of a given number. A more effective method for clarity when thinking and reasoning mathematically is to list all of the factors.

Factorising and Expanding

The distributive law is commonly used in algebra. It is often referred to as either β€˜expanding the brackets’ or β€˜removing the brackets’.
β€’ Expanding involves taking what is outside of the brackets and moving it through the brackets. For example:
3(π‘₯+2) can be expanded to 3Γ—π‘₯+3Γ—2 which is 3π‘₯+6
This means that: 3(π‘₯+2)=3π‘₯+6
β€’ Working in the opposite direction is called factorisation and this process is slightly more difficult.
β€’ Factorisation requires finding the highest common factor. (The HCF then sits outside of the brackets.)

Let’s look at 3π‘₯+6, we can factorise this back again from the expanded form.
Β° First we need to find a common β€˜factor’; we can see that π‘₯ can be divided by 3 and 6 can be divided by 3. Our common factor is 3. Now we put this 3 outside of the brackets (because we have divided each of the terms by three) to be multiplied by everything inside of the brackets.
Β° So we have 3(π‘₯+2)
β€’ Another factorisation example:

Function Notation

Most of the equations we have worked with so far have included a single variable. For example, 2π‘₯+5=11, π‘₯ is the single variable. Single variable equations can be solved easily. Yet, in real life you will often see questions that have two variables:
The cost of fuel is equal to the amount of fuel purchased and the price per litre $1.50/L.
The mathematical equation for this situation would be:
𝐢ost=𝐴amount ×𝑃𝑃L(price per litre)
∴𝐢=𝐴×1.5
If the amount of fuel is 20 litres, the cost can be calculated by using this formula:
𝐢=20Γ—1.5
∴𝐢=30
Another way to describe the above scenario is to use function notation. The cost is a function of amount of fuel.
If we let π‘₯ represent the amount, then the cost of fuel can be represented as 𝑓𝑓(π‘₯)

Therefore: 𝑓(π‘₯)=1.5π‘₯. We can now replace π‘₯ with any value.
If 𝑓(π‘₯)=1.5π‘₯ then solve for 𝑓(3)
This means replace π‘₯π‘₯ with the number 3.
𝑓(3)=1.5 Γ—3
βˆ΄π‘“(3)=4.5

Example Problems:

ASSIGNMENT : Equations Assignment MARKS : 10  DURATION : 50 minutes

 

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