Sec 1 Linear Equations: Worked Examples & Practice

Secondary 1 Mathematics · G3 focus · Study guide

Understand why equation steps work, handle brackets and fractions accurately, and turn word problems into equations you can check.

Concept → worked examples → practice9 min read · practice linked below

What does it mean to solve an equation?

A linear equation in one unknown asks for a value that makes two expressions equal. In $2x+6=18$ , the unknown is $x$ . Solving means finding its value and showing why it works. The equals sign means “has the same value as”. It does not mean “the answer comes next”.

Before you start

Be comfortable with negative numbers, equivalent fractions and collecting like terms. Remember: $3x+2x=5x$ , but $3x+2$ cannot be combined into one term. Multiplication applies to every term in brackets: $2(x+4)=2x+8$ .

An expression, such as $3x+2$ , has no equals sign. An equation, such as $3x+2=14$ , states an equality. A solution is a value that makes that statement true. In this chapter, “linear” means the variable has power one after simplification, with no squares or products of unknowns.

The balance principle: do the same to both sides

Imagine two equal quantities on a balance. Removing the same amount from both sides preserves equality. So does adding the same amount, or multiplying or dividing both sides by the same non-zero number. These steps produce equivalent equations: equations with the same solutions.

From

2x+6=18

, subtract 6 from both sides to get

2x=12

. Divide both sides by 2 to obtain

x=6

. Each block is labeled by its value, not drawn to scale.

The aim is to isolate the unknown: leave one $x$ alone on one side. Undo addition or subtraction first, then multiplication or division. Write one justified step per line so that a sign error is easy to spot.

Worked example 1: unknowns on both sides

Solve $5x+3=2x+15$ .

Subtract $2x$ from both sides to collect the unknown terms: $5x+3-2x=2x+15-2x$ so $3x+3=15.$
Subtract 3 from both sides: $3x=12.$
Divide both sides by 3: $\boxed{x=4}.$

Check in the original equation: the left side is $5(4)+3=23$ , and the right side is $2(4)+15=23$ . The two values agree. Choosing to subtract $2x$ kept the coefficient of $x$ positive, making the arithmetic easier.

Explore one equation step at a time

Follow the complete sequences below. At each step, identify the operation applied to both sides.

This balance represents equality of expressions. Negative values are allowed in an equation, even though physical weights cannot be negative.

Two operations undo two operations

$2x+6=18$
Start with two equal quantities. First undo the addition of 6.
$2x=12$
Subtract 6 from both sides. Two copies of the unknown now total 12.
$x=6$
Divide both sides by 2. Check: twice 6 plus 6 equals 18.

Collect the unknown terms first

$5x+3=2x+15$
Start by subtracting the smaller unknown term from both sides.
$3x+3=15$
Subtract $2x$ from both sides. Equality is preserved even when the quantity subtracted contains an unknown.
$3x=12$
Subtract 3 from both sides to leave only the unknown term.
$x=4$
Divide both sides by 3. Substitution gives 23 on each side of the original equation.

A negative coefficient is still a multiplier

$7-3x=19$
The coefficient of $x$ , meaning its numerical multiplier, is $-3$ . First remove the constant 7.
$-3x=12$
Subtract 7 from both sides. The negative sign belongs to the coefficient of $x$ .
$x=-4$
Divide both sides by $-3$ . Check: $7-3(-4)=7+12=19$ .

Brackets and negative signs

Expand brackets before collecting terms. A minus sign outside brackets changes the sign of every term inside. For example, $-(x-4)=-x+4$ . For $-3(2x-5)$ , multiply both terms by $-3$ , giving $-6x+15$ .

Worked example 2: a negative multiplier

Solve $8-3(2x-5)=x+2$ .

Multiply both terms inside the bracket by $-3$ . In particular, $(-3)(-5)=+15$ : $8-6x+15=x+2.$
Collect the numerical terms on the left: $23-6x=x+2.$
Add $6x$ to both sides: $23=7x+2.$
Subtract $2$ from both sides, then divide both sides by $7$ : $21=7x\quad\Rightarrow\quad\boxed{x=3}.$

Check: the original left side is $8-3(6-5)=5$ . The right side is $3+2=5$ . The crucial sign is $(-3)(-5)=+15$ .

Fractions: multiply every term by the LCM

For numerical denominators, multiply the whole equation by their lowest common multiple (LCM). This removes the fractions without changing the solution. A fraction bar groups the entire numerator: $\frac{x+1}{3}$ means $(x+1)\div3$ .

Worked example 3: two different denominators

Solve $\frac{x+1}{3}-\frac{x-2}{4}=2$ .

The LCM of 3 and 4 is 12. Multiply every term, including the right side, by 12:

12\left(\frac{x+1}{3}\right)-12\left(\frac{x-2}{4}\right)=12(2).

Use $12\div3=4$ and $12\div4=3$ , keeping each numerator grouped: $4(x+1)-3(x-2)=24.$
Expand both brackets. The minus sign belongs to the whole second term, so $-3(x-2)=-3x+6$ : $4x+4-3x+6=24.$
Collect like terms: $x+10=24.$
Subtract $10$ from both sides: $\boxed{x=14}.$

Check: $\frac{15}{3}-\frac{12}{4}=5-3=2$ . Two common mistakes are leaving the right side as 2, and expanding $-3(x-2)$ as $-3x-6$ .

This method is safer than trying to cross-multiply a sum of fractions. Cross-multiplication applies directly when each side is a single fraction. When the variable occurs in a denominator, excluded values must also be considered.

Turn a word problem into an equation

Define one unknown with its unit. Express the other quantities using it, then identify the equality in the question: total cost, equal lengths, or equal amounts. Solve the equation and translate the result back into a sentence.

Worked example 4: two related prices

A large notebook costs S$3 more than a small notebook. Two small notebooks and three large notebooks cost S$39 altogether. Find the price of each type.

Let $x$ dollars be the price of one small notebook. One large notebook costs $(x+3)$ dollars.

Each large notebook contributes a base price

x

and an extra S$3. The three extra amounts total S$9. The bar is schematic.

The two small notebooks cost $2x$ dollars. Each of the three large notebooks costs $x+3$ dollars, so their combined cost is $3(x+3)$ dollars. Add these costs to form the equation:

2x+3(x+3)=39.

Expand the bracket: $2x+3x+9=39.$
Collect the $x$ -terms and subtract $9$ from both sides: $5x+9=39\quad\Rightarrow\quad5x=30.$
Divide both sides by $5$ : $x=6.$
Interpret the unknown: $x$ is the small notebook's price, so the large notebook costs $x+3=9$ dollars.

Answer: a small notebook costs S$6 and a large notebook costs S$9. Check both conditions: $9-6=3$ , and $2(6)+3(9)=39$ . Writing $2x+3x+3=39$ would charge the extra S$3 only once.

Common errors and how to catch them

Use the reason to repair the line of working
Error	Correction and reason
Changing only one side	From $x+7=12$ , subtract 7 from both sides to obtain $x=5$ .
Losing a negative sign	From $-4x=20$ , divide by $-4$ , giving $x=-5$ .
Expanding only the first term	$-2(x-3)=-2x+6$ . Multiply both terms inside the brackets.
Checking a changed equation only	Substitute into the original equation. An earlier incorrect step may have changed the solution.

Practise with our topic questions

Apply the methods in this guide using our Sec 1 linear equations practice questions. The topic bank brings the questions and their existing worked solutions together, so you can attempt a question before checking the method.

Keep both sides balanced, apply a minus sign to the whole bracket, and substitute your answer into the original equation.

Frequently asked questions

Why does a term change sign when it moves across the equals sign?

“Moving” is shorthand for an operation on both sides. From $x+5=12$ , subtracting 5 gives $x+5-5=12-5$ . The result is $x=7$ . Think about the operation first, especially when brackets or negative terms are involved.

Can the solution be negative or a fraction?

Yes. For example, $2x=3$ gives $x=\frac32$ . The equation determines the value. A word problem may impose extra conditions: a number of notebooks must be a non-negative integer, while a temperature can be negative.

Must I always collect the unknown on the left?

No. $21=7x$ and $7x=21$ mean the same thing. Choose the side that makes the working simpler, often the side that leaves a positive coefficient. Both valid routes must give the same solution.

What if the unknown disappears?

If you obtain a false statement such as $0=5$ , there is no solution. If you obtain an always-true statement such as $0=0$ , every permitted value is a solution. For instance, $2(x+1)=2x+2$ holds for every real $x$ .

Revision checklist

Brackets present: expand every term, including its sign.
Numerical denominators present: multiply every term by their LCM.
Unknowns on both sides: collect them using the same operation on each side.
Word problem: define the unknown and unit before forming the equation.
Answer found: substitute into the original equation and interpret the result.