# Solving Linear Equation

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 x^2.

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)

## 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. But 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

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

({x+5}/3) 3 = (3)3

x + 5 = 9

Subtract 5,

x = 4

• 3(x - 1) = 27
• Notice, there are two different ways we could solve this, both correct, we can use the distributive property to multiply out the brackets then do inverse operations,

or,

we can start by dividing by 3, this looks to be the easier way,

First, divide by 3,

(3(x - 1)) / 3 = (27) / 3

x - 1 = 9

x - 1 + 1 = 9 + 1

x = 10

## Question 1c

{-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,

(-5/3) x + 3 = 5,

Firstly,

Subtract 3,

(-5/3) x + 3 - 3 = 5 - 3

(-5/3) x = 2,

We can solve this a couple ways,

We can it write this like,

{(-5) x}/3 = 2,

Then multiply by 3,

({(-5) x}/3)3 = (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,

(-5/3) x = 2

((-5/3) x) / (-5/3) = (2)/ (-5/3)

x = 2/ (-5/3)

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

x = 2(3/-5)

x = 6/-5, or x = -6/5

## Question 1d

{3(x - 5)}/ 6 + 7 = 5

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

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

1) Subtract 7

2) Multiply 6

3) Divide 3

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)6 = (-2)6

3(x - 5) = -12

3) Divide by 3

(3(x - 5))/3 = (-12)/3

x - 5 = -4