Solving Linear Equations (Edexcel IGCSE Maths A (Modular): Higher Unit 1): 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

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

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

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

• This equation is now easier to solve

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

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

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

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

• The solution to the equation is x = 4

• Adding 1 was undone by subtracting 1

• Multiplying by 2 was undone by dividing by 2

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

  • Multiplication and division are also inverse operations

Does the order of steps matter?

• As long as each step is applied correctly, the order in which inverse operations are applied does not matter

  • Applying the operations in one order may be easier than another

• Consider $4x+8=12$

  • It is easier to first subtract 8 from both sides

$4x=4$

• Then divide both sides by 4

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

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

  • This is incorrect as 8 has not been divided by 4

  • You must divide every term 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

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

How do I solve linear equations with negative numbers?

• For example, $2-3x=10$

  • Subtract 2 from both sides and simplify

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

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

$\begin{array}{rcl}-3x& =& 8\end{array}$

• Then divide both sides by -3 and simplify

$\begin{array}{rcl}-3x& =& 8\end{array}$

$\begin{array}{rcl}& & (÷-3)\end{array}\begin{array}{rcl}& & (÷-3)\end{array}$

$\begin{array}{rcl}x& =& -\frac{8}{3}\end{array}$

• Some people prefer to write $2-3x=10$ as $-3x+2=10$, which is equivalent

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

How do I solve linear equations with brackets?

• If a linear equation involves brackets, expand the brackets first

• For example, solve $2\left(x-3\right)=10$

  • Expand the brackets 

$2x-6=10$

• Then solve as shown previously

  • Add 6 then divide by 2

$\begin{array}{rcl}2x& =& 16\\ x& =& 8\end{array}$

• Expanding brackets first will always work, but you can also divide first

  • Dividing both sides of $2\left(x-3\right)=10$ by 2 gives $\left(x-3\right)=5$

    • which gives $x=8$

  • This method works but can lead to harder fractions

How do I solve linear equations with fractions?

• If a linear equation contains fractions, multiply both sides by the lowest common denominator

• For example, $\frac{x}{5}+4=\frac{9}{2}$

  • The lowest common denominator of 5 and 2 is 10

  • Multiply all terms on both sides by 10

$\begin{array}{rcl}\left(10\times \frac{x}{5}\right)+\left(10\times 4\right)& =& 10\times \frac{9}{2}\\ \frac{10x}{5}+40& =& \frac{90}{2}\end{array}$

• Simplify the fractions

$\begin{array}{rcl}2x+40& =& 45\end{array}$

• Now solve as before, by subtracting 40, then dividing by 2

$\begin{array}{rcl}2x& =& 5\\ x& =& \frac{5}{2}\end{array}$

• Unless the question specifies otherwise, you can leave the answer like this

  • A decimal or mixed number would also be accepted

What if the unknown is on the denominator?

• For example $\frac{4}{x-2}=3$

  • Multiply both sides of the equation by the denominator

$\frac{4}{x-2}\times \left(x-2\right)=3\left(x-2\right)$

• Simplify the fractions, and expand any brackets

$$\begin{gathered} 4=3\left(x-2\right) \\ 4=3x-6 \end{gathered}$$

• Now solve as before, by adding 6 to both sides, then dividing by 3

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

How do I solve linear equations with x terms on both sides?

• Collect the x terms (or whichever variable is involved) together on one side 

  • To do this, remove all the x terms from one side

    • It is easiest to remove the smallest x term to avoid negatives

• For example, $4x-7=11+x$

• Remove the x term on the right-hand side, by subtracting x from both sides

$\begin{array}{rcl}4x-7& =& 11+x\end{array}$

$\begin{array}{rcl}& & \left(-x\right)\left(-x\right)\end{array}$

$\begin{array}{rcl}3x-7& =& 11\end{array}$

• There are no longer any x terms on the right

• This now has the same form as previously seen

  • Solve by adding 7 then dividing by 3         

$\begin{array}{rcl}3x-7& =& 11\\ 3x& =& 18\\ x& =& 6\end{array}$