Mathematical problem solving

Mathematical problem solving means making progress when the route is not stated. The required techniques may be familiar, but you must decide how to represent the information, which structure matters, and whether your result answers the original question.

A reliable cycle is:

1. Understand: identify the target, data, conditions and domain.
2. Represent: use a diagram, table, graph, equation or simpler equivalent form.
3. Plan: connect the representation to a known definition or result.
4. Carry out: write a valid chain of mathematics, keeping exact values where possible.
5. Check and reflect: test the answer, interpret it, and consider whether every case has been found.

The cycle is not rigid. A failed attempt can reveal a better representation, so you may return from carrying out a plan to understanding the problem more precisely.

Prerequisites

You should be able to:

• manipulate algebraic expressions and solve basic equations;
• interpret graphs, tables and diagrams;
• use exact values, approximations and units correctly;
• explain a short chain of reasoning.

Review algebraic fluency if rearranging expressions requires too much attention. Review proof and reasoning if implication, counterexamples or general arguments are unfamiliar.

1. Understand before calculating

Separate four parts of the question.

Part	Question to ask
Target	What exactly must be found, proved or decided?
Data	Which values, equations or facts are given?
Conditions	What restrictions or assumptions apply?
Domain	Which values or objects are allowed?

Underline command words such as exact, positive, integer, maximum and show that. They change what counts as a complete answer.

Worked example: conditions decide the solution

A positive integer is three more than another positive integer. Their product is $40$. Find the integers.

Let the smaller integer be $n$. Then the larger is $n+3$, so

$$n(n+3)=40.$$

Rearrange and factorise:

$$\begin{aligned} n^2+3n-40&=0,\\ (n+8)(n-5)&=0. \end{aligned}$$

Thus $n=-8$ or $n=5$. The algebra has two solutions, but the domain requires a positive integer. Therefore the smaller integer is $5$ and the larger is $8$.

Substitution checks the result:

$8=5+3,\qquad 5\times8=40.$

The discarded root is not an algebraic error. It is a valid solution of the equation but not of the original problem.

Misconception: every number is useful data

A problem may contain redundant information. Do not force every number into a calculation. Work backwards from the target and ask which facts can connect to it.

2. Choose a useful representation

The same information can look difficult in words and simple in another form.

• Use a diagram for geometry, forces, motion or linked lengths.
• Use a table for a pattern, recurrence or several cases.
• Use a graph for intersections, signs, ranges and qualitative behaviour.
• Use an equation for an unknown quantity constrained by a relationship.
• Use a new variable when one expression occurs repeatedly.

Name variables with their meaning and units. Writing $t=$ time in seconds is more useful than introducing an unexplained $x$.

Worked example: draw and label

A rectangle has perimeter $28\text{ cm}$. Its length is $3\text{ cm}$ greater than its width. Find its area.

Let the width be $w\text{ cm}$. A labelled sketch would show dimensions

$w\quad\text{and}\quad w+3.$

The perimeter condition gives

$$2w+2(w+3)=28.$$

Hence

$$4w+6=28,$$

so

$w=\frac{11}{2}.$

The length is

$w+3=\frac{17}{2}.$

Therefore the area is

$$A=\frac{11}{2}\times\frac{17}{2} =\boxed{\frac{187}{4}\text{ cm}^2}.$$

A common error is to write $w+(w+3)=28$. That sum is only half the perimeter. The diagram makes the two pairs of equal sides visible.

Worked example: expose a hidden quadratic

Solve

$$2^{2x}-5(2^x)+4=0.$$

The expression $2^x$ repeats, and

$2^{2x}=(2^x)^2.$

Let $u=2^x$. Then

$$u^2-5u+4=0,$$

so

$$(u-1)(u-4)=0.$$

Thus $u=1$ or $u=4$. Return to the original variable:

$$2^x=1\Rightarrow x=0,$$

or

$$2^x=4\Rightarrow x=2.$$

Therefore $\boxed{x=0\text{ or }x=2}$. The substitution did not finish the problem because the target was $x$, not $u$.

3. Try simpler cases strategically

A simpler case can reveal a pattern, test a conjecture or remove irrelevant complexity. It does not by itself prove a general claim.

Worked example: find and justify a pattern

The first $n$ odd positive integers are

$1,3,5,\ldots,2n-1.$

Find their sum.

Calculate small cases:

$$\begin{array}{c|c} n&1+3+\cdots +(2n-1)\\ \hline 1&1=1^2\\ 2&1+3=4=2^2\\ 3&1+3+5=9=3^2\\ 4&1+3+5+7=16=4^2 \end{array}$$

This suggests

$$1+3+\cdots +(2n-1)=n^2.$$

To justify it, use the arithmetic series formula. There are $n$ terms, with first term $1$ and last term $2n-1$:

$$\begin{aligned} S_n&=\frac n2\bigl(1+(2n-1)\bigr)\\ &=\frac n2(2n)\\ &=n^2. \end{aligned}$$

The examples discovered the conjecture. The algebra proves it for every positive integer $n$.

Misconception: checking many cases proves the rule

No finite list proves a statement about infinitely many values. For example, $n^2-n+41$ is prime for many small positive integers, but at $n=41$ it equals

$41^2-41+41=41^2,$

which is not prime. Use examples to explore, then prove or disprove the resulting claim.

4. Work backwards from the target

When the first step is unclear, ask what would be sufficient to reach the final line. Then ask what would produce that intermediate result.

Worked example: build a chain backwards

The curve

$y=x^2+\frac{4}{x},\qquad x>0,$

has a stationary point. Find its coordinates and determine its nature.

The target needs three things:

1. the $x$ coordinate, found from $dy/dx=0$;
2. the $y$ coordinate, found by substitution;
3. classification as a maximum or minimum.

Differentiate:

$$\frac{dy}{dx}=2x-\frac{4}{x^2}.$$

At a stationary point,

$$2x-\frac{4}{x^2}=0.$$

Since $x>0$, multiplying by $x^2$ is valid:

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

so

$x^3=2\quad\Rightarrow\quad x=\sqrt[3]{2}.$

Now substitute into $y$. Since $x^3=2$, we have $4/x=2x^2$. Therefore

$$y=x^2+2x^2=3x^2=3\sqrt[3]{4}.$$

Finally,

$$\frac{d^2y}{dx^2}=2+\frac{8}{x^3}.$$

At $x=\sqrt[3]{2}$,

$$\frac{d^2y}{dx^2}=2+\frac82=6>0.$$

Hence the stationary point is the minimum

$$\boxed{\left(\sqrt[3]{2},3\sqrt[3]{4}\right)}.$$

Notice that solving $dy/dx=0$ alone does not answer a request for coordinates and nature.

5. Look for structure, not surface detail

Useful structures include symmetry, parity, factors, repeated expressions, conservation laws and quantities that remain unchanged.

Worked example: use symmetry

Without expanding fully, solve

$$(x-2)^2=(x+4)^2.$$

Equal squares mean

$x-2=x+4$

or

$x-2=-(x+4).$

The first equation gives $-2=4$, which is impossible. The second gives

$$x-2=-x-4,$$

so

$2x=-2\quad\Rightarrow\quad x=-1.$

Geometrically, $|x-2|=|x+4|$ asks for the point equally distant from $2$ and $-4$. Their midpoint is

$\frac{2+(-4)}2=-1.$

Both viewpoints give $\boxed{x=-1}$.

Worked example: introduce an invariant

Starting with the pair $(3,8)$, one move adds $2$ to both numbers. Can repeated moves produce the pair $(20,25)$?

The difference between the numbers is unchanged by every move:

$$(b+2)-(a+2)=b-a.$$

Initially the difference is

$8-3=5.$

The target pair also has difference

$25-20=5,$

so this invariant does not rule it out. However, after $k$ moves the first number is

$3+2k.$

To equal $20$, we would need

$3+2k=20\Rightarrow k=\frac{17}{2},$

which is not a whole number. Therefore the target cannot be reached.

The difference condition was necessary but not sufficient. A second invariant, the parity of each number, settles the question: $3+2k$ is always odd, so it can never equal $20$.

6. Manage a stuck attempt

Being stuck usually means that the current representation has stopped producing useful information. Do something diagnostic:

• restate the target in symbols;
• list the conditions you have not yet used;
• sketch or tabulate the situation;
• factorise, complete the square or substitute a repeated expression;
• solve a simpler version;
• work backwards one step;
• test an extreme or boundary case;
• ask whether a theorem’s conditions are satisfied.

Do not erase a failed attempt immediately. Mark why it failed. For example, “this creates two unknowns but only one equation” is useful knowledge.

Worked example: recover from an unhelpful method

Find the minimum value of

$$x+\frac{9}{x},\qquad x>0.$$

Trying values may suggest a minimum near $x=3$, but a table cannot prove it. Rewrite the gap above $6$:

$$\begin{aligned} x+\frac9x-6 &=\frac{x^2-6x+9}{x}\\ &=\frac{(x-3)^2}{x}. \end{aligned}$$

Because $x>0$ and $(x-3)^2\geq0$,

$$\frac{(x-3)^2}{x}\geq0.$$

Therefore

$$x+\frac9x\geq6.$$

Equality occurs when $(x-3)^2=0$, so $x=3$. The minimum value is $\boxed{6}$.

The condition $x>0$ is essential. Without it, dividing by $x$ would not preserve a non-negative expression.

7. Check on several levels

Checking is part of solving, not an optional final glance.

Algebraic check

Substitute solutions into the original equation, especially after squaring, cancelling, taking logarithms or multiplying by an expression containing a variable.

Numerical check

Estimate the sign and size. If $19.8^2/0.21$ is entered into a calculator, an estimate

$$\frac{20^2}{0.2}=2000$$

will expose an output such as $18.7$ as an entry error.

Context check

Check units, domain and physical meaning. A probability must be in $[0,1]$. A negative length is invalid. A modelled answer may need an integer even if the equation gives a decimal.

Completeness check

Ask whether there could be another case. The symbol $\pm$, endpoints, repeated roots and periodic trigonometric solutions are common sources of missing answers.

Worked example: reject an introduced solution

Solve

$$\sqrt{x+5}=x-1.$$

The left side is non-negative, so any solution must satisfy $x-1\geq0$, giving $x\geq1$. Square both sides:

$$x+5=(x-1)^2.$$

Hence

$$x+5=x^2-2x+1,$$

so

$$x^2-3x-4=0.$$

Factorising gives

$$(x-4)(x+1)=0,$$

so the candidates are $x=4$ and $x=-1$.

Check in the original equation:

$$x=4:\quad \sqrt9=3=4-1,$$

but

$$x=-1:\quad \sqrt4=2\ne-2.$$

Therefore the only solution is $\boxed{x=4}$. Squaring preserves every original solution but can introduce extra ones because $a^2=b^2$ does not imply $a=b$.

Self-check

1. A number is doubled and then $7$ is added, giving $31$. Form and solve an equation.
2. Solve $3^{2x}-10(3^x)+9=0$.
3. Explain why the values $2,6,12,20$ suggest, but do not prove, that the $n$th term is $n(n+1)$.
4. Find the minimum value of $x+16/x$ for $x>0$ by writing its difference from $8$ as a non-negative expression.
5. Solve $\sqrt{2x+3}=x$ and check all candidates.
6. A process adds $4$ to an integer on every move. Starting from $7$, can it reach $30$? Give an invariant argument.

Answers

1. Let the number be $x$. Then $2x+7=31$, so $2x=24$ and $x=12$.
2. Let $u=3^x>0$. Then $u^2-10u+9=(u-1)(u-9)=0$, so $u=1$ or $u=9$. Hence $x=0$ or $x=2$.
3. The formula gives the four listed values when $n=1,2,3,4$, but infinitely many later terms remain unchecked. A definition, recurrence or proof is needed to establish the general term.
4. For $x>0$, $$x+\frac{16}{x}-8=\frac{(x-4)^2}{x}\geq0.$$ Therefore the minimum is $8$, attained at $x=4$.
5. Since the square root is non-negative, $x\geq0$. Squaring gives $2x+3=x^2$, so $(x-3)(x+1)=0$. The candidates are $3$ and $-1$, but only $x=3$ satisfies the original equation.
6. No. Adding $4$ preserves the remainder modulo $4$. Every reachable number has the form $7+4k$ and is congruent to $3$ modulo $4$, whereas $30$ is congruent to $2$ modulo $4$.

A compact routine for unfamiliar questions

Before writing a long solution, note:

$$\boxed{\text{target}\quad\text{facts}\quad\text{conditions}\quad\text{representation}}$$

Then choose one justified next step. If it fails, record what the failure teaches and change the representation. At the end, check the original problem rather than only your final equation.

Continue with mathematical modelling to learn how assumptions turn real situations into mathematics. Use calculator and technology skills to investigate and verify results responsibly. For exam questions, choosing a method and checking answers develop this process further.