Linear Equations (SQA National 5 Maths): Revision Note

Solving linear equations with integer coefficients

What are linear equations?

• A linear equation is one that can be written (or rewritten) 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 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 usual

  • Add 6 then divide by 2

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

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

    • This will be the most negative x term, if one or both of the x terms have a minus sign in front of them

• 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 can now be solved as usual

  • Add 7 then divide by 3         

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

Solving linear equations with fractional coefficients

How do I solve linear equations with fractions?

• If a linear equation contains fractions (including algebraic 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 as a fraction

  • An equivalent decimal or mixed number would usually also be accepted

What if the unknown is in 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 usual

  • Add 6 then divide by 3

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