Solving Linear Equations

Understand what an equation is, use the balance method and inverse operations to solve one-step and two-step linear equations, and verify every solution by substitution.

Before you start

An algebraic expression is formed by combining letters and numbers with operations: for example, $3x + 5$ and $2y - 7$ . An expression has no equals sign.
A variable is a letter that stands for a number. In $4x$ , $4$ is the coefficient and $x$ is the variable. In an equation, the variable whose value we must find is called the unknown.
Inverse operations undo each other: addition undoes subtraction ( $+a$ and $-a$ ), and multiplication undoes division ( $\times a$ and $\div a$ ).
Working with negative numbers: $5 - 9 = -4$ , $-3 \times 4 = -12$ , and $\frac{-18}{-3} = 6$ .

What is a Linear Equation?

An equation is a mathematical statement that two expressions are equal. It always contains an equals sign ( $=$ ).

A linear equation in one variable is an equation where the variable has an index (power) of $1$ . It can be written in the form $ax + b = c$ , where $a$ , $b$ and $c$ are numbers and $a \neq 0$ .

The expression to the left of the equals sign is the left-hand side (LHS), and the expression to the right is the right-hand side (RHS).

Expression vs Equation

Algebraic Expression: $6 + x$ (a quantity that can take different values; no equals sign).

Linear Equation: $6 + x = 9$ (a statement that $6 + x$ equals $9$ ; has an equals sign).

To solve an equation means to find the numerical value of the variable that makes the equation true. That value is called the solution or root of the equation.

Think of an equation as a balanced scale. The equals sign means both pans are in perfect balance.

Whatever operation you perform on the left pan, you must perform the exact same operation on the right pan to keep the scale balanced.

The Balance Scale Model

The scale is balanced: $x + 5 = 12$ . Removing $5$ from both sides leaves $x$ balanced with $7$ , so $x = 7$ .

Example 1: Verifying Solutions by Substitution

Determine whether each given value is a solution to the equation $3x - 4 = 11$ :
(a) $x = 4$
(b) $x = 5$

Solution

(a) Substitute $x = 4$ into the left-hand side:

$\text{LHS} = 3(4) - 4 = 12 - 4 = 8$

$\text{RHS} = 11$

Since $\text{LHS} \neq \text{RHS}$ ( $8 \neq 11$ ), $x = 4$ is not a solution.

(b) Substitute $x = 5$ into the left-hand side:

$\text{LHS} = 3(5) - 4 = 15 - 4 = 11$

$\text{RHS} = 11$

Since $\text{LHS} = \text{RHS}$ ( $11 = 11$ ), $x = 5$ is the correct solution.

$x = 4$ is not a solution; $x = 5$ is the solution to $3x - 4 = 11$ .

Solving One-Step Equations

In a one-step equation, a single operation is applied to the variable. We isolate the variable by applying the inverse operation to both sides of the equation.

Equation Type	Operation on Variable	Inverse Operation to Apply
$x + a = b$	Adding $a$	Subtract $a$ from both sides: $x = b - a$
$x - a = b$	Subtracting $a$	Add $a$ to both sides: $x = b + a$
$ax = b$	Multiplying by $a$	Divide both sides by $a$ : $x = \frac{b}{a}$
$\frac{x}{a} = b$	Dividing by $a$	Multiply both sides by $a$ : $x = ab$

Example 2: One-Step Equations with Addition and Subtraction

Solve each equation and check the solution:
(a) $x + 19 = 12$
(b) $y - 8 = 4$

Solution

(a) Solve $x + 19 = 12$ :

Subtract $19$ from both sides to isolate $x$ :

$x + 19 - 19 = 12 - 19$

$x = -7$

Check: $\text{LHS} = -7 + 19 = 12 = \text{RHS} \quad \checkmark$

(b) Solve $y - 8 = 4$ :

Add $8$ to both sides to isolate $y$ :

$y - 8 + 8 = 4 + 8$

$y = 12$

Check: $\text{LHS} = 12 - 8 = 4 = \text{RHS} \quad \checkmark$

(a) $x = -7$ ; (b) $y = 12$ .

Example 3: One-Step Equations with Multiplication and Division

Solve each equation and check the solution:
(a) $4y = 20$
(b) $\frac{m}{4} = 3$
(c) $\frac{2}{5}x = 7$

Solution

(a) Solve $4y = 20$ :

Divide both sides by $4$ :

$\frac{4y}{4} = \frac{20}{4}$

$y = 5$

Check: $\text{LHS} = 4(5) = 20 = \text{RHS} \quad \checkmark$

(b) Solve $\frac{m}{4} = 3$ :

Multiply both sides by $4$ :

$\frac{m}{4} \times 4 = 3 \times 4$

$m = 12$

Check: $\text{LHS} = \frac{12}{4} = 3 = \text{RHS} \quad \checkmark$

Multiply both sides by $5$ :

$2x = 35$

Divide both sides by $2$ :

$x = \frac{35}{2} = 17\frac{1}{2}$

Check: $\text{LHS} = \frac{2}{5} \times \frac{35}{2} = \frac{70}{10} = 7 = \text{RHS} \quad \checkmark$

(a) $y = 5$ ; (b) $m = 12$ ; (c) $x = 17\frac{1}{2}$ (or $\frac{35}{2}$ ).

Solving Two-Step Equations

A two-step equation involves two operations, such as $3x + 7 = 19$ .

To isolate the variable, undo the operations in reverse order: undo the last operation done to the unknown first. In $3x + 7 = 19$ , $x$ was first multiplied by $3$ and then $7$ was added, so we undo the addition first and then the multiplication. For any equation of the form $ax + b = c$ , this means undoing the addition or subtraction first, then the multiplication or division.

Systematic Two-Step Method

Step 1: Remove the constant term by adding or subtracting the same number on both sides.

Step 2: Remove the coefficient of the variable by dividing or multiplying both sides.

Step 3: Check your answer by substituting back into the original equation.

Example 4: Standard Two-Step Equations

Solve each equation, showing all balance steps:
(a) $3x + 7 = 19$
(b) $5y - 3 = 17$

Solution

(a) Solve $3x + 7 = 19$ :

Step 1: Subtract $7$ from both sides:

$3x + 7 - 7 = 19 - 7$

$3x = 12$

Step 2: Divide both sides by $3$ :

$\frac{3x}{3} = \frac{12}{3}$

$x = 4$

Check: $\text{LHS} = 3(4) + 7 = 12 + 7 = 19 = \text{RHS} \quad \checkmark$

(b) Solve $5y - 3 = 17$ :

Step 1: Add $3$ to both sides:

$5y - 3 + 3 = 17 + 3$

$5y = 20$

Step 2: Divide both sides by $5$ :

$\frac{5y}{5} = \frac{20}{5}$

$y = 4$

Check: $\text{LHS} = 5(4) - 3 = 20 - 3 = 17 = \text{RHS} \quad \checkmark$

(a) $x = 4$ ; (b) $y = 4$ .

Example 5: Two-Step Equations with Negative Numbers

Solve each equation and check the solution:
(a) $5 - p = 3$
(b) $5 - 2m = 11$
(c) $2m + 9 = 5$

Solution

(a) Solve $5 - p = 3$ :

Subtract $5$ from both sides:

$5 - p - 5 = 3 - 5$

$-p = -2$

Divide both sides by $-1$ :

$p = 2$

Check: $\text{LHS} = 5 - 2 = 3 = \text{RHS} \quad \checkmark$

(b) Solve $5 - 2m = 11$ :

Subtract $5$ from both sides:

$5 - 2m - 5 = 11 - 5$

$-2m = 6$

Divide both sides by $-2$ (a positive divided by a negative gives a negative):

$\frac{-2m}{-2} = \frac{6}{-2}$

$m = -3$

Check: $\text{LHS} = 5 - 2(-3) = 5 + 6 = 11 = \text{RHS} \quad \checkmark$

Subtract $9$ from both sides:

$2m + 9 - 9 = 5 - 9$

$2m = -4$

Divide both sides by $2$ :

$m = -2$

Check: $\text{LHS} = 2(-2) + 9 = -4 + 9 = 5 = \text{RHS} \quad \checkmark$

(a) $p = 2$ ; (b) $m = -3$ ; (c) $m = -2$ .

Common Mistakes

Operating on only one side: Subtracting a number from the left side without subtracting it from the right side breaks the balance. Whatever is done to one side must be done to the other.
Sign errors when dividing by a negative number: In $-2m = 6$ , dividing both sides by $-2$ gives $m = -3$ . A common mistake is writing $m = 3$ . Remember that $\frac{\text{positive}}{\text{negative}} = \text{negative}$ .
Muddling expressions and equations: An expression such as $6 + x$ cannot be "solved" because there is no equals sign. Only equations such as $6 + x = 9$ have roots.
Forgetting the negative sign on a term: In $5 - p = 3$ , subtracting $5$ leaves $-p = -2$ , not $p = -2$ . The minus sign belongs to the term $p$ .

Practice Questions

Work each question on paper first using the balance method, then check your solution.

Question 1

Solve $x + 14 = 9$ .

ShowHide solution

Subtract $14$ from both sides:

$x + 14 - 14 = 9 - 14$

$x = -5$

Check: $\text{LHS} = -5 + 14 = 9 = \text{RHS} \quad \checkmark$

$x = -5$

Question 2

Solve $y - 15 = -6$ .

ShowHide solution

Add $15$ to both sides:

$y - 15 + 15 = -6 + 15$

$y = 9$

Check: $\text{LHS} = 9 - 15 = -6 = \text{RHS} \quad \checkmark$

$y = 9$

Question 3

Solve $6k = 42$ .

ShowHide solution

Divide both sides by $6$ :

$\frac{6k}{6} = \frac{42}{6}$

$k = 7$

Check: $\text{LHS} = 6(7) = 42 = \text{RHS} \quad \checkmark$

$k = 7$

Question 4

Solve $\frac{n}{5} = -4$ .

ShowHide solution

Multiply both sides by $5$ :

$\frac{n}{5} \times 5 = -4 \times 5$

$n = -20$

Check: $\text{LHS} = \frac{-20}{5} = -4 = \text{RHS} \quad \checkmark$

$n = -20$

Question 5

Solve $2x + 11 = 25$ .

ShowHide solution

Subtract $11$ from both sides:

$2x = 25 - 11 = 14$

Divide both sides by $2$ :

$x = \frac{14}{2} = 7$

Check: $\text{LHS} = 2(7) + 11 = 14 + 11 = 25 = \text{RHS} \quad \checkmark$

$x = 7$

Question 6

Solve $4y - 9 = 15$ .

ShowHide solution

Add $9$ to both sides:

$4y = 15 + 9 = 24$

Divide both sides by $4$ :

$y = \frac{24}{4} = 6$

Check: $\text{LHS} = 4(6) - 9 = 24 - 9 = 15 = \text{RHS} \quad \checkmark$

$y = 6$

Question 7

Solve $7 - 3a = 28$ .

ShowHide solution

Subtract $7$ from both sides:

$-3a = 28 - 7 = 21$

Divide both sides by $-3$ :

$a = \frac{21}{-3} = -7$

Check: $\text{LHS} = 7 - 3(-7) = 7 + 21 = 28 = \text{RHS} \quad \checkmark$

$a = -7$

Question 8

Solve $\frac{3}{4}x + 5 = 17$ .

ShowHide solution

Subtract $5$ from both sides:

$\frac{3}{4}x = 17 - 5 = 12$

Multiply both sides by $4$ :

$3x = 12 \times 4 = 48$

Divide both sides by $3$ :

$x = \frac{48}{3} = 16$

Check: $\text{LHS} = \frac{3}{4}(16) + 5 = 12 + 5 = 17 = \text{RHS} \quad \checkmark$

$x = 16$

Question 9

Solve $\frac{2x - 5}{3} = 7$ .

ShowHide solution

Here $x$ was multiplied by $2$ , then $5$ was subtracted, and the whole of $2x - 5$ was divided by $3$ last. Undo the last operation first: multiply both sides by $3$ :

$\frac{2x - 5}{3} \times 3 = 7 \times 3$

$2x - 5 = 21$

Add $5$ to both sides:

$2x = 21 + 5 = 26$

Divide both sides by $2$ :

$x = \frac{26}{2} = 13$

Check: $\text{LHS} = \frac{2(13) - 5}{3} = \frac{26 - 5}{3} = \frac{21}{3} = 7 = \text{RHS} \quad \checkmark$

$x = 13$

Question 10 (Setting: Public Transport)

Wei Ming's stored-value card has a balance of $x$ dollars. After he adds $20 at an MRT station ticket machine, his card balance is $34. Form an equation in $x$ and find his original card balance.

ShowHide solution

Original balance: $x$ dollars.

After topping up $20, balance is $(x + 20)$ dollars.

Form the equation:

$x + 20 = 34$

Subtract $20$ from both sides:

$x + 20 - 20 = 34 - 20$

$x = 14$

Check: $14 + 20 = 34$ dollars $\quad \checkmark$

The equation is $x + 20 = 34$ ; Wei Ming's original balance was $14.

Question 11 (Setting: Hawker Centre)

Siti bought 3 bowls of laksa at a hawker stall for $k$ dollars each, and 1 lime juice for $2. Her total bill came to $17. Form a linear equation in $k$ and find the cost of one bowl of laksa.

ShowHide solution

Cost of 3 bowls of laksa at $k$ dollars each: $3k$ dollars.

Total cost including the $2 drink: $(3k + 2)$ dollars.

Form the equation:

$3k + 2 = 17$

Subtract $2$ from both sides:

$3k = 17 - 2 = 15$

Divide both sides by $3$ :

$k = \frac{15}{3} = 5$

Check: $3(5) + 2 = 15 + 2 = 17$ dollars $\quad \checkmark$

The equation is $3k + 2 = 17$ ; each bowl of laksa cost $5.

Question 12 (Setting: School Sports Day)

At a secondary school sports day, the Red House scored $p$ points. The Blue House scored 15 points fewer than the Red House. If the Blue House scored 63 points, form an equation in $p$ and calculate the points scored by the Red House.

ShowHide solution

Let Red House points be $p$ .

Blue House scored 15 points fewer: $(p - 15)$ points.

Form the equation:

$p - 15 = 63$

Add $15$ to both sides:

$p - 15 + 15 = 63 + 15$

$p = 78$

Check: $78 - 15 = 63$ points $\quad \checkmark$

The equation is $p - 15 = 63$ ; the Red House scored 78 points.

Key Points

A linear equation states that two expressions are equal; it has an equals sign ( $=$ ) and the variable has power $1$ .
The solution (or root) is the number that makes LHS equal RHS when substituted into the equation.
An equation stays balanced only if the exact same operation is applied to both sides.
Undo the last operation done to the unknown first.
Use inverse operations: addition undoes subtraction, multiplication undoes division.
For two-step equations $ax + b = c$ , undo addition and subtraction first, then multiplication and division.
Always verify your answer by substituting it back into the original equation.