Linear equations and systems

One rule for every linear equation

An equation stays true if you do the same thing to both sides. Solving a linear equation is a sequence of such moves, each chosen to bring $x$ closer to being alone:

1. Clear fractions by multiplying every term by the common denominator.
2. Expand brackets, taking care with a minus sign in front of a bracket.
3. Collect the $x$ terms on one side and the numbers on the other.
4. Divide by the coefficient of $x$.
5. Substitute the answer back into the original equation.

Take $\frac{x + 2}{3} - \frac{x - 1}{4} = 2$. Multiply every term by 12: $4(x + 2) - 3(x - 1) = 24$. Expand: $4x + 8 - 3x + 3 = 24$; note that $-3 \times (-1) = +3$. Collect: $x + 11 = 24$, so $x = 13$. Check: $\frac{15}{3} - \frac{12}{4} = 5 - 3 = 2$.

The substitution check in step 5 costs a few seconds and catches nearly every sign slip, which is the most common way to lose a mark on an equation you can actually solve.

From words to equations

Word problems are equations written in sentences. A reliable routine:

• Name the unknowns with letters that remind you what they are ($a$ for adult tickets, $c$ for child tickets), and write down what each letter stands for.
• Find as many independent facts as unknowns. Typical facts are a total count, a total value, a difference, or a ratio.
• Write each fact as an equation, keeping units consistent (all in euros, all in minutes).
• Answer the question asked. If you solved for $a$ but the question wants $c$, take the last step.

"Twice as many A as B" is $A = 2B$, not $2A = B$; test such a sentence with small numbers if unsure. "5 less than $x$" is $x - 5$, not $5 - x$.

Two equations, two unknowns

Substitution suits a system where one equation already gives a variable by itself, or nearly: from $a + c = 40$, write $c = 40 - a$ and substitute.

Elimination suits a system where the coefficients line up. Multiply one or both equations so a variable has opposite (or equal) coefficients, then add (or subtract) to remove it. For $3x + 2y = 17$ and $2x - y = 2$, double the second to get $4x - 2y = 4$ and add: $7x = 21$, so $x = 3$, and then $y = 2x - 2 = 4$. Check: $9 + 8 = 17$ and $6 - 4 = 2$.

Look at what is asked before solving. If a question wants $x + y$, adding or subtracting the equations may give it directly. From $3x + 2y = 17$ and $2x + 3y = 13$, adding gives $5x + 5y = 30$, so $x + y = 6$ without finding $x$ or $y$.

When a system has no single answer

Two linear equations are two straight lines. Usually they cross once and there is one solution. But:

Case	What you see	Solutions
lines cross	coefficients not proportional	exactly one
parallel lines	coefficients proportional, constants not	none
same line	whole equations proportional	infinitely many

For $2x + 4y = 10$ and $x + 2y = k$: halving the first gives $x + 2y = 5$. If $k = 5$, the two equations are the same line and every point on it works; for any other $k$, the lines are parallel and nothing works. When elimination leaves a statement like $0 = 0$, you have the same line; when it leaves $0 = 3$, there is no solution.

Keeping arithmetic light

Without a calculator, choose moves that keep numbers small. Divide an equation through by a common factor first ($6x + 9y = 42$ becomes $2x + 3y = 14$). Eliminate the variable whose coefficients are easier to match. And if the numbers become ugly, recheck the setup: multiple-choice problems solved by hand often come out in whole numbers or simple fractions.