Linear Inequations (SQA National 5 Maths): Revision Note

When solving inequations, remember:

• Do not change the inequality sign to an equals sign

  • In an exam you can lose marks for doing this 

• Reverse the direction of the inequality sign when multiplying or dividing both sides by a negative number!

Solve, algebraically, the inequation $4\left(x-3\right)+7<9x+10$.

Answer:

Expand the brackets and simplify

$\begin{array}{rcl}4x-12+7& <& 9x+10\\ 4x-5& <& 9x+10\end{array}$

Subtract 4x from both sides

$\begin{array}{rcl}-5& <& 9x+10-4x\\ -5& <& 5x+10\end{array}$

Subtract 10 from both sides

$\begin{array}{rcl}-5-10& <& 5x\\ -15& <& 5x\end{array}$

Divide both sides by 5

• 5 is not a negative number, so there is no need to flip the inequality sign

$\begin{array}{rcl}\frac{-15}{5}& <& x\\ -3& <& x\end{array}$

That is the answer you are looking for

• Although you could also 'flip the inequation over' to get x on the left

$-3<x\mathrm{or}x>-3$ 

That Worked Example was solved in a way to avoid having to divide by a negative number. Instead the solution could have proceeded like this:

$\begin{array}{rcl}4x-5& <& 9x+10\\ 4x-5-9x& <& 10\\ -5x-5& <& 10\\ -5x& <& 10+5\\ -5x& <& 15\end{array}$

That can be solved by dividing by -5, but then you need to remember to flip the inequality sign:

$\begin{array}{rcl}x& \mathbf{>}& \frac{15}{-5}\\ x& >& -3\end{array}$

Solve, algebraically, the inequation $\frac{x-1}{4}+3>\frac{6x}{5}$.

Answer:

Multiply both sides by 4 to get rid of the fraction on the left

• 4 is not a negative number, so there is no need to flip the inequality sign

$\begin{array}{rcl}\frac{\cancel{4}\times \left(x-1\right)}{\cancel{4}}+4\times 3& >& \frac{4\times 6x}{5}\\ \left(x-1\right)+12& >& \frac{24x}{5}\\ x+11& >& \frac{24x}{5}\end{array}$

Multiply both sides by 5 to get rid of the fraction on the right

• 5 is not a negative number, so there is no need to flip the inequality sign

$\begin{array}{rcl}5\left(x+11\right)& >& \frac{\cancel{5}\times 24x}{\cancel{5}}\\ 5x+55& >& 24x\end{array}$

Subtract 5x from both sides

$\begin{array}{rcl}55& >& 24x-5x\\ 55& >& 19x\end{array}$

Divide both sides by 19

• 19 is not a negative number, so there is no need to flip the inequality sign

$\frac{55}{19}>x$

That is the answer you are looking for

• Although you could also 'flip the inequation over' to get x on the left

$\frac{55}{19}>x\mathrm{or}x<\frac{55}{19}$ 

If the question does not say x is an integer, do not assume x is an integer!

• x > 3 actually means any value greater than 3

  • 3.1 is possible

  • $\pi$ = 3.14159... is possible

List all the integer values of $x$ that satisfy 

$-4\le x<2$

Answer:

Integer values are whole numbers 

• -4 ≤ x shows that x includes -4, so this is the first integer

x = -4

x < 2 shows that x does not include 2

• Therefore the last integer is x = 1

x = 1

For the answer, list all the integers from -4 to 1

• Remember integers can be zero and negative

$x=-4,-3,-2,-1,0,1$