Words → operation → formula
Total costfixed charge + days × daily rate
→\(C=f+nd\)
Chapter 23
Equations
Build equations from words, solve linear and simultaneous equations accurately, master all three quadratic methods, rearrange formulas, and handle equations containing algebraic fractions.
Writing formulasWriting equationsSolving simple linear equationsSolving equations with a bracketSolving equations with the unknown on both sidesSolving equations involving fractionsSolving harder equations involving fractionsSolving linear simultaneous equationsSolving harder simultaneous equationsSolving quadratic equations of the form \(x^2+bx+c=0\) by factorisationSolving quadratic equations of the form \(ax^2+bx+c=0\) by factorisationSolving quadratic equations by completing the squareSolving quadratic equations using the quadratic formulaChanging the subject of formulasChanging the subject of harder formulasSolving equations involving algebraic fractions
By the end of this chapter, you should be able to
- construct formulas and equations from written information;
- solve linear equations, including brackets, fractions and unknowns on both sides;
- solve simultaneous equations by elimination and substitution;
- solve quadratic equations by factorisation, completing the square and the quadratic formula;
- rearrange formulas, including harder formulas where the new subject occurs more than once or is raised to a power;
- solve equations containing algebraic fractions.
1
Subtopic 1
Writing formulas
Exam-ready method
Choose the letter representing the total or required value as the subject. Attach units to quantities, not algebraic symbols.
A formula is a rule written with letters. Start by deciding what calculation you would do if the quantities were numbers, then replace the quantities by the letters given.
Mean
Meantotal ÷ number
→\(M=\frac hn\)
Worked example · Easy
Rectangle area
The length of a rectangle is \(l\) cm and the width is \(w\) cm. Write a formula for the area \(A\).
\[A=l\times w=\boxed{lw}\]
Worked example · Medium
Fixed charge plus usage
A taxi journey costs a fixed charge of \(f\) dollars plus \(p\) dollars for each of \(n\) kilometres. Write a formula for the total cost \(C\).
Distance charge = \(pn\).
\[\boxed{C=f+pn}\]New worked example · Exam-style
Fixed charge and two rates
A delivery service charges 6 dollars plus 2 dollars for each small parcel and 5 dollars for each large parcel. Write a formula for total cost \(C\) when there are \(s\) small and \(l\) large parcels.
Worked method — show your mathematical steps and check restrictions or units.
\[C=\text{fixed charge}+\text{small-parcel charge}+\text{large-parcel charge}\]
\[\boxed{C=6+2s+5l}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Area formula · Build confidence[2 marks]
Write a formula for the area \(A\) of a triangle with base \(b\) and height \(h\).
Show Answer
Step-by-step solution
\[A=\frac12\times\text{base}\times\text{height}\]
\[\boxed{A=\frac12bh}\]
2Pay plus bonus · Develop skill[3 marks]
A worker earns \(p\) dollars per hour for \(h\) hours and receives a bonus \(B\). Express weekly pay \(W\).
Show Answer
Step-by-step solution
\[\text{Basic pay}=ph\]
\[\boxed{W=ph+B}\]
3Geometry model · Exam-style[4 marks]
A rectangle has width \(w\) and length \(w+5\). Write its perimeter \(P\) in terms of \(w\).
Show Answer
Step-by-step solution
\[P=2w+2(w+5)\]
\[\boxed{P=4w+10}\]
2
Subtopic 2
Writing equations
Exam-ready method
Define the unknown, translate each statement into an expression, form an equality and check the answer against the context.
An equation states that two expressions are equal. Translate each statement carefully, then solve the equation you have formed.
length\(a\)
=
3 × width\(3(a-15)\)
Worked example · Medium
Ages
Mia is \(x\) years old. Her brother is 5 years older. Their ages total 31. Find both ages.
Brother’s age is \(x+5\).
\[x+(x+5)=31\]\[2x+5=31\]\[2x=26\]\[x=13\]Mia is \(\boxed{13}\) and her brother is \(\boxed{18}\).
Worked example · Exam-style
Geometry equation
A rectangle has width \(x\) cm and length \(x+7\) cm. Its perimeter is 42 cm. Find its dimensions.
\[2x+2(x+7)=42\]\[2x+2x+14=42\]\[4x=28\]\[x=7\]
Width \(=\boxed{7\text{ cm}}\), length \(=\boxed{14\text{ cm}}\).
New worked example · Exam-style
Two connected ages
Mina is twice Samir’s age. Their total is 36 years. Find their ages.
Worked method — show your mathematical steps and check restrictions or units.
\[\text{Let Samir}=x,\quad \text{Mina}=2x\]
\[x+2x=36\Rightarrow3x=36\Rightarrow x=12\]
\[\boxed{12\text{ and }24\text{ years}}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Number puzzle · Build confidence[2 marks]
Five less than three times a number equals 22. Find the number.
Show Answer
Step-by-step solution
\[3x-5=22\]
\[3x=27\Rightarrow\boxed{x=9}\]
2Consecutive integers · Develop skill[3 marks]
The sum of two consecutive integers is 57. Find both.
Show Answer
Step-by-step solution
\[x+(x+1)=57\]
\[2x=56\Rightarrow\boxed{28,29}\]
3Perimeter problem · Exam-style[4 marks]
A rectangle’s length is 4 cm greater than its width. Its perimeter is 44 cm. Find the dimensions.
Show Answer
Step-by-step solution
\[2w+2(w+4)=44\]
\[4w=36\Rightarrow w=9\]
\[\boxed{9\text{ cm by }13\text{ cm}}\]
3
Subtopic 3
Solving simple linear equations
Exam-ready method
Use inverse operations and keep both sides balanced; substitute your answer to check it.
Think of an equation as a balanced scale. To keep it balanced, do the same operation to both sides.
Worked example · Easy
Two inverse operations
Solve \(7x+5=26\).
\[7x+5=26\]\[7x=21\quad\text{(subtract 5)}\]\[\boxed{x=3}\quad\text{(divide by 7)}\]
New worked example · Exam-style
Negative coefficient
Solve \(7-4x=27\).
Worked method — show your mathematical steps and check restrictions or units.
\[-4x=27-7=20\]
\[\boxed{x=-5}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Addition and multiplication · Build confidence[2 marks]
Solve \(6x+7=31\).
Show Answer
Step-by-step solution
\[6x=24\]
\[\boxed{x=4}\]
2Negative unknown · Develop skill[3 marks]
Solve \(5-3x=-10\).
Show Answer
Step-by-step solution
\[-3x=-15\]
\[\boxed{x=5}\]
3Decimal coefficient · Exam-style[4 marks]
Solve \(0.4x+2.2=6.2\).
Show Answer
Step-by-step solution
\[0.4x=4\]
\[\boxed{x=10}\]
4
Subtopic 4
Solving equations with a bracket
Exam-ready method
You can first divide by the coefficient outside a single bracket, or expand the bracket; both give the same answer.
You can either expand the bracket first or, when the whole bracket is multiplied by one number, divide first. Both methods are valid.
Method 1: expand\(4(x-3)=20\)\(4x-12=20\)\(x=8\)
Method 2: divide first\(4(x-3)=20\)\(x-3=5\)\(x=8\)
Worked example · Medium
Bracket equation
Solve \(5(2x-3)=35\).
Divide both sides by 5:
\[2x-3=7\]\[2x=10\]\[\boxed{x=5}\]New worked example · Exam-style
Bracket with negative expression
Solve \(5(2x-3)=3x+20\).
Worked method — show your mathematical steps and check restrictions or units.
\[10x-15=3x+20\]
\[7x=35\]
\[\boxed{x=5}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Divide first · Build confidence[2 marks]
Solve \(4(x-2)=20\).
Show Answer
Step-by-step solution
\[x-2=5\]
\[\boxed{x=7}\]
2Expand first · Develop skill[3 marks]
Solve \(3(2x+1)=27\).
Show Answer
Step-by-step solution
\[6x+3=27\Rightarrow6x=24\]
\[\boxed{x=4}\]
3Negative bracket · Exam-style[4 marks]
Solve \(-2(x-4)=18\).
Show Answer
Step-by-step solution
\[x-4=-9\]
\[\boxed{x=-5}\]
5
Subtopic 5
Solving equations with the unknown on both sides
Exam-ready method
Expand if needed, collect all unknown terms on one side and constants on the other, then divide.
Expand brackets if necessary, then collect all the unknown terms on one side and constants on the other. It is usually convenient to keep the final coefficient of the unknown positive.
Worked example · Medium
Unknown on both sides
Solve \(5x-7=2x+11\).
\[5x-7=2x+11\]\[3x-7=11\quad\text{(subtract }2x\text{)}\]\[3x=18\]\[\boxed{x=6}\]
Worked example · Exam-style
Brackets on both sides
Solve \(3(2x+1)=2(x+9)\).
\[6x+3=2x+18\]\[4x=15\]\[\boxed{x=\frac{15}{4}=3.75}\]
New worked example · Exam-style
Nested bracket equation
Solve \(3(2x-1)=2(x+9)+x\).
Worked method — show your mathematical steps and check restrictions or units.
\[6x-3=3x+18\]
\[3x=21\]
\[\boxed{x=7}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Simple collection · Build confidence[2 marks]
Solve \(8x-5=3x+20\).
Show Answer
Step-by-step solution
\[5x=25\]
\[\boxed{x=5}\]
2Unknown with a negative solution · Develop skill[3 marks]
Solve \(2x+9=5x+21\).
Show Answer
Step-by-step solution
\[-3x=12\]
\[\boxed{x=-4}\]
3Two brackets · Exam-style[4 marks]
Solve \(4(x+2)=2(3x-1)\).
Show Answer
Step-by-step solution
\[4x+8=6x-2\]
\[10=2x\]
\[\boxed{x=5}\]
6
Subtopic 6
Solving equations involving fractions
Exam-ready method
Multiply every term of the equation by the lowest common denominator; do not omit constant terms.
The first goal is to remove the fractions. Multiply every term by a common denominator.
\(\frac{x}{3}+\frac{x}{2}=5\)LCD = 6
→
\(2x+3x=30\)fractions eliminated
→
\(x=6\)solve normally
Worked example · Medium
Clear numerical denominators
Solve \(\frac{x}{4}+\frac{x}{6}=10\).
The LCD of 4 and 6 is 12. Multiply every term by 12:
\[3x+2x=120\]\[5x=120\]\[\boxed{x=24}\]Worked example · Medium
Unknown in the denominator
Solve \(\frac{360}{x}=12\).
\[360=12x\]\[\boxed{x=30}\]
New worked example · Exam-style
Three denominators
Solve \(\frac{x}{2}+\frac{x}{3}-\frac{x}{6}=12\).
Worked method — show your mathematical steps and check restrictions or units.
\[\text{Multiply all terms by }6:\quad3x+2x-x=72\]
\[4x=72\Rightarrow\boxed{x=18}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1One fractional unknown · Build confidence[2 marks]
Solve \(\frac{x}{4}+3=8\).
Show Answer
Step-by-step solution
\[\frac{x}{4}=5\]
\[\boxed{x=20}\]
2Two fractions · Develop skill[3 marks]
Solve \(\frac{x}{3}+\frac{x}{6}=7\).
Show Answer
Step-by-step solution
\[\frac{2x+x}{6}=7\]
\[3x=42\Rightarrow\boxed{x=14}\]
3Fraction with constant · Exam-style[4 marks]
Solve \(\frac{2x-1}{5}=3\).
Show Answer
Step-by-step solution
\[2x-1=15\]
\[\boxed{x=8}\]
7
Subtopic 7
Solving harder equations involving fractions
Exam-ready method
Place brackets around any numerator multiplied by the common denominator; expand after clearing every denominator.
When a numerator is an expression, keep it in brackets after multiplying by the common denominator. This prevents sign errors.
Worked example · Exam-style
Two fractional expressions
Solve \(\frac{x-1}{3}-\frac{x+2}{4}=2\).
The LCD is 12:
\[4(x-1)-3(x+2)=24\]\[4x-4-3x-6=24\]\[x-10=24\]\[\boxed{x=34}\]Sign warning
In \(-3(x+2)\), the negative 3 multiplies both terms. Writing the bracket before expanding is the safest method.
New worked example · Exam-style
Brackets in both numerators
Solve \(\frac{2x+1}{3}-\frac{x-2}{4}=5\).
Worked method — show your mathematical steps and check restrictions or units.
\[4(2x+1)-3(x-2)=60\]
\[8x+4-3x+6=60\]
\[5x=50\Rightarrow\boxed{x=10}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Subtraction with brackets · Build confidence[2 marks]
Solve \(\frac{x+2}{3}-\frac{x-1}{6}=4\).
Show Answer
Step-by-step solution
\[2(x+2)-(x-1)=24\]
\[x+5=24\Rightarrow\boxed{x=19}\]
2LCD of 4 and 6 · Develop skill[3 marks]
Solve \(\frac{3x+2}{4}+\frac{x-1}{6}=5\).
Show Answer
Step-by-step solution
\[3(3x+2)+2(x-1)=60\]
\[11x+4=60\]
\[\boxed{x=\frac{56}{11}}\]
3Fractional result · Exam-style[4 marks]
Solve \(\frac{x-3}{2}+\frac{x+1}{5}=4\).
Show Answer
Step-by-step solution
\[5(x-3)+2(x+1)=40\]
\[7x-13=40\]
\[\boxed{x=\frac{53}{7}}\]
8
Subtopic 8
Solving linear simultaneous equations
Exam-ready method
Eliminate one variable by adding or subtracting equations, then substitute back and check both original equations.
Two linear equations in two unknowns normally have one common solution. Graphically, it is the point where their two straight lines intersect; algebraically, elimination finds that same point exactly.
Same signssubtract equations
Opposite signsadd equations
Worked example · Medium
Eliminate directly
Solve \(2x+3y=13\) and \(2x-y=5\).
Subtract the second equation from the first:
\[(2x+3y)-(2x-y)=13-5\]\[4y=8\]\[y=2\]Substitute into \(2x-y=5\):
\[2x-2=5\]\[2x=7\]\[\boxed{x=3.5,\ y=2}\]New worked example · Exam-style
Scale then eliminate
Solve \(2x+3y=17\) and \(3x-2y=6\).
Worked method — show your mathematical steps and check restrictions or units.
\[4x+6y=34,\quad9x-6y=18\]
\[13x=52\Rightarrow x=4\]
\[8+3y=17\Rightarrow\boxed{x=4,\ y=3}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Add equations · Build confidence[2 marks]
Solve \(x+y=11\), \(x-y=3\).
Show Answer
Step-by-step solution
\[2x=14\Rightarrow x=7\]
\[\boxed{x=7,\ y=4}\]
2Eliminate a variable · Develop skill[3 marks]
Solve \(2x+y=13\), \(x+y=8\).
Show Answer
Step-by-step solution
\[\text{Subtract: }x=5\]
\[\boxed{x=5,\ y=3}\]
3Substitution · Exam-style[4 marks]
Solve \(y=3x-2\) and \(x+y=14\).
Show Answer
Step-by-step solution
\[x+3x-2=14\Rightarrow x=4\]
\[\boxed{x=4,\ y=10}\]
9
Subtopic 9
Solving harder simultaneous equations
Exam-ready method
Multiply one or both equations to create matching coefficients, or substitute a simple expression. Interpret both unknowns in context.
If the coefficients do not match, multiply one or both equations first. If one equation already has \(x\) or \(y\) as the subject, substitution may be quicker.
Eliminationmatch one pair of coefficientsadd or subtractsubstitute back
Substitutionuse an equation with \(x=\dots\) or \(y=\dots\)replace that lettersolve and substitute back
Worked example · Exam-style
Multiply both equations
Solve \(3x+2y=8\) and \(5x-3y=7\).
Multiply the first by 3 and the second by 2:
\[9x+6y=24\]\[10x-6y=14\]Add:
\[19x=38\Rightarrow x=2\]Substitute into \(3x+2y=8\):
\[6+2y=8\Rightarrow y=1\]\(\boxed{x=2,\ y=1}\).
Worked example · Medium
Substitution
Solve \(y=3x-4\) and \(2x+y=11\).
\[2x+(3x-4)=11\]\[5x=15\Rightarrow x=3\]\[y=3(3)-4=5\]
\(\boxed{x=3,\ y=5}\).
New worked example · Exam-style
Two prices from total cost
Three adult tickets and two child tickets cost 46 dollars. Two adult tickets and three child tickets cost 44 dollars. Find each price.
Worked method — show your mathematical steps and check restrictions or units.
\[3a+2c=46,\quad 2a+3c=44\]
\[9a+6c=138,\quad4a+6c=88\]
\[5a=50\Rightarrow a=10\]
\[3(10)+2c=46\Rightarrow\boxed{a=10,\ c=8}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Two unknowns · Build confidence[2 marks]
Solve \(3x+2y=19\), \(2x+3y=21\).
Show Answer
Step-by-step solution
\[9x+6y=57,\quad4x+6y=42\]
\[5x=15\Rightarrow x=3\]
\[\boxed{x=3,\ y=5}\]
2Negative coefficient · Develop skill[3 marks]
Solve \(4x-y=5\), \(2x+3y=27\).
Show Answer
Step-by-step solution
\[y=4x-5\]
\[2x+3(4x-5)=27\Rightarrow14x=42\]
\[\boxed{x=3,\ y=7}\]
3Word problem · Exam-style[4 marks]
Two pens and three rulers cost 13 dollars; four pens and one ruler cost 11 dollars. Find the prices.
Show Answer
Step-by-step solution
\[2p+3r=13,\quad 4p+r=11\]
\[r=11-4p\Rightarrow2p+33-12p=13\]
\[p=2\Rightarrow r=3\]
\[\boxed{\text{Pen }2,\ \text{ruler }3}\]
10
Subtopic 10
Solving quadratic equations of the form \(x^2+bx+c=0\) by factorisation
Exam-ready method
Rearrange to zero, find two numbers with the given sum and product, factorise and set each factor equal to zero.
First rearrange so one side is zero. Factorise, then use the zero-product rule: if \(AB=0\), then \(A=0\) or \(B=0\).
\((x-2)(x-5)=0\)
splits into\(x-2=0\)or\(x-5=0\)
\(x=2\) or \(x=5\)Worked example · Easy
Two distinct roots
Solve \(x^2-7x+10=0\).
\[(x-5)(x-2)=0\]\[x-5=0\quad\text{or}\quad x-2=0\]
\(\boxed{x=5\text{ or }x=2}\).
Worked example · Medium
Repeated root
Solve \(x^2-6x+9=0\).
\[(x-3)^2=0\]
So \(\boxed{x=3}\), a repeated solution.
New worked example · Exam-style
Quadratic with two negative roots
Solve \(x^2+11x+28=0\).
Worked method — show your mathematical steps and check restrictions or units.
\[x^2+11x+28=(x+4)(x+7)\]
\[(x+4)(x+7)=0\]
\[\boxed{x=-4\text{ or }x=-7}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Positive roots · Build confidence[2 marks]
Solve \(x^2-8x+15=0\).
Show Answer
Step-by-step solution
\[(x-3)(x-5)=0\]
\[\boxed{x=3\text{ or }x=5}\]
2Opposite signs · Develop skill[3 marks]
Solve \(x^2+x-12=0\).
Show Answer
Step-by-step solution
\[(x+4)(x-3)=0\]
\[\boxed{x=-4\text{ or }x=3}\]
3Rearrange first · Exam-style[4 marks]
Solve \(x^2+2x=24\).
Show Answer
Step-by-step solution
\[x^2+2x-24=(x+6)(x-4)=0\]
\[\boxed{x=-6\text{ or }x=4}\]
11
Subtopic 11
Solving quadratic equations of the form \(ax^2+bx+c=0\) by factorisation
Exam-ready method
Factorise \(ax^2+bx+c\) by checking factor pairs of \(a\) and \(c\). Expand your proposed factors to verify the middle term.
Factorise exactly as in Chapter 20, then set each bracket equal to zero.
Worked example · Exam-style
Coefficient of \(x^2\) is not 1
Solve \(2x^2+7x+3=0\).
\[(2x+1)(x+3)=0\]\[2x+1=0\quad\text{or}\quad x+3=0\]\[\boxed{x=-\frac12\text{ or }x=-3}\]
New worked example · Exam-style
Non-unit leading coefficient
Solve \(6x^2+7x-3=0\).
Worked method — show your mathematical steps and check restrictions or units.
\[6x^2+7x-3=(3x-1)(2x+3)\]
\[(3x-1)(2x+3)=0\]
\[\boxed{x=\frac13\text{ or }x=-\frac32}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Positive constants · Build confidence[2 marks]
Solve \(2x^2+7x+3=0\).
Show Answer
Step-by-step solution
\[(2x+1)(x+3)=0\]
\[\boxed{x=-\frac12\text{ or }x=-3}\]
2Difference signs · Develop skill[3 marks]
Solve \(3x^2-5x-2=0\).
Show Answer
Step-by-step solution
\[(3x+1)(x-2)=0\]
\[\boxed{x=-\frac13\text{ or }x=2}\]
3Take out a common factor · Exam-style[4 marks]
Solve \(4x^2-12x=0\).
Show Answer
Step-by-step solution
\[4x(x-3)=0\]
\[\boxed{x=0\text{ or }x=3}\]
12
Subtopic 12
Solving quadratic equations by completing the square
Exam-ready method
Take half the coefficient of \(x\), square it, then compensate for that term. Use both positive and negative square roots.
Completing the square rewrites a quadratic so one side becomes a perfect square. For \(x^2+bx\), take half of \(b\), then square it.
\(x^2+6x\)half of 6 is 3
→
add \(3^2=9\)\(x^2+6x+9\)
→
\((x+3)^2\)perfect square
Worked example · Exam-style
Solve by completing the square
Solve \(x^2-6x+2=0\), giving exact answers.
\[x^2-6x=-2\]
Half of \(-6\) is \(-3\); add 9 to both sides:
\[(x-3)^2=7\]\[x-3=\pm\sqrt7\]\[\boxed{x=3\pm\sqrt7}\]Worked example · Medium
Completed-square form and minimum
Write \(x^2+8x+3\) in completed-square form and state its minimum value.
\[x^2+8x+3=(x+4)^2-16+3=(x+4)^2-13\]
Since a square cannot be negative, the minimum value is \(\boxed{-13}\), when \(x=-4\).
New worked example · Exam-style
A quadratic with no real roots
Solve \(x^2+6x+13=0\) over the real numbers.
Worked method — show your mathematical steps and check restrictions or units.
\[(x+3)^2+4=0\]
\[(x+3)^2=-4\]
\[\boxed{\text{No real solutions}}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Simple completing square · Build confidence[2 marks]
Solve \(x^2+4x-5=0\).
Show Answer
Step-by-step solution
\[(x+2)^2-9=0\]
\[x+2=\pm3\]
\[\boxed{x=1\text{ or }x=-5}\]
2Decimal roots · Develop skill[3 marks]
Solve \(x^2-6x+1=0\).
Show Answer
Step-by-step solution
\[(x-3)^2=8\]
\[x=3\pm\sqrt8\]
\[\boxed{x=3\pm2\sqrt2}\]
3Leading coefficient · Exam-style[4 marks]
Solve \(2x^2+8x-10=0\) by completing the square.
Show Answer
Step-by-step solution
\[x^2+4x-5=0\]
\[(x+2)^2=9\]
\[\boxed{x=1\text{ or }x=-5}\]
13
Subtopic 13
Solving quadratic equations using the quadratic formula
Exam-ready method
Identify \(a,b,c\) including their signs. Use the discriminant \(b^2-4ac\) to check the number of real roots and round only at the end.
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
identify \(a\)identify \(b\)identify \(c\)substitute with signs
For \(ax^2+bx+c=0\), substitute the values of \(a\), \(b\) and \(c\) carefully. The whole numerator is divided by \(2a\).
Worked example · Exam-style
Use the formula
Solve \(2x^2+3x-4=0\), correct to 2 decimal places.
Here \(a=2\), \(b=3\), \(c=-4\).
\[x=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{4}=\frac{-3\pm\sqrt{41}}4\]So \(\boxed{x\approx0.85\text{ or }x\approx-2.35}\).
Formula errors to avoid
Keep the sign of \(c\), especially when it is negative, and put the entire expression \(-b\pm\sqrt{b^2-4ac}\) over \(2a\).
New worked example · Exam-style
Two irrational roots
Solve \(2x^2-3x-1=0\) exactly.
Worked method — show your mathematical steps and check restrictions or units.
\[a=2,\ b=-3,\ c=-1\]
\[x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-1)}}{2(2)}\]
\[\boxed{x=\frac{3\pm\sqrt{17}}4}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Read coefficients · Build confidence[2 marks]
Solve \(x^2-x-6=0\) using the quadratic formula.
Show Answer
Step-by-step solution
\[x=\frac{1\pm\sqrt{1+24}}{2}\]
\[\boxed{x=3\text{ or }x=-2}\]
2Repeated root · Develop skill[3 marks]
Solve \(x^2+6x+9=0\).
Show Answer
Step-by-step solution
\[\Delta=6^2-4(1)(9)=0\]
\[x=\frac{-6}{2}\]
\[\boxed{x=-3}\]
3No real root · Exam-style[4 marks]
How many real solutions does \(x^2+2x+5=0\) have?
Show Answer
Step-by-step solution
\[\Delta=2^2-4(1)(5)=-16\]
\[\boxed{\text{No real roots since }\Delta<0}\]
14
Subtopic 14
Changing the subject of formulas
Exam-ready method
Undo operations in reverse order until the required letter is alone.
Treat a formula like an equation. Undo operations one at a time until the required letter is isolated.
\(T=3(a+c)\)
expand\(T=3a+3c\)
subtract \(3a\)\(T-3a=3c\)
divide by 3\(c=\frac{T-3a}{3}\)
Worked example · Medium
Make a letter the subject
Make \(r\) the subject of \(P=q+\frac{r}{s}\).
\[P-q=\frac rs\]\[s(P-q)=r\]
\(\boxed{r=s(P-q)}\).
Worked example · Exam-style
Subject inside a bracket
Make \(w\) the subject of \(P=2(l+w)\).
\[\frac P2=l+w\]\[\boxed{w=\frac P2-l}\]
New worked example · Exam-style
Rearrange a formula with brackets
Make \(x\) the subject of \(y=4(3x-2)\).
Worked method — show your mathematical steps and check restrictions or units.
\[\frac{y}{4}=3x-2\]
\[\frac{y}{4}+2=3x\]
\[\boxed{x=\frac{y+8}{12}}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Linear formula · Build confidence[2 marks]
Make \(t\) the subject of \(v=u+at\).
Show Answer
Step-by-step solution
\[v-u=at\]
\[\boxed{t=\frac{v-u}{a}}\]
2Perimeter formula · Develop skill[3 marks]
Make \(w\) the subject of \(P=2(l+w)\).
Show Answer
Step-by-step solution
\[\frac P2=l+w\]
\[\boxed{w=\frac P2-l}\]
3Circle circumference · Exam-style[4 marks]
Make \(r\) the subject of \(C=2\pi r\).
Show Answer
Step-by-step solution
\[\frac{C}{2\pi}=r\]
\[\boxed{r=\frac{C}{2\pi}}\]
15
Subtopic 15
Changing the subject of harder formulas
Exam-ready method
If the subject appears twice, collect its terms then factorise. If the subject is squared or cubed, apply the appropriate root after isolating the power.
Harder formulas may contain a power/root of the new subject, or the new subject may appear more than once. If it appears more than once, collect all its terms together and factorise.
Worked example · Medium
Cube root required
Make \(r\) the subject of \(V=\frac43\pi r^3\).
\[3V=4\pi r^3\]\[r^3=\frac{3V}{4\pi}\]\[\boxed{r=\sqrt[3]{\frac{3V}{4\pi}}}\]
Worked example · Exam-style
Subject appears twice
Make \(x\) the subject of \(a=\frac{x+cx}{d}\).
\[ad=x+cx\]\[ad=x(1+c)\]\[\boxed{x=\frac{ad}{1+c}}\]
New worked example · Exam-style
Subject occurs on both sides
Make \(x\) the subject of \(p=\frac{x}{a}+bx\).
Worked method — show your mathematical steps and check restrictions or units.
\[ap=x+abx=x(1+ab)\]
\[\boxed{x=\frac{ap}{1+ab}}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Square root subject · Build confidence[2 marks]
Make \(r\) the subject of \(A=\pi r^2\), assuming \(r\geq0\).
Show Answer
Step-by-step solution
\[r^2=\frac A\pi\]
\[\boxed{r=\sqrt{\frac A\pi}}\]
2Cube root subject · Develop skill[3 marks]
Make \(r\) the subject of \(V=\frac43\pi r^3\).
Show Answer
Step-by-step solution
\[r^3=\frac{3V}{4\pi}\]
\[\boxed{r=\sqrt[3]{\frac{3V}{4\pi}}}\]
3Factor common subject · Exam-style[4 marks]
Make \(x\) the subject of \(ax+bx=c\).
Show Answer
Step-by-step solution
\[x(a+b)=c\]
\[\boxed{x=\frac c{a+b}}\]
16
Subtopic 16
Solving equations involving algebraic fractions
Exam-ready method
Write denominator restrictions before multiplying by an algebraic common denominator. Check candidate roots in the original equation to reject forbidden values.
Find a common denominator containing the algebraic factors, multiply every term by it, then expand and solve the resulting linear or quadratic equation.
1factor denominators if needed
2find common denominator
3multiply every term
4expand, collect and solve
Worked example · Exam-style
Algebraic denominators
Solve \(\frac{2}{x}+\frac{1}{x-1}=3\).
The common denominator is \(x(x-1)\).
\[2(x-1)+x=3x(x-1)\]\[3x-2=3x^2-3x\]\[3x^2-6x+2=0\]This does not factorise simply, so use the quadratic formula:
\[x=\frac{6\pm\sqrt{36-24}}6=1\pm\frac{\sqrt3}{3}\]\(\boxed{x=1\pm\frac{\sqrt3}{3}}\), noting that original denominators require \(x\ne0,1\).
New worked example · Exam-style
Reject an extraneous root
Solve \(\frac{x}{x-2}=\frac{2}{x-2}\).
Worked method — show your mathematical steps and check restrictions or units.
\[x\ne2,\quad \text{multiply by }x-2\]
\[x=2\text{ is the only candidate}\]
\[\boxed{\text{No solution because }x=2\text{ is excluded}}\]
Try it yourself — from confidence to examination level
Attempt each question before opening its step-by-step solution.
1Simple reciprocal · Build confidence[2 marks]
Solve \(\frac{12}{x}=3\).
Show Answer
Step-by-step solution
\[x\ne0,\quad12=3x\]
\[\boxed{x=4}\]
2Algebraic denominator · Develop skill[3 marks]
Solve \(\frac{x+1}{x-1}=3\).
Show Answer
Step-by-step solution
\[x\ne1,\quad x+1=3(x-1)\]
\[2x=4\]
\[\boxed{x=2}\]
3Quadratic after clearing · Exam-style[4 marks]
Solve \(x+\frac6x=5\).
Show Answer
Step-by-step solution
\[x\ne0,\quad x^2+6=5x\]
\[x^2-5x+6=(x-2)(x-3)=0\]
\[\boxed{x=2\text{ or }x=3}\]
Chapter 23 practice
The questions move from forming equations and linear solving to simultaneous equations, quadratics, rearranging formulas and algebraic fractions. Keep each answer closed until you have attempted the question.
A. Forming and solving linear equations
1Write a formula[2 marks]
The total cost \(C\) is a fixed charge \(f\) plus \(n\) items at \(p\) dollars each. Write a formula for \(C\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[\boxed{C=f+np}\]2Form an equation[4 marks]
One number is 7 greater than another. Their total is 41. Let the smaller number be \(x\). Form an equation and find both numbers.
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[x+(x+7)=41\]\[2x=34\]\[x=17\]The numbers are \(\boxed{17\text{ and }24}\).
3Bracket equation[3 marks]
Solve \(4(2x-1)=28\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[2x-1=7\]\[2x=8\]\[\boxed{x=4}\]4Unknown on both sides[3 marks]
Solve \(7x-4=3x+20\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[4x=24\]\[\boxed{x=6}\]5Fraction equation[4 marks]
Solve \(\frac{x}{3}+\frac{x}{4}=14\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
Multiply every term by 12:
\[4x+3x=168\]\[7x=168\]\[\boxed{x=24}\]6Harder fractions[5 marks]
Solve \(\frac{x+2}{5}-\frac{x-1}{3}=1\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
Multiply by 15:
\[3(x+2)-5(x-1)=15\]\[3x+6-5x+5=15\]\[-2x=4\]\[\boxed{x=-2}\]B. Simultaneous equations
7Direct elimination[4 marks]
Solve \(x+y=9\) and \(x-y=3\).
Show Answer
Add the equations:
\[2x=12\Rightarrow x=6\]Then \(6+y=9\), so \(y=3\).
\(\boxed{x=6,\ y=3}\).
8Multiply before eliminating[5 marks]
Solve \(2x+3y=12\) and \(3x-2y=5\).
Show Answer
Multiply first equation by 2 and second by 3:
\[4x+6y=24\]\[9x-6y=15\]Add: \(13x=39\), so \(x=3\).
Then \(6+3y=12\), so \(y=2\).
\(\boxed{x=3,\ y=2}\).
9Substitution[5 marks]
Solve \(y=2x+1\) and \(3x+y=16\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[3x+(2x+1)=16\]\[5x=15\Rightarrow x=3\]\[y=2(3)+1=7\]\(\boxed{x=3,\ y=7}\).
C. Quadratic equations
10Simple factorisation[4 marks]
Solve \(x^2-9x+20=0\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[(x-4)(x-5)=0\]\(\boxed{x=4\text{ or }x=5}\).
11General quadratic factorisation[5 marks]
Solve \(3x^2-x-2=0\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[(3x+2)(x-1)=0\]\(\boxed{x=-\frac23\text{ or }x=1}\).
12Complete the square[5 marks]
Solve \(x^2+4x-1=0\) exactly by completing the square.
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[x^2+4x=1\]\[(x+2)^2=5\]\[\boxed{x=-2\pm\sqrt5}\]13Quadratic formula[5 marks]
Solve \(3x^2+2x-7=0\), correct to 2 decimal places.
Show Answer
\[x=\frac{-2\pm\sqrt{2^2-4(3)(-7)}}{6}=\frac{-2\pm\sqrt{88}}6\]
\(\boxed{x\approx1.23\text{ or }x\approx-1.90}\).
14Quadratic from geometry[6 marks]
A rectangle is 6 m longer than it is wide. Its area is 40 m². Find its dimensions.
Show Answer
Let width be \(x\), so length is \(x+6\).
\[x(x+6)=40\]\[x^2+6x-40=0\]\[(x+10)(x-4)=0\]Dimensions must be positive, so \(x=4\).
\(\boxed{4\text{ m by }10\text{ m}}\).
D. Changing the subject and algebraic fractions
15Change the subject[4 marks]
Make \(t\) the subject of \(v=u+at\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[v-u=at\]\[\boxed{t=\frac{v-u}{a}}\]16Formula with a square[5 marks]
Make \(r\) the subject of \(A=\pi r^2\), assuming \(r>0\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[r^2=\frac A\pi\]\[\boxed{r=\sqrt{\frac A\pi}}\]17Subject appears twice[6 marks]
Make \(x\) the subject of \(p=ax+bx\).
Show Answer
Method: Translate the question into an algebraic statement, apply the same operation to both sides or use the stated algebraic method, then verify the result.
\[p=x(a+b)\]\[\boxed{x=\frac{p}{a+b}}\]18Algebraic fraction equation[7 marks]
Solve \(\frac{1}{x}+\frac{1}{x+2}=\frac34\).
Show Answer
Multiply by \(4x(x+2)\):
\[4(x+2)+4x=3x(x+2)\]\[8x+8=3x^2+6x\]\[3x^2-2x-8=0\]\[(3x+4)(x-2)=0\]\(\boxed{x=-\frac43\text{ or }x=2}\). Both are allowed because \(x\ne0,-2\).
Chapter 23 high-grade checklist
Forming equations
I translate words into algebra carefully before solving.
Balance
I apply the same operation to both sides of an equation.
Fractions
I multiply every term by the common denominator and keep expressions in brackets.
Simultaneous equations
I can choose elimination or substitution and always substitute back to find the second unknown.
Quadratics
I can select factorisation, completing the square or the quadratic formula and remember that quadratics normally have two roots.
Changing the subject
If the subject occurs twice, I collect its terms and factorise; if it is powered, I use the appropriate inverse operation.