National 5 Maths
Linear Equations

The letter \(x\) appears only once in this simple equation. So we just perform the necessary operations to make \(x\) the subject of the equation.

First, we will add 2 to both sides, to get \(3x\) on its own. Then we will divide by 3.

$$ \begin{eqnarray} 3x-2 &=& 5\\[6pt] 3x &=& 5+2\\[6pt] 3x &=& 7\\[6pt] x &=& \small\frac{7}{3}\normalsize \end{eqnarray} $$

Note that this solution should be left as an improper (top-heavy) fraction. Mixed numbers and decimals should usually be avoided.

Video solution by Clelland Maths 

There are several ways to solve this equation.

If we wanted to avoid negative number arithmetic as much as possible, we could add \(4a\) to both sides and then subtract 1 from each side, but that would take more steps than is necessary.

The most efficient method is as follows:

$$ \begin{eqnarray} 9-4a &=& 1 \\[6pt] -4a &=& 1-9 \\[6pt] -4a &=& -\!8\\[6pt] a &=& \small\frac{-8}{-4}\normalsize \\[6pt] a &=& 2 \end{eqnarray} $$

Video solution by Clelland Maths 

We are going to bring the \(t\) terms to the left and the constant terms (the pure numbers) to the right.

So we add \(5t\) to both sides (to get rid of the \(-5t\) on the right) and subtract 8 from both sides (to get rid of the +8 on the left).

$$ \begin{eqnarray} 7t+8 &=& 32-5t \\[6pt] 7t+5t &=& 32-8 \\[6pt] 12t &=& 24\\[6pt] t &=& \small \frac{24}{12} \normalsize \\[6pt] t &=& 2 \end{eqnarray} $$

Video solution by Clelland Maths 

Our first step should be to expand the right hand side to get rid of the brackets.

Then we take the necessary steps to bring the \(n\) terms to one side and the constant terms to the other.

$$ \begin{eqnarray} 4n-3 &=& \overparen{\overparen{3(3n}-2})\\[6pt] 4n-3 &=& 9n-6\\[6pt] 4n-9n &=& -6+3\\[6pt] -5n &=& -\!3\\[6pt] n &=& \small \frac{3}{5} \normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

The "over 3" needs to be eliminated, so we multiply the equation through by 3.

$$ \begin{eqnarray} \small \frac{x}{3} \normalsize -2 &=& x \\[6pt] x-6 &=& 3x \\[6pt] x-3x &=& 6 \\[6pt] -2x &=& 6 \\[6pt] x &=& -\!3 \end{eqnarray} $$

Video solution by Clelland Maths 

When you get a "fraction equals fraction" equation like this, you can cross-multiply.

Alternatively, give both fractions a common denominator and multiply through by it to get rid of the fractions:

$$ \begin{eqnarray} \small\frac{x-2}{3}\normalsize &=& \small\frac{2x+1}{2}\normalsize \\[8pt] \small\frac{2(x-2)}{6}\normalsize &=& \small\frac{3(2x+1)}{6}\normalsize \\[8pt] 2(x-2) &=& 3(2x+1)\\[8pt] 2x-4 &=& 6x+3\\[8pt] 2x-6x &=& 3+4\\[6pt] -4x &=& 7\\[6pt] x &=& -\small \frac{7}{4} \normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

This is another "fraction equals fraction" equation, so we will cross-multiply.

You need to know how to handle the negative sign on the right hand side. What we do is to move it either to the top or the bottom of the fraction before cross-multiplying. In the solution below, we have moved it to the top.

$$ \begin{gather} \small\frac{x+2}{5}\normalsize = -\small\frac{3x}{4}\normalsize \\[8pt] \small\frac{x+2}{5}\normalsize = \small\frac{-3x}{4}\normalsize \\[8pt] 5(-3x) = 4(x+2)\\[8pt] -15x = 4x+8 \\[8pt] -15x-4x = 8\\[6pt] -19x = 8\\[6pt] x = -\small \frac{8}{19} \normalsize \end{gather} $$

Video solution by Clelland Maths 

This nasty example stretches the limits of what it would be fair to ask you to do at National 5.

The key to solving this is rewriting the entire equation with a common denominator. The lowest common factor of 6 and 8 is 24. Don't just use 6\(\times\)8\(=\)48 because that would make the numbers bigger than necessary.

Note that even the whole number in the equation should be forced to be over 24. We will use \(2=\frac{48}{24}\).

Once we have each term written over 24, we will just multiply the whole equation through by 24 to get rid of all the fractions. Then it will be easy.

$$ \begin{eqnarray} \small\frac{5}{6}\normalsize a-2 &=& \small\frac{3a+4}{8}\normalsize \\[8pt] \small\frac{5}{6}\normalsize a-2 &=& \small\frac{3}{8}\normalsize a + \small\frac{4}{8}\normalsize \\[8pt] \small\frac{20}{24}\normalsize a-\small\frac{48}{24}\normalsize &=& \small\frac{9}{24}\normalsize a + \small\frac{12}{24}\normalsize \\[10pt] 20a-48 &=& 9a+12\\[6pt] 20a-9a &=& 12+48\\[6pt] 11a &=& 60\\[6pt] a &=& \small \frac{60}{11} \normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

Here, we need to expand the bracket and collect like terms before solving. The SQA has never set a question of this type in a Nat 5 exam, but it is compatible with the course specification .

$$ \begin{eqnarray} 6-2(3x-5) &=& -\!10x\\[6pt] 6-6x+10 &=& -\!10x\\[6pt] -6x+16 &=& -\!10x\\[6pt] -6x+10x &=& -\!16\\[6pt] 4x &=& -\!16\\[6pt] x &=& -\!4 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{2x}{3}\normalsize - \small\frac{5}{6}\normalsize &=& 2x \\[8pt] \small\frac{4x}{6}\normalsize - \small\frac{5}{6}\normalsize &=& 2x \\[8pt] \small\frac{4x-5}{6}\normalsize &=& 2x \\[8pt] 4x-5 &=& 12x \\[6pt] 4x-12x &=& 5 \\[6pt] -8x &=& 5 \\[6pt] x &=& -\!\small\frac{5}{8}\normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

This equation arose within a similar shapes question in the 2017 non-calculator paper. Here is a link to the full question.

The inclusion of decimal numbers in a non-calculator equation was surprising, but the arithmetic is fair.

Because this is a "fraction equals fraction" equation, we can cross-multiply to begin.

Note that \(5\times 2.6\) can be done mentally as it's half of \(10\times 2.6\small.\)

$$ \begin{eqnarray} \small\frac{7}{5}\normalsize &=& \small\frac{x+2.6}{x}\normalsize \\[8pt] 7x &=& 5(x+2.6)\\[6pt] 7x &=& 5x+13\\[6pt] 7x-5x &=& 13\\[6pt] 2x &=& 13\\[6pt] x &=& \small \frac{13}{2} \normalsize \end{eqnarray} $$

As this question was to do with the lengths of the sides of triangles, it probably makes more sense to give the final answer as \(6.5\small,\) but either form is acceptable.

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{x}{2}\normalsize -1 &=& \small\frac{3-x}{5}\normalsize \\[8pt] \small\frac{5x}{10}\normalsize -1 &=& \small\frac{2(3-x)}{10}\normalsize \\[8pt] \small\frac{5x}{10}\normalsize -1 &=& \small\frac{6-2x}{10}\normalsize \\[8pt] \small\frac{5x}{10}\normalsize - \small\frac{10}{10}\normalsize &=& \small\frac{6-2x}{10}\normalsize \\[8pt] \small\frac{5x-10}{10}\normalsize &=& \small\frac{6-2x}{10}\normalsize \\[8pt] 5x-10 &=& 6-2x \\[6pt] 5x+2x &=& 6+10 \\[6pt] 7x &=& 16 \\[6pt] x &=& \small\frac{16}{7}\normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

This equation arose in context within this 2022 non-calculator question.

There are several ways to solve it, but perhaps the easiest is to get rid of the fraction right at the beginning, by multiplying both sides by 2.

$$ \begin{eqnarray} \small\frac{3}{2}\normalsize (x+12) &=& 6(8-x) \\[8pt] 3(x+12) &=& 12(8-x)\\[6pt] 3x+36 &=& 96-12x\\[6pt] 3x+12x &=& 96-36\\[6pt] 15x &=& 60\\[6pt] x &=& \small\frac{60}{15}\normalsize\\[6pt] x &=& 4 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} \small\frac{5x+1}{2}\normalsize &=& \small\frac{4x}{3}\normalsize +1 \\[8pt] \small\frac{3(5x+1)}{6}\normalsize &=& \small\frac{2(4x)}{6}\normalsize + \small\frac{6}{6} \\[8pt] 15x+3 &=& 8x+6 \\[6pt] 15x-8x &=& 6-3 \\[6pt] 7x &=& 3 \\[6pt] x &=& \small\frac{3}{7}\normalsize \end{eqnarray} $$