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## Solving basic equations

### What is an equation?

An equation is a way of showing that two mathematical expressions are equal to each other.  You will recognise them by the presence of an equals sign ‘=’.  In this tutorial we will look at equations that contain only one unknown (letter) and use various techniques to find out the value of the letter.

### How do we keep an equation equal?

If one side of an equals sign has a different value to the other then the equation is no longer valid.  To ensure that we are always dealing with an equation we must always perform the same operation on both sides of the equals sign.  So if we add 5 to one side we add 5 to the other, if we square root one side then we must square root the other etc.

### How can we cancel an operation?

If I was to walk ten kilometres south, I can cancel what I have just done by walking ten kilometres north (doing the opposite).

If I give someone $10, I can cancel that process by taking$10 off that person (doing the opposite)

In short, if I do something and then do the opposite, the two operations cancel  each other out.

Table of opposites

 + – $\times$ $\div$ ${}^2$ $\sqrt{}$

### Stage 1

Definition:

An equation where only one operation is happening to the unknown.

Process:

1. Consider the side of the equation where the unknown is
2. Determine what operation is being applied to the unknown
3. Perform the opposite unknown to both sides of the equation

Examples

Solve $x+5=7$

We are adding 5 to the $x$ so we must cancel it by subtracting 5 from both sides

So $x=7-5$

$\implies x=2$

Solve $x-7=4$ Cancel by adding 7

So $x=4+7$

$\implies x=11$

Solve $3x=12$ Cancel by dividing by 3

So $x=\frac{12}{3}$

$\implies x=4$

Solve $\frac{x}{9}=3$ Cancel by multiplying by 9

So $x=3\times 9$

$x=27$

### Stage 2

Definition:

An equation where two operations are happening to the unknown.  One operation will be addition or subtraction and the other operation will be either multiplication or division.

Process:

1. Consider the side of the equation where the unknown is
2. Determine what operation is being applied to the term that the unknown is part of
3. Perform the opposite unknown to both sides of the equation
4. You should now have a stage 1 problem that you can solve as above, but you will only have one term that needs to be split

Examples

Solve $3x+5=14$

$-5\text{ from both sides}\implies 3x=9$

Split $\implies x=\frac{3}{9}=3$

Answer $x=3$

Solve $7x-3=32$ $+3$ to both sides$\implies 7x=35$

Split $\implies x=\frac{35}{7}=5$

Solve $\frac{x}{3}+4=6$ $-4\text{ from both sides}\implies \frac{x}{3}=2$

Split $\implies x=2\times 3=6$

Answer $x=6$

Solve $\frac{x}{2}-3=4$ $+3$ to both sides$\implies \frac{x}{2}=7$

Split $\implies x=7\times 2=14$

Answer $x=14$

### Stage 3

Definition:

An equation where the unknown is present on both sides of the equals sign.

Process:

1. Decide which of the terms with the unknown is the smallest
2. Subtract this term from both sides of the equation and subtract any term without the unknown from the side with the unknown we are not moving (Group)
3. You should now have a stage 2 problem that you can solve as above

Examples

Solve $4x-2=3x+5$

Group $\implies 4x-3x=5+2$

$\implies x=7$

Solve $2x+8=5x-7$Group $\implies 15=3x$ Split $\implies 5=x$

Answer $x=5$

Solve $6x+5=45-2x$Group $8x=40$ Split $x=\frac{40}{8}=5$

Answer $x=5$

Solve $3x+5=23-4x$(Remember $-4<3$) Group $\implies 7x=18$

Split $\implies x=\frac{18}{7}$

Answer $x=\frac{18}{7}$

### Brackets

Definition:

These equations have brackets present in one or more places.

Process:

1. Multiply out any brackets that are present
2. Simplify both sides of the equation
3. Decide whether you have a stage 1 or 2 problem and solve appropriately

Examples

Solve $3(x+5)=6(x-2)$

Expand $\implies 3x+15=6x-12$

Group ($-3x\text{ and } +12$ since we want to cancel -12)$\implies 27=3x$

Split $\implies 9=x$

Answer $x=9$

Solve $4(3-2x)-7=3(x+10)$Expand $\implies 12-8x-7=3x+20$ Simplify $\implies 5-8x=3x+20$

Group $\implies -15=11x$

Split $\implies \frac{-15}{11}=x$

Answer $x=-\frac{15}{11}$