Algebraic Equations
Grade 8
The difference between a number sentence and an equation
The statement 2 + 3 = 1 + 4 is called a number sentence. A number sentence is true if the left hand
side (LHS) is equal to the right hade side (RHS).
The number sentence 2 + 3 = 1 + 4 is true since LHS = RHS.
The number sentence 3 + 8 = 2 + 6 is not true since LHS RHS.
Exercise 1
Write down the appropriate number for ? which will make the following number sentences true:
(a)
4+?=6
(b)
3-?=1
(c)
7 – ? = -2
(d)
3–?=5
(e)
? – 6 = -3
(f)
? – 5 = -12
When a number sentence has a variable in it, we call the number sentence an equation. For example,
the number sentence 2x + 3 = 11 is called an equation because it contains the variable x in it. The
variable x is treated just the same as the question mark.
In the equation 2x + 3 = 11, the coefficient of x is 2 and the numbers 3 and 11 are constants. If x
takes on a value of 4, then the equation will be a true statement since 2(4) + 3 = 11. The value of this
variable (4) is called the solution of the equation and we write the solution as x = 4.
Let us now try to solve the equation in the examples below:
(a) x + 3 = 10
(b) 13 – a = 8
(d) -4 + t = 10
(e) 2x – 4 = -6
Solutions
(a) 7 + 3 = 10
∴x=7
(d) -4 + 14 = 10 ∴ t = 14
(b)
(e)
13 – 5 = 8
∴a=5
2(-1) – 4 = -6 ∴ x = -1
(c)
5
(f) -5 + b = -10
(c)
(f)
5
∴ y = 20
-5 + (-5) = -10, ∴b = -5
We have just solved all of the above equations by inspection.
Equations get quite complicated and it is not always possible to solve by inspection. We will now
solve equations using and algebraic method.
Solve for x: 2x + 4 = 6. This equation has a LHS(left-hand side) and a RHS(right-hand side)
2x + 4 = 6
LHS
RHS
The main aim is to solve the equation for the unknown variable, in this case x. We do this by isolating
the variable x. This is done using inverse operations.
1
This means:
+ becomes –
- becomes +
x becomes
becomes x
So whatever is done to the LHS of the equation is done on the RHS of the equation. The whole point is
to get the unknown(x) to be on its own.
Examples: (Adding and Subtracting)
e.g. 1
x+6=9
We’re adding 6 to x. So do the inverse of +, that is –
(-6) x + 6 – 6 = 9 – 6 This step you can do mentally
x=3
e.g. 2
x–3=5
(+3) x – 3 + 3 = 5 + 3
We’re subtracting 3 from x. So do the inverse of - that is +
This step you can do mentally
x=8
N.B. In your solution, try to keep the = in a straight line, one below the other.
Exercise 2
Solve the following equations
(2.1)
x+4=2
(2.2)
k–5=3
(2.3)
a – 4 = -3
(2.4)
x – 6 = -5
(2.5)
3=x–4
(2.6)
y+7=7
Examples: (Multiplying and Dividing)
e.g. 3
4x = -8
We’re multiplying x by 4. So do the inverse of x, that is
( 4
This step you can do mentally
x = -2
e.g.4
(x 3)
13
.
.
1 31
5
We’re dividing x by 3. So do the inverse of
5 .3
3 can be written as .
= 15
15
Divide the numerator and denominator by 3
is just x
x = 15
Exercise 3
Solve the following equations
(3.1)
2a = 10
(3.2)
3b = -18
(3.3)
(3.4)
-7x = 14
(3.5)
4l = 100
(3.6)
2
10
, that is x.
Examples: (Equations requiring more than 1 operation)
When an equation requires more than one operation, always remove the constant first. Do the inverse
for the constant, then the inverse for the variable.
e.g. 5
In the example below, +4 is the constant.
2x + 4 = 6
(-4)
Notice we’re adding 4. So do the inverse of +, which is –
2x + 4 – 4 = 6 – 4 This you can do mentally
2x = 2
( 2
Now we’re multipying x by 2. So do the inverse of x, i.e.
This you can do mentally
x=1
e.g. 6
In the next example, -8 is the constant
-x – 8 = 2
(+8)
We’re subtracting 8 from -x, so do the inverse of -, which is +
-x – 8 + 8 = 2 + 8 This you can do mentally
-x = 10
The coefficient of x (the number in front of x) is -1.
We have -1 . x,
so do the inverse of x, that is
+
x = -10
e.g. 7
19 = -5 + 3x
-5 is the constant, so do the inverse of -, which is +
(+5) 5 + 19 = +5 – 5 + 3x This you can do mentally
24 = 3x
( 3
We’re multiplying x by 3, so do the inverse of x, that is
8=x
Examples: (Variables and constants on both sides of the equation sign)
When an equation has both variables and constants, ‘move’ the variables to one side and the constants
to the other side.
e.g. 8
(-3x)
4x = 3x + 5
4x – 3x = 3x – 3x + 5
‘Move’ 3x to LHS. The sign in front of 3x is +, so do a –
This step you can do mentally
x=5
e.g. 9
6x = -4x + 10
(+4x) 6x + 4x = -4x + 4x + 10
10x = 10
( 10
‘Move’ -4x to the left. The sign in front of -4x is -, so do a +
This step you can do mentally
We’re multiplying x by 10. So the inverse of x is
x =1
3
e.g. 10
x + 3 = -2x – 9
‘Move’ -2x to LHS, by doing the inverse of -2x, that is +2x
(+2x) x + 2x + 3 = -2x + 2x – 9
3x + 3 = -9
We’re adding 3, so do the inverse, that -3
(-3) 3x + 3 – 3 = -9 – 3
This step can be done mentally
3x = -12
( 3
We’re multiplying 3 and x. So do the inverse,
by 3
x = -4
Exercise 4
(4.1)
6x = 5x + 6
(4.2)
12x – 5 = 13x – 9
(4.3)
6x + 2 = 4x – 10
(4.4)
2 – 3x = 5 – x + 3
(4.5)
2y + 6 = 9 – y
(4.6)
3 – 4x = 9 + 2x
Examples: (Equations with brackets)
If an equation has brackets, remove the brackets first, by using the distributive law.
e.g. 11
3(x + 2) = 2(x + 1)
3(x + 2) = 2(x + 1)
Use the distributive law to distribute 3 and 2 into the brackets
3x + 6 = 2x + 2
‘Move’ the 2x to LHS. In front of the 2x is a +, therefore -2x
(-2x) 3x – 2x + 6 = 2x – 2x + 2
x+6=2
This step can be done mentally
We’re adding 6, therefore the inverse is -6
(-6) x + 6 – 6 = 2 – 6
x = -4
e.g. 12
3 – (y + 2) = 5y + 1
The number here is -1
3 – 1(y + 2) = 5y + 9
Use the distributive law to distribute -1 into the bracket
3 – y – 2 = 5y + 9
3 and -2 are Like terms, so simplify them, i.e. 3 – 2 =1
-y + 1 = 5y + 9
There is a + in front of 5y, so do the inverse of +5y, i.e. -5y
(-5y) -5y – y +1 = -5y + 5y + 9
-4y + 1 = 9
We’re adding 1, so do the inverse of + , i.e. –
(-1) -4y + 1 – 1 = 9 – 1
-4y = 8
( −4)
This step can be done mentally
We’re multiplying y by -4, so do the inverse, i.e.
-4
y = -2
Exercise 5
(5.1)
2(3 – x) = 8
(5.2)
1 – (x + 2) = 5 + x
(5.3)
5(x – 1) = 4(x – 2)
(5.4)
5(p – 1) – (1 – 2p) = 8
(5.5)
3(x – 4) = 2(x – 2) + 2 – x
(5.6)
10 – 3(a + 7) = a + 5
(5.8)
2(2x + 5) = -2( 1 + x)
(5.9)
2(2 – x) = 3(x + 3)
(5.7) 2 – 5(1 – y) = 3y + 5
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