Linear equations: method and worked examples

Mid-October 2026: in Year 1 of high school, whether it is an art, science or languages liceo or a technical institute, you have usually just finished calculations with polynomials and fractions and the teacher is starting linear (first-degree) equations — first plain numerical ones, then ones with fractions and literal ones. The test on this topic falls in four to seven weeks, so in November. It is the right moment to build your method, because equations are the foundation of all the algebra of the coming years: systems, inequalities, quadratic equations, functions.

The sentence I hear most often is “I know how to do them, but I always get something wrong”. It is almost always true: the procedure is short, and the mistakes are in the signs and the fractions, not in the idea. Here you will find the complete method, the special cases (impossible and indeterminate), the check, and three worked examples with every step.

What it means to solve an equation

A linear equation in the unknown $x$ is an equality that, after doing the calculations, reduces to

with $a$ and $b$ numbers. Solving it means finding the value of $x$ that makes the equality true. If $a\ne 0$ there is exactly one solution:

The rest of the exercise is bringing the equation to that form without breaking the equality. For that there are two rules, the equivalence principles:

1. First principle: you can add or subtract the same number (or the same expression) to both sides. It is the rule that translates into “I move it over and change the sign”.
2. Second principle: you can multiply or divide both sides by the same number different from zero. It is the rule that lets you get rid of the denominators and divide by the coefficient.

The “sign change” is not magic: it is the first principle. If you have $x+5=9$, you subtract $5$ from both sides and get $x=4$. Understanding this avoids the mistake of moving a term without changing it, and the opposite one of changing the sign of everything, even of what does not move.

The procedure, in six steps

With this sequence you can solve any linear equation. Learn the order, and do it the same way every time.

1. Remove the brackets by applying the distributive property (and watch the signs in front of the brackets).
2. Remove the denominators: calculate the LCM of the denominators and multiply both sides by that value.
3. Simplify each side by collecting like terms.
4. Bring the $x$ terms to the left and the numbers to the right, changing the sign of whatever crosses the equals sign.
5. Collect like terms again, until you reach $ax=b$.
6. Divide by the coefficient of $x$ and check.

The step nobody wants to do is the last one, and yet it is the one that gets you the mark: we will see it further on.

Whole equations with brackets

The critical point is the sign in front of the bracket. If there is a minus in front, all the signs inside change:

If there is a number in front, multiply every term:

A classic mistake is writing $3(x-2)=3x-2$, that is, multiplying only the first term. Another is writing $-2(1-x)=-2-2x$: the sign of the second term must become $+2x$, because $(-2)\cdot (-x)=+2x$. If the signs in multiplications still give you trouble, revise products with polynomials in special products: formulas and exercises, where the distributive property is used all the time.

Equations with fractions: the LCM does all the work

When there are numerical denominators, the second principle lets you get rid of them in one go: you multiply all the terms, left and right, by the LCM of the denominators. If calculating the LCM is your weak point, GCD and LCM with prime factorisation explains it with exercises.

Example:

The LCM of $2$ and $3$ is $6$. I multiply every term by $6$:

There are two typical mistakes. The first is multiplying by the LCM only the terms with a fraction and forgetting the whole-number term (here the $5$ becomes $30$). The second is getting the sign wrong when there is a minus in front of the fraction: the fraction has to be put in brackets before removing the denominator. With

I multiply by $2$ and write

If I had written $2x-x-1=4$ I would have found $x=5$, which is wrong. Check with $x=3$: $3-\frac{2}{2}=2$. It works.

When the unknown is in the denominator

In Year 1, rational equations (with $x$ in the denominator) sometimes appear at the end of the chapter. One step is added before anything else: the existence conditions, that is, the values of $x$ that make a denominator zero and so are not allowed. At the end, if the solution you found coincides with one of those values, it has to be discarded.

Existence condition: $x\ne 2$. I multiply by $x-2$ and get $x=2$, which however is exactly the excluded value. The equation is impossible. Whoever forgets the conditions writes "$x=2$" and loses the exercise.

Impossible and indeterminate equations

After reducing, it can happen that the $x$ terms disappear. You have not made a mistake: you are in one of the two special cases.

Impossible: you get a false equality, of the type $0\cdot x=b$ with $b\ne 0$. No value of $x$ makes it true.

The answer is written “impossible equation”, solution set $\varnothing$.

Indeterminate: you get an always-true equality, $0\cdot x=0$. Any value of $x$ is a solution: the equation is an identity.

Here the answer is “indeterminate equation”, solutions: all numbers.

The trick to avoid mixing them up: look at what is left when the $x$ terms cancel. If what is left is false ($2=1$) it is impossible; if what is left is true ($0=0$) it is indeterminate. Never write "$x=0$" in these cases: $x=0$ is a genuine solution, whereas here the solutions are none or infinitely many.

With a parameter (literal equations)

When the coefficients contain a letter, for example $a$, the equation is discussed: you have to say what happens for each value of the parameter. The typical case is

If $a\ne 2$ you can divide and find $x=\frac{4}{a-2}$. If $a=2$ the equation becomes $0\cdot x=4$, so it is impossible. The question “for which values of $a$ can I divide?” is always the first to ask: it is the same attention to a zero divisor that you had for the second principle.

The check: thirty seconds that are worth a mark

To check a solution you just substitute it into the starting equation (not the one you have already simplified) and calculate the two sides separately. They must give the same number. If they do not match, the mistake is almost always in one of the first three steps of the procedure, so redo those.

Do it carefully above all with fractions: that is where signs escape. An equation solved and checked in a test earns the full answer; a wrong solution that is not checked leaves you with a result you yourself do not know is right.

Three worked examples

Try to solve them on your own before reading the solution.

Exercise 1: whole equation with brackets

I remove the brackets:

I collect like terms on the left:

I bring the $x$ terms to the left and the numbers to the right:

Check: on the left $3\cdot 11-2\cdot (-12)=33+24=57$; on the right $4\cdot 13+5=57$. It works.

To train on a wider range of Year 1 tests, there are the practice tests: they are calibrated on the high school syllabus, so use them as training on the topics common to all tracks.

Exercise 2: with fractions

The LCM of $2$, $3$ and $6$ is $6$. I multiply every term by $6$, keeping the fractions in brackets:

I remove the brackets (the minus in front of the second term changes both signs):

I bring the $x$ terms to the left and the numbers to the right:

I divide by $-2$:

Check with $x=\frac{7}{2}$: the first term equals $\frac{9/2}{2}=\frac{9}{4}$, the second $\frac{6}{3}=2$, so on the left $\frac{9}{4}-2=\frac{1}{4}$. On the right $\frac{3/2}{6}=\frac{1}{4}$. It works.

A fractional solution is not an alarm bell: in linear equations it is perfectly normal.

Exercise 3: a problem solved with an equation

A book has a certain number of pages. On Monday I read $\frac{2}{5}$ of them, on Tuesday $\frac{1}{3}$ of the total, and I have $28$ left to read. How many pages does the book have?

I choose the unknown: $x$ is the total number of pages. On Monday I read $\frac{2}{5}x$, on Tuesday $\frac{1}{3}x$, and $28$ remain. The total is the sum of the three parts:

The LCM of $5$ and $3$ is $15$. I multiply every term by $15$:

I collect and bring the $x$ terms to the right:

Check against the text: on Monday $\frac{2}{5}\cdot 105=42$ pages, on Tuesday $\frac{1}{3}\cdot 105=35$ pages, $77$ in all; $105-77=28$. It matches the data of the problem.

Word problems: how to turn a text into an equation

The step that throws most students is not solving the equation, it is writing it. The method I use in lessons:

1. Read the text twice and underline what is asked.
2. Choose the unknown and write it out in full (”$x$ = number of pages”). It is usually the quantity requested, or the one from which the others can be obtained.
3. Express the other quantities in terms of $x$. If one side exceeds the other by $4$ cm, the two sides are $x$ and $x+4$.
4. Look for the relation that the text gives you: a total sum, a perimeter, two equal quantities.
5. Write the equation, solve it, check against the text, not just in the equation. A negative side or a non-whole number of people tells you the set-up is wrong.

A short example: a rectangle has a perimeter of $56$ cm and the base exceeds the height by $4$ cm. With $h$ the height and $h+4$ the base:

So the height is $12$ cm and the base $16$ cm: the perimeter is $2\cdot (12+16)=56$. If you need to factorise an expression in order to simplify before solving, the route is in factorising polynomials: common factor and trinomial.

The mistakes that cost marks

• Signs in front of brackets. The minus changes everything inside. It is mistake number one, I see it in tests of every level.
• Multiplying only some terms by the LCM. When you remove the denominators, every term, even the whole-number one, has to be multiplied.
• Forgetting the sign change when moving a term across the equals sign, or changing it twice.
• Dividing by the wrong coefficient: if $-2x=-7$ the solution is $\frac{7}{2}$ and not $-\frac{7}{2}$. The minus signs cancel.
• Writing "$x=0$" in an impossible or indeterminate equation, instead of saying what it is.
• Not checking. Substitution catches almost all sign mistakes in thirty seconds.
• Putting $=$ between expressions that are not equal in the intermediate steps: every line must be equivalent to the previous one. If you need to start again, start again, do not overwrite.
• In word problems, forgetting to write what the unknown represents and to give the answer with the unit of measurement.

How to prepare for the test

In four to seven weeks there is time to do it properly, without hurry. The plan I recommend:

1. This week: redo the three exercises by hand without looking at the solution, changing the numbers. If the result does not match the check, find the mistake before going on.
2. When fractions arrive in class: do five equations with different denominators, and for each one write the LCM as the first step.
3. Two weeks before the test: write three impossible and three indeterminate equations, inventing them yourself. Whoever can build them can recognise them.
4. In the last week: solve two problems a day with the five-step method. It is what weighs most in the test and what is trained least.

If you want to connect these equations to what comes next, in equations, inequalities and systems in Year 2 you will find where they end up (systems and inequalities) with a more hands-on slant. For the general overview of the topic there is the Maths page.

If the signs keep betraying you

Usually, when a Year 1 student gets equations wrong, the problem is not the algebra: it is a writing habit (skipped steps, muddled lines, calculations done in the head). In lessons I work on exactly that: a tidy sheet, one step per line, always check. If you like, write to me from the contact page and we will do a free introductory call: send me a test or two exercises that do not work out and I will tell you where the procedure jams.