Solving Linear Equations

Solving equations is more or less what algebra is all about. In this lesson we will practice solving a variety of linear equations.

Note: Linear Equations are equations that have an unknown variable that is to the power of 1, i.e. x1 i.e. x, so an equation like 2x + 1 = 5x - 3. It must have no variables like x2.

Solve the following linear equations. Note that some answers may not be whole numbers.

A quick reminder of how to solve these questions:

  1. First, see if we can simplify the equation through Commutative, Associative or Distributive Properties,
  2. Use Inverse Operations, which is doing the opposite of the order of operations (the opposite of what has been done to x)

The solutions in the above animation are repeated with a more detailed explanation below.

Question 1a

2x - 6 = 10

We will start to “speed up” our Inverse Operations as we get better at them. If you are unsure, try going through the previous lesson. However, for this first one, we will go over it again.

2x - 6 = 10

Identify what has been done to x on the left-hand side following the order of operations (BEDMAS)

1) Multiplied by 2

2) Subtracted by 6

Now, Inverse Operations tells us we do the opposite order with the opposite operation.

1) Add 6

2) Divide by 2

So,

Add 6 to both sides (remember the Additive Properties of Equality and the Multiplicative Properties of Equality).

2x - 6 + 6 = 10 + 6

2x = 10 + 6

2x = 16

Divide both sides by 2

2x
2
  =  
16
2

x = 8

Question 1b

x + 5
3
  = 3

Multiply by 3

opening bracket
x + 5
3
closing bracket3 = 3(3)

x + 5 = 9

Subtract 5

x = 4

Question 1c

3(x - 5)
6
  + 7 = 5

When dealing with a larger equation, it might help to go over your Inverse Operations.

Identify what has been done to x, on the side of the equation that has x.

BEDMAS!

1) Brackets first, subtract 5

2) Multiply and Divide, the order does not matter for multiplication and division, but let us choose multiply, so, Multiply by 3

3) Divide by 6

4) Add 7

Now, we do the opposite of this order, with the opposite operations

1) Subtract 7

2) Multiply 6

3) Divide 3

4) Add 5

Now apply this to our equation,  
3(x - 5)
6
  + 7 = 5

1) Subtract 7

3(x - 5)
6
  + 7 - 7 = 5 - 7
3(x - 5)
6
  = -2

2) Multiply by 6,

3(x - 5)
6
 x 6 = (-2)6

3(x - 5) = -12

3) Divide by 3

3(x - 5)
3
  =  
-12
3

x - 5 = -4

4) Add 5

x - 5 + 5 = -4 + 5

x = 1

Question 1d

-5
3
 x + 3 = 5

This might be a bit daunting seeing a fraction as a coefficient. However, stick to our rules; it might help to write a bracket around the -5/3 to help is see it as a coefficient multiplying our variable.

opening bracket
-5
3
closing bracketx + 3 = 5

Firstly, subtract 3

opening bracket
-5
3
closing bracketx + 3 - 3 = 5 - 3
opening bracket
-5
3
closing bracketx = 2

We can solve this a couple ways.

We can it write this like,

(-5)x
3
 = 2

Then multiply by 3

opening bracket
-5x
3
closing bracket3 = (2)3

-5x = 6

Then divide by - 5

-5x
-5
  =  
6
-5
x =  -
6
5

OR

We can solve by dividing by the coefficient
-5
3
  on both sides, in one step.
opening bracket
-5
3
closing bracketx = 2
opening bracket
-5
3
closing bracketx ÷
-5
3
 = 2 ÷ 
-5
3
x = 2 ÷  
-5
3

Something divided by a fraction is the same as multiplying something by the inverse (or flipped version) of that fraction i.e. 2 ÷ 1/2 = 2 x 2/1

x = 2 x  
3
-5
x =  
6
-5
 , or x =  -
6
5

Recap

  1. Identify the order of operations - make a list of the order of operations that you would do to the variable.
  2. Inverse order of operations and reverse each operation itself - reverse the order of operations, and do the inverse operation in each step (i.e. subtract 2 becomes add 2, divide by 3 becomes multiply by 3).
  3. Apply the Inverse operations and solve for the variable - Follow all the correct steps and you will be left with your variable isolated giving you the answer.