Solving Linear Equations (Cambridge International IGCSE Maths)

Solving linear equations

What are linear equations?

• A linear equation is one that can be written in the form $ax+b=c$

  • $a,b,$ and $c$* *are numbers and $x$is the variable

    • 2x + 3 = 5
    • 3x + 4 = 1
    • x - 5 = -3

• The greatest power of x is 1

  • There are no terms like x²

How do I solve linear equations?

• You need to use operations like adding, subtracting, multiplying and dividing to get x on its own

• Any operation you do to one side of the equation must also be done to the other side

• For example, to solve $2x+1=9$ look at the +1 on the left

  • Undo this by subtracting 1 from both sides and simplifying

$\begin{array}{rcl} 2 x + 1 & = 9 \end{array}$

$\begin{gathered} ( - 1 ) \end{gathered} \begin{gathered} ( - 1 ) \end{gathered}$

$\begin{gathered} 2 x + 1 - 1 & = 9 - 1 \\ 2 x & = 8 \end{gathered}$

• This equation is now easier to solve
• 2x is 2 × x so undo this by dividing both sides by 2 and simplifying

$\begin{gathered} 2 x & = 8 \end{gathered}$

$\begin{gathered} ( \div 2 ) \end{gathered} \begin{gathered} ( \div 2 ) \end{gathered}$

$\begin{gathered} \frac{2 x}{2} & = \frac{8}{2} \\ x = 4 \end{gathered}$

• The solution to the equation is x = 4

• Note that +1 was undone by subtracting 1

  • addition and subtraction are said to be inverse(opposite) operations

• Similarly, multiplying by 2 was undone by dividing by 2

  • multiplication and division are also inverse operations

Does the order of operations matter?

• Take $4x+8=12$

  • It is easier to first subtract 8 from both sides

$\begin{gathered} 4 x + 8 - 8 & = 12 - 8 \\ 4 x & = 4 \end{gathered}$

• then divide both sides by 4

$\begin{gathered} \frac{4 x}{4} & = \frac{4}{4} \\ x = 1 \end{gathered}$

• If you want to first divide by 4, a common mistake is to write $x+8=3$

  • You cannot divide just the 4x and the 12 by 4

    • You must also divide the 8 by 4

$\begin{gathered} \frac{4 x}{4} + \frac{8}{4} & = \frac{12}{4} \\ x + 2 & = 3 \end{gathered}$

• Then subtract 2 from both sides

$\begin{gathered} x + 2 - 2 & = 3 - 2 \\ x = 1 \end{gathered}$

How do I solve linear equations with negative numbers?

• For example, solve the equation $2-3x=10$
• Add $3x$ to both sides to get rid of the negative term on the left-hand side

$\begin{gathered} 2 - 3 x & = 10 \end{gathered}$

$\begin{gathered} ( + 3 x ) \end{gathered} \begin{gathered} ( + 3 x ) \end{gathered}$

$\begin{gathered} 2 = 10 + 3 x \end{gathered}$

• You can then rewrite the equation as $3x+10=2$

  • And solve in the usual way

    • Subtract 10 from both sides $\begin{gathered} 3 x & = - 8 \end{gathered}$
    • Divide both sides by 3 $x=-\frac{8}{3}$

• An alternative way to think about $2-3x=10$ is by reordering the left-hand side

  • $2-3x$ is the same as $-3x+2$

  • Reordering like this does not change the right-hand side

    • So the equation is $-3x+2=10$

  • Some people prefer to see it like this

    • You then subtract 2 and divide by -3 as before

Exam Hint In the exam, substitute your answer back into the original equation to check you got it right!

Worked Example Solve the equation

$5x-8=22$

Answer:

Add 8 to both sides of the equation

$\begin{gathered} 5 x & = 22 + 8 \end{gathered}$

Work out 22 + 8

$5x=30$

Divide both sides by 5

$x=\frac{30}{5}$

Work out 30 ÷ 5 Keep the x on the left-hand side

$x=6$