Solving Linear Equations (WJEC GCSE Maths & Numeracy (Double Award): Foundation): Revision Note

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

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

$\begin{array}{rcl}& & (-1)\end{array}\begin{array}{rcl}& & (-1)\end{array}$

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

• This equation is now easier to solve

• 2x is 2 × x  so undo this by dividing both sides by 2 and simplifying

$\begin{array}{rcl}2x& =& 8\end{array}$

$\begin{array}{rcl}& & (÷2)\end{array}\begin{array}{rcl}& & (÷2)\end{array}$

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

• 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

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

• then divide both sides by 4

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

• 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

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

• Then subtract 2 from both sides

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

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

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

$\begin{array}{rcl}& & (+3x)\end{array}\begin{array}{rcl}& & (+3x)\end{array}$

$\begin{array}{rcl}2& =& 10+3x\end{array}$

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

  • And solve in the usual way

    • Subtract 10 from both sides
      $\begin{array}{rcl}3x& =& -8\end{array}$

    • 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