Maths · 2 Algebra

2.5 Equations

Before you start

Collecting like terms and expanding brackets, including $- 3 (x - 2) = - 3 x + 6$ .
Factorising a quadratic such as $x^{2} + 5 x + 6$ into $(x + 2) (x + 3)$ .
Adding fractions with different denominators by finding a common denominator.

Syllabus objectives covered (7)

Construct expressions, equations and formulas.
Solve linear equations in one unknown.
Solve fractional equations with numerical and linear algebraic denominators.
Solve simultaneous linear equations in two unknowns.
Solve quadratic equations by factorisation, using a graphic display calculator and by use of the quadratic formula.
Change the subject of formulas.
Use a graphic display calculator to solve equations, including those which may be unfamiliar.

Part 1 of 7

Turning words into algebra

Before you can solve anything you have to write it down. This is the skill people skip, and then lose marks on in every applied question later.
Start by naming the unknown explicitly: 'let $n$ be the smaller number'. Write that sentence on the page. If you don't define your letter, an examiner cannot follow your working and neither can you, three lines later.
Then translate phrase by phrase. 'Two more than $n$ ' is $n + 2$ . 'Two consecutive even numbers' are $n$ and $n + 2$ , so their product is $n (n + 2)$ .
Notice the difference between an expression, an equation and a formula. $n (n + 2)$ is an expression — it has no equals sign and nothing to solve. $n (n + 2) = 48$ is an equation. $A = n (n + 2)$ is a formula, which gives one quantity in terms of others.
For a word problem, the equation comes from whatever the question says two things are equal to. 'The perimeter is 34 cm' gives you $2 (l + w) = 34$ . Look for the sentence containing 'is', 'equals', or 'total'.

Part 2 of 7

Linear equations in one unknown

A linear equation has the unknown to the power 1 only — no $x^{2}$ , no $x$ in a denominator. There is one governing idea: whatever you do to one side, do to the other, and the equation stays true.
Work in reverse order to how the expression was built. If $x$ was multiplied by 5 and then had 3 added, undo the $+ 3$ first and the $\times 5$ second.
With brackets, expand first: $5 - 2 x = 3 (x + 7)$ becomes $5 - 2 x = 3 x + 21$ .
$5 - 2 x = 3 x + 21$
Then gather the $x$ terms on the side where they end up positive. Adding $2 x$ to both sides gives $5 = 5 x + 21$ , which avoids the negative-coefficient slips that cost so many marks.
Subtract 21: $- 16 = 5 x$ , so $x = - \frac{16}{5}$ . A fraction or a negative is a perfectly good answer — don't assume you have gone wrong.
$x = - \frac{16}{5} = - 3.2$
Always substitute your answer back into the original equation. It takes fifteen seconds and catches almost every arithmetic error. Checking against your rearranged version only confirms your own mistake.

Part 3 of 7

Fractional equations

These look worse than they are. The single move that solves all of them: multiply every term by the denominators until no fractions remain.
With a numerical denominator, multiply through by the lowest common multiple. For $\frac{x}{3} + \frac{x}{4} = 7$ , multiply everything by 12 to get $4 x + 3 x = 84$ , so $x = 12$ .
With an algebraic denominator, the same idea applies but the denominator is a bracket. For $\frac{x}{2 x + 1} = 4$ , multiply both sides by $(2 x + 1)$ .
$x = 4 (2 x + 1)$
Now it is linear: $x = 8 x + 4$ , so $- 7 x = 4$ and $x = - \frac{4}{7}$ .
Two algebraic denominators need both. For $\frac{2}{x + 2} + \frac{3}{2 x - 1} = 1$ , multiply every term by $(x + 2) (2 x - 1)$ — and multiply the right-hand side too, which is the step people forget.
$2 (2 x - 1) + 3 (x + 2) = (x + 2) (2 x - 1)$
Expand both sides carefully and you get a quadratic, which you then solve by the methods in the next section. A fractional equation quietly becoming a quadratic is normal, not a sign of error.
One check that matters: a value making any denominator zero is not a solution, however cleanly it drops out of the algebra. In the example above, $x = - 2$ and $x = \frac{1}{2}$ would both have to be rejected.

Part 4 of 7

Simultaneous linear equations

Two unknowns need two equations. Geometrically you are finding where two straight lines cross, which is worth picturing — it tells you immediately why there is normally exactly one answer.
Elimination is usually quickest. Make the coefficients of one letter match, then add or subtract the equations to make that letter vanish.
For $3 x + 2 y = 16$ and $x - 2 y = 0$ , the $y$ coefficients are already $+ 2$ and $- 2$ . Adding the equations eliminates $y$ straight away: $4 x = 16$ , so $x = 4$ , and then $y = 2$ .
The rule that saves you: same signs subtract, different signs add. Coefficients $+ 2$ and $- 2$ have different signs, so you add.
Substitution is better when one equation already has a letter on its own. From $y = 2 x - 1$ , replace every $y$ in the other equation and solve the single-unknown equation that results.
Always find both unknowns and state them as a pair. Half the marks live in the second value, and an answer of ' $x = 4$ ' alone is an incomplete solution.
Check by substituting into the equation you did not use to find the second unknown — that tests both values at once.

Part 5 of 7

Quadratic equations: three routes to the same answer

First, rearrange so one side is zero: $x^{2} + 5 x = - 6$ becomes $x^{2} + 5 x + 6 = 0$ . Nothing works until you have done this.
Route 1 — factorisation. If it factorises, this is fastest. $x^{2} + 5 x + 6 = (x + 2) (x + 3) = 0$ . A product is zero only if one of the factors is zero, so $x = - 2$ or $x = - 3$ .
That zero-product step is the actual mathematics; the factorising is just preparation. It is also why you must have zero on one side — $(x + 2) (x + 3) = 12$ tells you nothing useful.
Route 2 — the quadratic formula. It always works, and the syllabus tells you it is given on the formula list, so you do not need to memorise it.
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Substitute carefully with brackets around every negative: for $2 x^{2} - 7 x + 3 = 0$ , that is $a = 2$ , $b = - 7$ , $c = 3$ , giving $x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}$ , so $x = 3$ or $x = \frac{1}{2}$ .
If $b^{2} - 4 a c$ is not a perfect square, leave the answer in surd form when the question asks for it — the syllabus says you may be expected to. $\frac{3 + \sqrt{5}}{2}$ is a better answer than $2.618$ if exactness was requested.
Route 3 — the graphic display calculator. Use the equation solver, or graph $y = 2 x^{2} - 7 x + 3$ and read off where it crosses the $x$ -axis. Both give the same two roots.
Choose by paper. In Paper 2 you have no calculator, so factorise or use the formula. In Papers 4 and 6 the calculator is fastest — but if the question says Show that, you must produce algebra, because a calculator answer shows nothing.

Part 6 of 7

Changing the subject

Rearranging a formula is solving an equation where the answer happens to be a letter instead of a number. Exactly the same moves, in exactly the same reverse order.
Make $r$ the subject of $A = π r^{2}$ : divide by $π$ to get $\frac{A}{π} = r^{2}$ , then take the square root, $r = \sqrt{\frac{A}{π}}$ . A power of the subject is undone by a root, and a root by a power.
The hard case, and the one 0607 examines: the subject appears twice. Then you cannot simply unpick it — you have to collect it.
Make $x$ the subject of $a x + b = c x + d$ . Gather the $x$ terms on one side: $a x - c x = d - b$ .
$a x - c x = d - b$
Now factorise out the $x$ , which is the move the whole method depends on: $x (a - c) = d - b$ , so $x = \frac{d - b}{a - c}$ .
If you can say to yourself 'collect, factorise, divide', you can do every twice-appearing rearrangement in the syllabus.

Part 7 of 7

Using the GDC on equations you don't recognise

Some equations have no algebraic method available to you — $2 x - 1 = \frac{1}{x^{3}}$ , for instance. This is not a trick; the syllabus expects you to reach for the calculator.
The reliable method is graphical. Enter the left-hand side as $Y_{1}$ and the right-hand side as $Y_{2}$ , then find the intersection.
Set a sensible window first. If nothing appears on screen, the roots are outside your view — zoom out before concluding there is no solution.
Look for every intersection. A curve and a line can cross more than once, and the calculator only reports the one you ask it about.
Give non-exact answers to 3 significant figures unless told otherwise, and do not round anything you are going to use again — carry the calculator's full value into later parts.
Write down what you solved and what you found, even when the calculator did the work: 'intersection of $y = 2 x - 1$ and $y = x^{- 3}$ at $x = 1.17$ '. Communication is examined in its own right on Paper 6.

Worked examples

Example 1. The length of a rectangle is $3$ cm more than twice its width. Its perimeter is $42$ cm. Work out the width.

Show the working

Name the unknown first: let the width be $w$ cm.
Translate the first sentence. 'Three more than twice the width' is $2 w + 3$ , so the length is $2 w + 3$ cm.
The perimeter of a rectangle is $2 (length + width)$ , and the question tells us it equals 42.
$2 (2 w + 3 + w) = 42$
Simplify inside the bracket, then expand.
$2 (3 w + 3) = 42 \Rightarrow 6 w + 6 = 42$
Subtract 6 and divide by 6.
$6 w = 36 \Rightarrow w = 6$
Check against the original wording: width 6, length $2 (6) + 3 = 15$ , perimeter $2 (6 + 15) = 42$ . Correct.

Answer: The width is $6$ cm.

Example 2. Solve $\frac{x}{x + 2} = \frac{3}{x - 6}$ .

Show the working

Cross-multiply — which is just multiplying both sides by $(x + 2) (x - 6)$ .
$x (x - 6) = 3 (x + 2)$
Expand both sides.
$x^{2} - 6 x = 3 x + 6$
Bring everything to one side so the quadratic is set equal to zero.
$x^{2} - 9 x - 6 = 0$
This does not factorise, so use the quadratic formula with $a = 1$ , $b = - 9$ , $c = - 6$ .
$x = \frac{9 \pm \sqrt{81 + 24}}{2} = \frac{9 \pm \sqrt{105}}{2}$
Neither root makes a denominator zero — they would have to be $- 2$ or $6$ — so both are valid.
Give the values to 3 significant figures as well, since the question did not ask for an exact form.

Answer: $x = \frac{9 + \sqrt{105}}{2} = 9.62$ or $x = \frac{9 - \sqrt{105}}{2} = - 0.624$ (3 s.f.).

Example 3. Solve the simultaneous equations $4 x + 3 y = 19$ and $2 x - y = 3$ .

Show the working

Look for the easiest match. Doubling the second equation gives $4 x - 2 y = 6$ , matching the $4 x$ in the first.
The $x$ coefficients are now the same sign, so subtract the equations.
$(4 x + 3 y) - (4 x - 2 y) = 19 - 6$
The $x$ terms vanish, leaving $5 y = 13$ , so $y = 2.6$ .
Substitute back into the simpler original equation, $2 x - y = 3$ : $2 x - 2.6 = 3$ , so $2 x = 5.6$ and $x = 2.8$ .
Check in the equation you have not yet used: $4 (2.8) + 3 (2.6) = 11.2 + 7.8 = 19$ . Correct.

Answer: $x = 2.8$ , $y = 2.6$ .

Example 4. Make $t$ the subject of $\frac{t + 1}{t - 2} = k$ .

Show the working

Clear the fraction by multiplying both sides by $(t - 2)$ .
$t + 1 = k (t - 2)$
Expand the right-hand side so every $t$ is visible.
$t + 1 = k t - 2 k$
The subject appears twice, so collect the $t$ terms on one side and everything else on the other.
$t - k t = - 2 k - 1$
Factorise out $t$ — this is the step that makes the whole method work.
$t (1 - k) = - 2 k - 1$
Divide by the bracket. Multiplying numerator and denominator by $- 1$ tidies the signs.
Collect, factorise, divide: the same three moves every time the subject appears twice.

Answer: $t = \frac{- 2 k - 1}{1 - k} = \frac{2 k + 1}{k - 1}$ .

Where people go wrong

Mistake: Multiplying only some terms when clearing a fraction, e.g. turning $\frac{x}{3} + 2 = 5$ into $x + 2 = 5$ .

Instead: Multiply every term on both sides: $x + 6 = 15$ .

The equation stays balanced only if the whole of each side is multiplied. Missing the constant term is the single most common error in fractional equations.

Mistake: Solving $(x + 2) (x + 3) = 12$ by writing $x + 2 = 12$ or $x + 3 = 12$ .

Instead: Expand, set equal to zero, then factorise: $x^{2} + 5 x - 6 = 0$ , giving $x = 1$ or $x = - 6$ .

The zero-product rule only works when the product is zero. There are many pairs of numbers multiplying to 12, so no factor is determined.

Mistake: Substituting a negative $b$ into the quadratic formula without brackets, e.g. writing $- b = - 7$ when $b = - 7$ .

Instead: $b = - 7$ gives $- b = + 7$ . Write brackets around every substituted negative.

A single sign error changes both roots, and because the answer still looks reasonable there is nothing to alert you.

Mistake: Finding one unknown in a pair of simultaneous equations and stopping.

Instead: Always substitute back to find the second, and state both.

A solution is a pair of values. Mark schemes reserve marks for each, so half an answer scores about half the marks at best.

Mistake: Accepting a solution that makes a denominator zero.

Instead: Check each root against the original denominators and reject any that make one zero.

The algebra of clearing fractions can introduce values that were never valid in the original equation.

Mistake: Rounding partway through and carrying the rounded value on.

Instead: Keep the full value in the calculator and round only the final answer, to 3 significant figures.

The 0607 accuracy rules are explicit about this, and early rounding loses accuracy marks even when the method is perfect.

Check yourself

A number is doubled and then 7 is subtracted, giving 19. Write down an equation and solve it.
Answer
Let the number be $n$ . Then $2 n - 7 = 19$ , so $2 n = 26$ and $n = 13$ .
Solve $4 (x - 3) = 2 x + 5$ .
Answer
$4 x - 12 = 2 x + 5$ , so $2 x = 17$ and $x = 8.5$ .
Solve $\frac{5}{x} + 1 = 3$ .
Answer
Multiply through by $x$ : $5 + x = 3 x$ , so $2 x = 5$ and $x = 2.5$ .
Solve $x + y = 10$ and $x - y = 4$ .
Answer
Adding gives $2 x = 14$ , so $x = 7$ and then $y = 3$ .
Solve $x^{2} - 7 x + 12 = 0$ by factorisation.
Answer
$(x - 3) (x - 4) = 0$ , so $x = 3$ or $x = 4$ .
Make $h$ the subject of $V = \frac{1}{3} π r^{2} h$ .
Answer
Multiply by 3 and divide by $π r^{2}$ : $h = \frac{3 V}{π r^{2}}$ .
Make $y$ the subject of $3 y + a = b y - 5$ .
Answer
Collect: $3 y - b y = - 5 - a$ . Factorise: $y (3 - b) = - 5 - a$ . Divide: $y = \frac{- 5 - a}{3 - b} = \frac{5 + a}{b - 3}$ .
How would you use a graphic display calculator to solve $x^{3} = 2 x + 1$ , and how many solutions should you expect to look for?
Answer
Graph $y = x^{3}$ and $y = 2 x + 1$ and find every intersection. A cubic and a line can meet up to three times, so check the whole shape of the curve, not just the first crossing you see.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A rectangle has width $x$ cm. Its length is $4$ cm greater than its width. The area of the rectangle is $96 {cm}^{2}$ .

(a) Show that $x^{2} + 4 x - 96 = 0$ . 2 marks

Mark scheme

M1 x(x + 4) = 96 seen oe
A1 expands and rearranges to the given result, with no errors; nfww

(b) Solve the equation to find the width of the rectangle. 3 marks

Mark scheme

M1 (x + 12)(x - 8) seen, or correct substitution into the quadratic formula
A1 x = 8 and x = -12
B1 width = 8 cm, with -12 rejected as a length cannot be negative; dep on both roots seen

Question 2. Two adult tickets and three child tickets for a museum cost $$ 31$ . One adult ticket and five child tickets cost $$ 27$ .

(a) Write two equations in terms of $a$ , the cost of an adult ticket, and $c$ , the cost of a child ticket. 2 marks No calculator

Mark scheme

B1 2a + 3c = 31
B1 a + 5c = 27

(b) Solve your equations to find the cost of each type of ticket. 3 marks No calculator

Mark scheme

M1 a correct method to eliminate one variable, e.g. doubling the second equation; FT their equations
A1 c = 3.40 oe
A1 a = 10

(c) Work out the cost of one adult ticket and four child tickets. 1 mark

Mark scheme

B1 23.60 cao; FT their (b)

Question 3. The formula for the period of a pendulum is $T = 2 π \sqrt{\frac{L}{g}}$ .

(a) Show that $L = \frac{g T^{2}}{4 π^{2}}$ . 3 marks No calculator

Mark scheme

M1 divides by 2π to isolate the root
M1 squares both sides
A1 reaches the given result with no errors; nfww

(b) Calculate $L$ when $T = 1.8$ and $g = 9.8$ , giving your answer correct to 3 significant figures. 2 marks

Mark scheme

M1 correct substitution into their rearranged formula; FT
A1 0.804 awrt; isw further rounding

Question 4. An equation is given as $x^{3} - 4 x = 2$ .

(a) Sketch the graphs you would use on a graphic display calculator to solve this equation. 2 marks

Mark scheme

B1 y = x³ - 4x, cubic shape with three x-intercepts indicated
B1 y = 2, a horizontal line

(b) Determine all solutions of the equation, giving each correct to 3 significant figures. 3 marks

Mark scheme

B1 -1.68 awrt
B1 -0.539 awrt
B1 2.21 awrt

Words to know

Expression: A collection of terms with no equals sign, such as $3 x + 5$ . It can be simplified but not solved.
Equation: A statement that two expressions are equal, such as $3 x + 5 = 11$ . It can be solved.
Formula: An equation giving one quantity in terms of others, such as $A = π r^{2}$ .
Subject of a formula: The letter standing alone on one side, which the formula gives you directly.
Simultaneous equations: Two or more equations that must hold at the same time, solved together for shared unknowns.
Surd form: An exact answer left with a root sign, such as $\frac{9 + \sqrt{105}}{2}$ , rather than rounded to a decimal.

All Maths lessons Next: 2.6 Inequalities