Download A2 Linear Equations

KS4 Mathematics
A2 Linear equations
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Contents
A2 Linear equations
A A2.1 Equations, formulae and identities
A A2.2 Balancing equations
A A2.3 Equations with brackets
A A2.4 Equations with fractions
A A2.5 Using equations to solve problems
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Expressions and identities
Remember in an algebraic expression we usually have a
combination of numbers, letters and operations.
For example, 3x + 5 + 2x – 6.
Expressions do not contain an equals sign.
Sometimes an expression is rewritten in an equivalent form.
3x + 5 + 2x – 6 = 5x – 1
This is an example of an identity. An identity is true for all
values of x. This is sometimes written with the symbol ≡ as,
3x + 5 + 2x – 6 ≡ 5x – 1
Is identically equal to
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Equations
An equation links an algebraic expression and a number, or
two algebraic expressions with an equals sign.
For example,
x + 7 = 13 is an equation.
In an equation the unknown usually has a particular value.
Finding the value of the unknown is called solving the
equation.
x + 7 = 13
x=6
When we solve an equation we should line up the equals signs.
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Formulae
A formula is a special type of equation that links two or more
physical variables.
For example in the formula,
P = 2(l + w)
P represents the perimeter of a rectangle and l and w
represent its length and width.
We can use this formula to work out the perimeter of any
rectangle given its length and width.
Physical variables such as length, time, mass and temperature
have units attached to them. These are not written in the
formula and so it is important that they are consistent.
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Contents
A2 Linear equations
A A2.1 Equations, formulae and identities
A A2.2 Balancing equations
A A2.3 Equations with brackets
A A2.4 Equations with fractions
A A2.5 Using equations to solve problems
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Linear equations
Linear equations are the easiest type of equation to solve
because the unknown is not raised to any power other than 1.
We can solve very simple linear equations by inspection.
For example,
19 – x = 8
x = 11
7x = 42
x=6
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We think of this as, “what number subtracted
from 19 gives us an answer of 8?”
We think of this as, “what number multiplied
by 7 gives us an answer of 42?”
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Linear equations
When more than one operation is preformed on the unknown
we need to solve the equation in several steps.
We can do this by performing the same operations to both
sides of the equals sign to keep the equation balanced.
For example,
4x + 5 = 29
–5
subtract 5 from both sides:
4x = 24
÷4
divide both sides by 4:
–5
÷4
x=6
Check that 4 × 6 + 5 is equal to 29 in the original equation.
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Equations with unknowns on both sides
In some cases the unknown appears on both sides of the
equals sign.
For this type of equation it is important to work systematically
to collect terms with unknowns on the left hand side of the
equation and numbers on the right.
Always aim to have a single unknown on the left equal to a
single number on the right.
For example,
unknowns
numbers
6x – 2 = 2x + 1
Here we’ll label the sides as ‘unknowns’ and ‘numbers’. We
are aiming to have the unknown on the left and a number on
the right.
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Equations with unknown on both sides
The equation can then be solved by performing the same
operations on both sides until the solution is found.
unknowns
numbers
8x – 2 = 2x + 1
add 2 to both sides:
+2
+2
8x = 2x + 3
subtract 2x from both sides:
– 2x – 2x
6x = 3
divide both sides by 6:
÷6
÷6
x = 0.5
Check by substituting x = 0.5 into the expressions in the original
equation. Both sides are equal to 2, so the solution is correct.
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Contents
A2 Linear equations
A A2.1 Equations, formulae and identities
A A2.2 Balancing equations
A A2.3 Equations with brackets
A A2.4 Equations with fractions
A A2.5 Using equations to solve problems
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Equations with brackets
Equations can contain brackets. For example,
2(3x – 5) = 4x
To solve this we can
multiply out the brackets:
6x –10 = 4x
+ 10
add 10 to both sides:
+ 10
6x = 4x + 10
- 4x - 4x
subtract 4x from both sides:
2x = 10
÷2
divide both sides by 2:
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÷2
x=5
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Equations with brackets
Sometimes we can solve equations such as
2(3x – 5) = 4x
by first dividing both sides by the number in front of the bracket:
divide both sides by 2:
3x – 5 = 2x
+5
add 5 to both sides:
+5
3x = 2x + 5
- 2x - 2x
subtract 2x from both sides:
x=5
In this example, dividing first means that there are fewer steps.
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Contents
A2 Linear equations
A A2.1 Equations, formulae and identities
A A2.2 Balancing equations
A A2.3 Equations with brackets
A A2.4 Equations with fractions
A A2.5 Using equations to solve problems
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Solving equations with fractional coefficients
Sometimes the coefficient of an unknown is a fraction. For
example,
3
x–5=9–x
4
We can remove the 4 from the denominator by multiplying
both sides of the equation by 4.
3
4( 4 x – 5) = 4(9 – x)
expand the brackets:
3x – 20 = 36 – 4x
add 4x to both sides:
7x – 20 = 36
add 20 to both sides:
7x = 56
divide both sides by 7:
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x=8
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Solving equations with fractional coefficients
If an equation contains more than one fraction, these can be
removed by multiplying throughout by the lowest common
multiple of the two denominators. For example,
2
1
x
=
x+1
3
2
What is the lowest common multiple of 3 and 2?
The lowest common multiple of 3 and 2 is 6. Multiplying both
sides by 6,
2
1
6( 3 x) = 6( 2 x + 1)
expand the brackets:
subtracting 3x from both sides:
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4x = 3x + 6
x=6
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Solving equations involving division
In this example the whole of one side of the equation is divided
by 5.
2x + 7
=x–1
5
To remove the 5 from the denominator we multiply both sides
of the equation by 5.
2x + 7 = 5(x – 1)
expand the brackets:
2x + 7 = 5x – 5
swap sides:
5x – 5 = 2x + 7
add 5 to both sides:
5x = 2x + 12
subtract 2x from both sides:
3x = 12
divide both sides by 3:
x=4
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Solving equations involving division
When both sides of an equation are divided by a number we
must remove these by multiplying both sides by the lowest
common multiple of the two denominators. For example,
5x – 3
2x – 1
=
4
3
What is the lowest common multiple of 4 and 3?
The lowest common multiple of 4 and 3 is 12. Multiplying every
term by 12 gives us:
3
4
12(5x – 3)
12(2x – 1)
=
41
31
which simplifies to:
3(5x – 3) = 4(2x – 1)
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Solving equations involving division
We can then solve the equation as usual.
3(5x – 3) = 4(2x – 1)
expand the brackets:
add 9 to both sides:
15x – 9 = 8x – 4
15x = 8x + 5
subtract 8x from both sides:
7x = 5
divide both sides by 7:
x=
5
7
Although this answer could be written as a rounded decimal,
it is more exact left as a fraction.
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Solving equations involving division
We have seen that
simplifies to
5x – 3
2x – 1
=
4
3
3(5x – 3) = 4(2x – 1)
How could we perform this simplification in one step?
Multiplying both sides by 4 cancels out the 4 on the left hand
side and multiplies the expression on the right-hand side by 4.
Multiplying both sides by 3 cancels out the 3 on the right hand
side and multiplies the expression on the left-hand side by 3.
Doing this in one step in often called cross-multiplication.
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Simplifying equations by cross-multiplication
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Solving equations involving division
Sometimes the unknowns appear in the denominator.
For example,
4
5
=
(x + 3)
(3x – 5)
In this example, we can multiply both sides by (x + 3) and
(3x – 5) in one step to give:
4(3x – 5) = 5(x + 3)
expand the brackets:
subtract 5x from both sides:
add 20 to both sides:
divide both sides by 7:
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12x – 20 = 5x + 15
7x – 20 = 15
7x = 35
x=5
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Equivalent equations
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Contents
A2 Linear equations
A A2.1 Equations, formulae and identities
A A2.2 Balancing equations
A A2.3 Equations with brackets
A A2.4 Equations with fractions
A A2.5 Using equations to solve problems
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Constructing an equation
I’m thinking of a number.
When I subtract 9 from the number and double it, I get the
same answer as dividing the number by 5.
What number am I thinking of?
Let’s call the unknown number n.
We can solve this problem by writing the equation:
n
2(n – 9) =
5
The number with 9
subtracted and doubled
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is the
same as
the number
divided by 5.
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Solving the equation
We can solve this equation by performing the same
operations on both sides,
n
2(n – 9) =
5
multiply both side by 5:
10(n – 9) = n
expand the bracket:
add 90 to both sides:
subtract n from both sides:
divide both sides by 9:
10n – 90 = n
10n = n + 90
9n = 90
n = 10
We can check the solution by substituting it back into the
original equation:
2 × (10 – 9) = 10 ÷ 5
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Using equations to solve problems
The sum of the ages of Ben and his daughter, Alice, is 66
years. In one year’s time Ben will be three times Alice’s age.
How old is Alice now?
Let’s call Alice’s age a.
Looking at the first piece of information we can write
Ben’s current age as 66 – a.
Using the second piece of information we can also write
Ben’s current age as 3(a + 1) – 1
This is three times Alice’s age in one year time, minus one to
give Ben’s age now. This simplifies to,
3(a + 1) – 1 = 3a + 3 – 1
= 3a + 2
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Using equations to solve problems
The expressions 66 – a and 3a + 2 both give Ben’s current
age.
This means that they are equal and so we can write them in
an equation:
66 – a = 3a + 2
swap sides:
3a + 2 = 66 – a
subtract 2 from each side:
3a = 64 – a
add a to both sides:
4a = 64
divide both sides by 4:
a = 16
Alice is 16, so Ben is 50. In one year Alice will be 17 and Ben
will be 51.
51 is three times 17, so our solution is correct.
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Find the value of x
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