Adding and Subtracting Linear Expressions

Learn how to identify like terms, keep track of signs, and combine linear terms with integer and fractional coefficients to simplify algebraic expressions.

Before you start

A term is a number, a letter, or numbers and letters multiplied together. In $3x - 4y + 7$ , the terms are $3x$ , $-4y$ and $7$ .
Every term takes the sign ( $+$ or $-$ ) directly in front of it. In $5a - 8b$ , the second term is $-8b$ , with coefficient $-8$ .
Adding and subtracting directed numbers: $3 - 5 = -2$ , and $-3 + 6 - 2 = 1$ .
Adding fractions: to add $\frac{2}{3}$ and $\frac{1}{5}$ , find the common denominator $15$ : $\frac{10}{15} + \frac{3}{15} = \frac{13}{15}$ .

1. What Is a Linear Expression?

In algebra, an expression is called linear if each of its terms contains at most one variable, and that variable is raised only to the power of $1$ . Constant terms (plain numbers with no letters) are also allowed.

Here are examples of linear expressions:

- $2x - 5$ (one variable, $x$ , raised to the power of $1$ )

- $m + 4n - 3$ (two variables, $m$ and $n$ , each raised to the power of $1$ )

- $\frac{a}{3} - \frac{2b}{5} + 8$ (variables with fractional coefficients)

If a term has a variable raised to a power other than $1$ (such as $x^2$ or $w^3$ ), the square root of a variable (such as $\sqrt{x}$ ), or two or more variables multiplied together (such as $2xy$ or $ab$ ), the expression is non-linear.

Expression	Linear or Non-linear?	Reason
$4x + 7$	Linear	The variable $x$ has power $1$ .
$w^3 + 3w - 2$	Non-linear	The term $w^3$ has variable $w$ raised to power $3$ .
$2a - 3b + c$	Linear	All variables ( $a, b, c$ ) are raised to power $1$ .
$5xy - 2$	Non-linear	The term $5xy$ is a product of two variables ( $x \times y$ ).
$\frac{x}{3} + 4$	Linear	Can be written as $\frac{1}{3}x + 4$ ; $x$ is to the power $1$ .
$a + b - 2ab$	Non-linear	The term $-2ab$ contains two variables multiplied together.
$2x - 2y + z - 9$	Linear	Four terms, each variable has power $1$ .

2. Like and Unlike Terms

The Definition of Like Terms

Like terms are algebraic terms that have the exact same variables raised to the exact same powers.

Their numerical coefficients can be different.

All constant terms (plain numbers such as $4, -9, 0.5$ ) are like terms with each other.

Unlike terms have different variables, or the same variable raised to different powers.

The Order of Factors Does Not Matter

Because multiplication is commutative ( $x \times y = y \times x$ ), the order in which variables are written does not change the term.

For example, $3xy$ and $-5yx$ are like terms because $xy = yx$ .

Similarly, $4mn$ and $-nm$ are like terms because $nm = mn$ .

Terms such as $xy$ and $mn$ are not linear, but the same rule decides whether they are like terms, so you will see a few of them in this lesson.

Worked Example 1: Grouping Like Terms

From the following list of algebraic terms, group all the like terms together and list any terms that have no like partners:
$5m, \; -3, \; 8mn, \; 2n, \; 2xy, \; -5x, \; 3nm, \; 17y, \; -9m, \; -xy, \; 7, \; -2yx$

Solution:

Identify terms sharing the exact same variable parts:

- Terms in $m$ : $5m$ and $-9m$ .

- Terms in $mn$ : $8mn$ and $3nm$ (since $nm = mn$ ).

- Terms in $xy$ : $2xy$ , $-xy$ and $-2yx$ (since $yx = xy$ ).

- Constant terms: $-3$ and $7$ .

- Terms with no like partners: $2n$ (no other term in $n$ alone), $-5x$ (no other term in $x$ alone), and $17y$ (no other term in $y$ alone).

Like groups: ( $5m, -9m$ ), ( $8mn, 3nm$ ), ( $2xy, -xy, -2yx$ ) and ( $-3, 7$ ). Terms with no like partners: $2n, -5x, 17y$ .

3. Adding and Subtracting Like Terms

In arithmetic, $3 \text{ apples} + 2 \text{ apples} = 5 \text{ apples}$ . In algebra, $3a + 2a = 5a$ .

When adding or subtracting like terms, we add or subtract their numerical coefficients and keep the variable part unchanged.

This works because $3a + 2a$ means 3 lots of $a$ and 2 more lots of $a$ , which is 5 lots of $a$ .

Only like terms can be combined. Unlike terms, such as $3x + 2y$ , cannot be simplified and must be left as they are.

Combining Coefficients

To combine like terms:

1. Identify the like terms.

2. Add or subtract their numerical coefficients according to the rules of directed numbers.

3. Attach the shared variable part to the resulting coefficient.

If the resulting coefficient is $1$ , we write $x$ , not $1x$ . If it is $-1$ , we write $-x$ , not $-1x$ . If it is $0$ , the term equals $0$ and drops out completely ( $0x = 0$ ).

Worked Example 2: Combining Like Terms with One Variable

Simplify each of the following expressions completely:
(a) $x + 2x - 5x$
(b) $-3y + 6y - 2y$
(c) $3a - 2a - a$
(d) $-4p - 7p + 15p$

Solution:

(a) The coefficients are $1, 2,$ and $-5$ :

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

(b) The coefficients are $-3, 6,$ and $-2$ :

$-3y + 6y - 2y = (-3 + 6 - 2)y = (3 - 2)y = 1y = y$

(c) The coefficients are $3, -2,$ and $-1$ :

$3a - 2a - a = (3 - 2 - 1)a = 0a = 0$

When the coefficient is zero, the entire term becomes zero: do not write $0a$ .

(d) The coefficients are $-4, -7,$ and $15$ :

$-4p - 7p + 15p = (-4 - 7 + 15)p = (-11 + 15)p = 4p$

(a) $-2x$ (b) $y$ (c) $0$ (d) $4p$

4. Expressions with Several Variables and Constants

When an algebraic expression contains multiple variables and constant numbers, we simplify it by grouping like terms together before combining their coefficients.

The 3-Step Strategy for Simplifying Linear Expressions

Step 1: Group like terms together. Move like terms next to each other. Crucial rule: Every term must carry the sign ( $+$ or $-$ ) that appears immediately in front of it.

Step 2: Order terms cleanly. By standard convention, write variable terms in alphabetical order ( $a, b, c, \dots, x, y, z$ ), and place constant numbers last.

Step 3: Combine each group. Add or subtract the coefficients of each like group.

Adding or Subtracting an Expression in Brackets

To add an expression in brackets, remove the brackets; no sign changes: $+(3x - 4) = 3x - 4$ .

To subtract an expression in brackets, subtract every term in it, so each sign inside the brackets changes: $-(3x - 4) = -3x + 4$ .

Taking away $(3x - 4)$ takes away $3x$ but gives back the $4$ , because $3x - 4$ is $4$ less than $3x$ .

Worked Example 3: Simplifying Expressions with Multiple Variables

Simplify each expression completely:
(a) $2m + 3n - 4m + 2$
(b) $5x - 3y + 7 - 2x + 8y - 12$
(c) $-3c + 4d + 2c - 3d - 5$
(d) $2a + 2b - 4a - (-4b)$
(e) $(5x + 2) - (3x - 4)$

Solution:

(a) Group $m$ -terms together, followed by $n$ -terms and constants:

$\begin{aligned} 2m + 3n - 4m + 2 &= (2m - 4m) + 3n + 2 \\ &= -2m + 3n + 2 \end{aligned}$

(b) Group $x$ -terms, $y$ -terms, and constant numbers:

$\begin{aligned} 5x - 3y + 7 - 2x + 8y - 12 &= (5x - 2x) + (-3y + 8y) + (7 - 12) \\ &= 3x + 5y - 5 \end{aligned}$

(c) Group $c$ -terms, $d$ -terms, and constants:

$\begin{aligned} -3c + 4d + 2c - 3d - 5 &= (-3c + 2c) + (4d - 3d) - 5 \\ &= -c + d - 5 \end{aligned}$

(d) Subtracting a negative term is equivalent to adding: $-(-4b) = +4b$ :

$\begin{aligned} 2a + 2b - 4a - (-4b) &= 2a + 2b - 4a + 4b \\ &= (2a - 4a) + (2b + 4b) \\ &= -2a + 6b \end{aligned}$

(e) Remove the first brackets as they are. Subtract every term of the second brackets: $-(3x) = -3x$ and $-(-4) = +4$ :

$\begin{aligned} (5x + 2) - (3x - 4) &= 5x + 2 - 3x + 4 \\ &= (5x - 3x) + (2 + 4) \\ &= 2x + 6 \end{aligned}$

(a) $-2m + 3n + 2$ (b) $3x + 5y - 5$ (c) $-c + d - 5$ (d) $-2a + 6b$ (e) $2x + 6$

5. Linear Expressions with Fractional Coefficients

Linear terms can have fractions as coefficients, such as $\frac{2a}{3}$ (which means $\frac{2}{3}a$ ) or $\frac{b}{5}$ (which means $\frac{1}{5}b$ ).

To combine like terms with fractional coefficients:

1. Group like terms together.

2. For each variable group, find the lowest common multiple (LCM) of the denominators to express the coefficients with a common denominator.

3. Add or subtract the numerators and simplify the fraction.

4. Leave a fractional coefficient as a proper or improper fraction in simplest form (such as $\frac{7a}{15}$ or $\frac{49t}{8}$ ), never as a mixed number. $\frac{7}{15}a$ and $\frac{7a}{15}$ mean the same.

Worked Example 4: Expressions with Fractional Coefficients

Simplify each expression completely:
(a) $\frac{2a}{3} + \frac{2b}{5} - \frac{a}{5} - \frac{3b}{2}$
(b) $(\frac{p}{2} + \frac{q}{3}) + (\frac{2p}{5} - \frac{3q}{6})$
(c) $(-\frac{7t}{8} - 2s + \frac{1}{3}) - (-7t + 8s - \frac{4}{5})$

Solution:

(a) Group $a$ -terms and $b$ -terms:

$(\frac{2a}{3} - \frac{a}{5}) + (\frac{2b}{5} - \frac{3b}{2})$

For the $a$ -terms, the common denominator of $3$ and $5$ is $15$ :

$\frac{2a}{3} - \frac{a}{5} = \frac{10a}{15} - \frac{3a}{15} = \frac{7a}{15}$

For the $b$ -terms, the common denominator of $5$ and $2$ is $10$ :

$\frac{2b}{5} - \frac{3b}{2} = \frac{4b}{10} - \frac{15b}{10} = -\frac{11b}{10}$

Combine the two results:

$= \frac{7a}{15} - \frac{11b}{10}$

(b) Remove brackets and group $p$ -terms and $q$ -terms. Note that $\frac{3q}{6} = \frac{q}{2}$ :

$(\frac{p}{2} + \frac{2p}{5}) + (\frac{q}{3} - \frac{q}{2})$

LCM for $p$ -denominators ( $2$ and $5$ ) is $10$ : $\frac{5p}{10} + \frac{4p}{10} = \frac{9p}{10}$ .

LCM for $q$ -denominators ( $3$ and $2$ ) is $6$ : $\frac{2q}{6} - \frac{3q}{6} = -\frac{q}{6}$ .

$= \frac{9p}{10} - \frac{q}{6}$

(c) Subtract every term of the second brackets: $-(-7t) = +7t$ , $-(+8s) = -8s$ , and $-(-\frac{4}{5}) = +\frac{4}{5}$ :

$-\frac{7t}{8} - 2s + \frac{1}{3} + 7t - 8s + \frac{4}{5}$

Group $s$ -terms, $t$ -terms, and numbers (alphabetical order puts $s$ before $t$ ):

$(-2s - 8s) + (-\frac{7t}{8} + 7t) + (\frac{1}{3} + \frac{4}{5})$

Calculate each part:

$-2s - 8s = -10s$

$-\frac{7t}{8} + \frac{56t}{8} = \frac{49t}{8}$

$\frac{1}{3} + \frac{4}{5} = \frac{5}{15} + \frac{12}{15} = \frac{17}{15}$

Combining all parts in alphabetical order:

$= -10s + \frac{49t}{8} + \frac{17}{15}$

(a) $\frac{7a}{15} - \frac{11b}{10}$ (b) $\frac{9p}{10} - \frac{q}{6}$ (c) $-10s + \frac{49t}{8} + \frac{17}{15}$

6. Real-World Applications and the Unit Rule

Algebraic expressions often represent real-world physical quantities such as lengths, perimeters, masses, and costs.

When writing an algebraic answer that has a unit of measurement:

The Unit Rule for Algebraic Expressions

Whenever an algebraic expression has more than one term and is followed by a unit, the entire expression must be enclosed in brackets.

Examples:

- Write $(3x + 5)\text{ cm}$ , not $3x + 5\text{ cm}$ (which looks like $3x$ plus $5\text{ cm}$ ).

- Write $(8p + 4q - 5)\text{ m}$ , not $8p + 4q - 5\text{ m}$ .

- Write $(4a + 7b)$ dollars, not $4a + 7b$ dollars. The $ sign is kept for a fixed amount, such as $84.

If the expression has only one term, brackets are not required: $10w\text{ seconds}$ or $5x\text{ kg}$ .

Worked Example 5: Grain Sacks and Total Cost

A farmer packs three sacks of wheat. Their masses are $(2u + 3v - 2)\text{ kg}$ , $(5u + 4v + 3)\text{ kg}$ , and $(u + 9v + 1)\text{ kg}$ .
(a) Find an expression in terms of $u$ and $v$ for the total mass of the three sacks.
(b) If $u = 2$ and $v = 4$ , find the mass of each sack and the total mass in kilograms.
(c) Wheat is sold at $1.25 per kilogram. Calculate the total selling price of the three sacks.

Solution:

(a) Add the masses of the three sacks:

$\text{Total mass} = (2u + 3v - 2) + (5u + 4v + 3) + (u + 9v + 1)$

Group like terms:

$= (2u + 5u + u) + (3v + 4v + 9v) + (-2 + 3 + 1)$

$= (8u + 16v + 2)\text{ kg}$

(b) Substitute $u = 2$ and $v = 4$ into each mass:

$\text{First sack} = 2(2) + 3(4) - 2 = 4 + 12 - 2 = 14\text{ kg}$

$\text{Second sack} = 5(2) + 4(4) + 3 = 10 + 16 + 3 = 29\text{ kg}$

$\text{Third sack} = 2 + 9(4) + 1 = 2 + 36 + 1 = 39\text{ kg}$

Every mass is positive, as a mass must be. Now substitute into the simplified expression for the total:

$\begin{aligned} \text{Total mass} &= 8(2) + 16(4) + 2 \\ &= 16 + 64 + 2 \\ &= 82\text{ kg} \end{aligned}$

Check: $14 + 29 + 39 = 82$ .

(c) Multiply the mass by the unit price of $1.25 per kg:

$\text{Total price} = 82 \times 1.25 = 102.50$

The total selling price is $102.50.

(a) $(8u + 16v + 2)\text{ kg}$ (b) $14\text{ kg}$ , $29\text{ kg}$ , $39\text{ kg}$ ; total $82\text{ kg}$ (c) $102.50

Common Mistakes to Avoid

Adding unlike terms: Writing $3x + 2y = 5xy$ . This is completely wrong; unlike terms cannot be combined. $3x + 2y$ is already in simplest form.
Dropping the negative sign when rearranging: Rearranging $7 - 4x$ to $4x - 7$ . The minus sign belongs to $4x$ , so the rearranged expression is $-4x + 7$ .
Changing only the first sign after a minus: Writing $(5x + 2) - (3x - 4) = 5x + 2 - 3x - 4$ . Every term of the subtracted brackets changes sign: $5x + 2 - 3x + 4 = 2x + 6$ .
Forgetting that $x$ means $1x$ : Writing $5x - x = 5$ . A single variable $x$ has an invisible coefficient of $1$ , so $5x - x = 5x - 1x = 4x$ .
Omitting brackets before units: Writing $8p + 4q - 5\text{ m}$ instead of $(8p + 4q - 5)\text{ m}$ . Without brackets, the unit applies only to the last term.

Practice Questions

Work out each problem on paper before clicking to reveal the solution. Questions 9 to 12 apply your algebra skills to real-world settings.

Question 1: Linear vs Non-linear Expressions

State whether each expression is linear or non-linear, giving a brief reason:
(a) $7x - 4$
(b) $3a^2 + 5a - 1$
(c) $2p - 5q + 3r$
(d) $4xy + 1$

ShowHide solution

(a) Linear. The only variable is $x$ , and its power is $1$ .

(b) Non-linear. The variable $a$ is raised to the power of $2$ in the term $3a^2$ .

(c) Linear. Each of the three variables ( $p, q, r$ ) is raised to the power of $1$ , and no two variables are multiplied together.

(d) Non-linear. The term $4xy$ contains two variables multiplied together.

(a) Linear (b) Non-linear (c) Linear (d) Non-linear

Question 2: Identifying Like Terms

From the list $4a, \; -7b, \; 3ab, \; -a, \; 2bc, \; 9b, \; 5ba$ , group all sets of like terms together.

ShowHide solution

- Terms in $a$ : $4a$ and $-a$ .

- Terms in $b$ : $-7b$ and $9b$ .

- Terms in $ab$ : $3ab$ and $5ba$ (since $ba = ab$ ).

- $2bc$ contains the letter $c$ as well, and no other term has exactly $bc$ , so it has no like partner.

Like groups: ( $4a, -a$ ), ( $-7b, 9b$ ), and ( $3ab, 5ba$ ). Term with no partner: $2bc$ .

Question 3: Basic Combining of Like Terms

Simplify each expression completely:
(a) $7m + 4m - 2m$
(b) $-5y - 8y + 3y$
(c) $9k - k - 8k$
(d) $-12w + 5w + 7w$

ShowHide solution

(a) $(7 + 4 - 2)m = 9m$ .

(b) $(-5 - 8 + 3)y = (-13 + 3)y = -10y$ .

(c) $(9 - 1 - 8)k = 0k = 0$ .

(d) $(-12 + 5 + 7)w = (-12 + 12)w = 0w = 0$ .

(a) $9m$ (b) $-10y$ (c) $0$ (d) $0$

Question 4: Combining Coefficients with Directed Numbers

Simplify each of the following:
(a) $4x - 9x + 2x$
(b) $-6p - (-2p) + 8p$
(c) $15t - 22t - 3t$
(d) $-a - a - a$

ShowHide solution

(a) $(4 - 9 + 2)x = -3x$ .

(b) $-(-2p) = +2p$ , so $(-6 + 2 + 8)p = 4p$ .

(c) $(15 - 22 - 3)t = -10t$ .

(d) $(-1 - 1 - 1)a = -3a$ .

(a) $-3x$ (b) $4p$ (c) $-10t$ (d) $-3a$

Question 5: Two Variables and Constant Terms

Simplify each expression:
(a) $3x + 5y - 7x + 2y$
(b) $8a - 4 - 3a + 9$
(c) $-5m + 2n - 3m - 7n + 10$

ShowHide solution

(a) Group like terms: $(3x - 7x) + (5y + 2y) = -4x + 7y$ .

(b) Group like terms: $(8a - 3a) + (-4 + 9) = 5a + 5$ .

(c) Group like terms: $(-5m - 3m) + (2n - 7n) + 10 = -8m - 5n + 10$ .

(a) $-4x + 7y$ (b) $5a + 5$ (c) $-8m - 5n + 10$

Question 6: Three Variables and Sign Reversals

Simplify each expression completely:
(a) $(3u - 4v + 3) - (6v + 3u + 6)$
(b) $(-8 + 2a) - (-3a + 7d)$
(c) $4p - 3q + 2r - (2p - 5q - 4r)$

ShowHide solution

(a) Subtract every term of the second brackets:
$3u - 4v + 3 - 6v - 3u - 6$
Group like terms:
$(3u - 3u) + (-4v - 6v) + (3 - 6) = 0 - 10v - 3 = -10v - 3$

(b) Subtract every term of the second brackets: $-8 + 2a + 3a - 7d$ .
Alphabetical order puts $a$ before $d$ , constants last:
$(2a + 3a) - 7d - 8 = 5a - 7d - 8$

(c) Subtract every term in the brackets: $4p - 3q + 2r - 2p + 5q + 4r$ .
Group like terms:
$(4p - 2p) + (-3q + 5q) + (2r + 4r) = 2p + 2q + 6r$

(a) $-10v - 3$ (b) $5a - 7d - 8$ (c) $2p + 2q + 6r$

Question 7: Single Variable with Fractional Coefficients

Simplify each of the following:
(a) $\frac{x}{2} + \frac{x}{3}$
(b) $\frac{3y}{4} - \frac{2y}{5}$
(c) $\frac{5m}{6} - m + \frac{m}{3}$

ShowHide solution

(a) Common denominator of $2$ and $3$ is $6$ :
$\frac{3x}{6} + \frac{2x}{6} = \frac{5x}{6}$

(b) Common denominator of $4$ and $5$ is $20$ :
$\frac{15y}{20} - \frac{8y}{20} = \frac{7y}{20}$

(c) Write $m$ as $\frac{6m}{6}$ . Common denominator is $6$ :
$\frac{5m}{6} - \frac{6m}{6} + \frac{2m}{6} = \frac{5 - 6 + 2}{6}m = \frac{1}{6}m = \frac{m}{6}$

(a) $\frac{5x}{6}$ (b) $\frac{7y}{20}$ (c) $\frac{m}{6}$

Question 8: Multiple Variables with Fractional Coefficients

Simplify completely:
(a) $\frac{3a}{4} + \frac{b}{2} - \frac{a}{3} + \frac{2b}{5}$
(b) $(\frac{2x}{3} - \frac{y}{4}) - (\frac{x}{6} - \frac{3y}{8})$

ShowHide solution

(a) Group $a$ -terms and $b$ -terms:
$(\frac{3a}{4} - \frac{a}{3}) + (\frac{b}{2} + \frac{2b}{5})$
For $a$ : $\frac{9a}{12} - \frac{4a}{12} = \frac{5a}{12}$ .
For $b$ : $\frac{5b}{10} + \frac{4b}{10} = \frac{9b}{10}$ .
Result: $\frac{5a}{12} + \frac{9b}{10}$ .

(b) Subtract every term of the second brackets:
$\frac{2x}{3} - \frac{y}{4} - \frac{x}{6} + \frac{3y}{8}$
For $x$ : $\frac{4x}{6} - \frac{x}{6} = \frac{3x}{6} = \frac{x}{2}$ .
For $y$ : $-\frac{2y}{8} + \frac{3y}{8} = \frac{y}{8}$ .
Result: $\frac{x}{2} + \frac{y}{8}$ .

(a) $\frac{5a}{12} + \frac{9b}{10}$ (b) $\frac{x}{2} + \frac{y}{8}$

Question 9: Triangle Perimeter in a Setting

The three sides of a triangle measure $(2x + 3)\text{ cm}$ , $(3x - 1)\text{ cm}$ , and $(x + 4)\text{ cm}$ .
(a) Write down an expression in terms of $x$ for the perimeter of the triangle in its simplest form.
(b) If $x = 5$ , find the length of each side and the total perimeter of the triangle in centimetres.

ShowHide solution

(a) The perimeter is the sum of the three side lengths:
$\text{Perimeter} = (2x + 3) + (3x - 1) + (x + 4)$
Group like terms:
$= (2x + 3x + x) + (3 - 1 + 4) = (6x + 6)\text{ cm}$
Remember to enclose the multi-term expression in brackets before the unit.

(b) Substitute $x = 5$ into each side length:
$\text{Side 1} = 2(5) + 3 = 10 + 3 = 13\text{ cm}$
$\text{Side 2} = 3(5) - 1 = 15 - 1 = 14\text{ cm}$
$\text{Side 3} = 5 + 4 = 9\text{ cm}$
Total perimeter $= 6(5) + 6 = 30 + 6 = 36\text{ cm}$ .
Check: $13 + 14 + 9 = 36\text{ cm}$ .

(a) $(6x + 6)\text{ cm}$ (b) Sides: $13\text{ cm}, 14\text{ cm}, 9\text{ cm}$ ; Perimeter $= 36\text{ cm}$

Question 10: School Bookshop Purchases in a Setting

At a school bookshop, each notebook costs $3 and each pen costs $2.
Mei Ling buys $m$ notebooks and $n$ pens. Junjie buys $(2m - 1)$ notebooks and $(n + 4)$ pens.
(a) Write an expression in terms of $m$ for the total number of notebooks bought by the two students.
(b) Write an expression in terms of $m$ and $n$ for the total number of items (notebooks and pens) bought by the two students.
(c) If $m = 4$ and $n = 6$ , find the total number of items bought and the total amount the two students paid.

ShowHide solution

(a) Add the notebooks:
$\text{Notebooks} = m + (2m - 1) = m + 2m - 1 = (3m - 1)\text{ notebooks}$

(b) First add the pens:
$\text{Pens} = n + (n + 4) = n + n + 4 = (2n + 4)\text{ pens}$
Then add the notebooks and the pens:
$\text{Items} = (3m - 1) + (2n + 4) = 3m + 2n - 1 + 4 = (3m + 2n + 3)\text{ items}$
The answer has more than one term and a unit, so it is written in brackets.

(c) Substitute $m = 4$ and $n = 6$ :
$\text{Items} = 3(4) + 2(6) + 3 = 12 + 12 + 3 = 27\text{ items}$
For the cost, find the number of each kind first:
$\text{Notebooks} = 3(4) - 1 = 11, \quad \text{Pens} = 2(6) + 4 = 16$
Check: $11 + 16 = 27$ items.
Notebooks: $11 \times 3 = 33$ , so $33. Pens: $16 \times 2 = 32$ , so $32.
Total paid: $33 + 32 = 65$ , so the two students paid $65.

(a) $(3m - 1)$ notebooks (b) $(3m + 2n + 3)$ items (c) $27$ items; $65

Question 11: The Fourth Side of a Garden Bed

A garden bed is in the shape of a quadrilateral. Its total perimeter is $(9p + 7q + 5)\text{ m}$ .
Three of its boundary sides measure $(2p + 3q + 1)\text{ m}$ , $(3p - q + 2)\text{ m}$ , and $(p + 2q + 1)\text{ m}$ .
(a) Find an expression in terms of $p$ and $q$ for the length of the fourth side in its simplest form.
(b) If $p = 4$ and $q = 2$ , calculate the length of each of the four sides and verify that every side has a positive length.

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(a) First, sum the three known side lengths:
$\begin{aligned} \text{Sum of 3 sides} &= (2p + 3q + 1) + (3p - q + 2) + (p + 2q + 1) \\ &= (2p + 3p + p) + (3q - q + 2q) + (1 + 2 + 1) \\ &= (6p + 4q + 4)\text{ m} \end{aligned}$
Subtract this sum from the total perimeter to find the fourth side:
$\begin{aligned} \text{Fourth side} &= (9p + 7q + 5) - (6p + 4q + 4) \\ &= 9p + 7q + 5 - 6p - 4q - 4 \\ &= (3p + 3q + 1)\text{ m} \end{aligned}$

(b) Substitute $p = 4$ and $q = 2$ into all four sides:
$\text{Side 1} = 2(4) + 3(2) + 1 = 8 + 6 + 1 = 15\text{ m}$
$\text{Side 2} = 3(4) - (2) + 2 = 12 - 2 + 2 = 12\text{ m}$
$\text{Side 3} = 4 + 2(2) + 1 = 4 + 4 + 1 = 9\text{ m}$
$\text{Fourth side} = 3(4) + 3(2) + 1 = 12 + 6 + 1 = 19\text{ m}$
All four lengths are greater than $0$ , as lengths must be.
Check total perimeter: $15 + 12 + 9 + 19 = 55\text{ m}$ .
Using the formula: $9(4) + 7(2) + 5 = 36 + 14 + 5 = 55\text{ m}$ .

(a) $(3p + 3q + 1)\text{ m}$ (b) Sides: $15\text{ m}, 12\text{ m}, 9\text{ m}, 19\text{ m}$ ; all are positive and total $55\text{ m}$ .

Question 12: Sports Tournament Scores in a Setting

In a school sports tournament, three teams scored points based on game results:

Team Alpha scored $(4x + 3y - 5)$ points.
Team Beta scored $(2x - y + 8)$ points.
Team Gamma scored $(3x + 2y - 1)$ points.
(a) Write an expression in terms of $x$ and $y$ for the total points scored by all three teams.
(b) If $x = 10$ and $y = 6$ , find the score of each team and identify the winning team.

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(a) Add the points of all three teams:
$\text{Total} = (4x + 3y - 5) + (2x - y + 8) + (3x + 2y - 1)$
Group like terms:
$= (4x + 2x + 3x) + (3y - y + 2y) + (-5 + 8 - 1)$
$= (9x + 4y + 2)\text{ points}$

(b) Substitute $x = 10$ and $y = 6$ into each team’s score expression:
$\text{Team Alpha} = 4(10) + 3(6) - 5 = 40 + 18 - 5 = 53\text{ points}$
$\text{Team Beta} = 2(10) - (6) + 8 = 20 - 6 + 8 = 22\text{ points}$
$\text{Team Gamma} = 3(10) + 2(6) - 1 = 30 + 12 - 1 = 41\text{ points}$
Total points $= 53 + 22 + 41 = 116\text{ points}$ .
Check with total expression: $9(10) + 4(6) + 2 = 90 + 24 + 2 = 116\text{ points}$ .
Team Alpha won with $53$ points.

(a) $(9x + 4y + 2)\text{ points}$ (b) Alpha: $53$ , Beta: $22$ , Gamma: $41$ . Team Alpha won.

Key Points to Remember

Linear Expressions: Contain variables raised only to the power of $1$ , with at most one variable per term. Plain constant numbers are permitted.
Like Terms: Have the exact same variables raised to the exact same powers (e.g. $4a$ and $-7a$ , or $4ab$ and $-7ba$ ). Only like terms can be added or subtracted.
Every Term Keeps Its Sign: When rearranging terms, always carry the $+$ or $-$ sign that sits immediately in front of each term.
Alphabetical Ordering: Write variable terms in alphabetical order ( $a, b, c, \dots, x, y, z$ ) with constant numbers placed last.
Zero Coefficients: When coefficients add to zero, the term vanishes ( $3a - 3a = 0$ ). Do not write $0a$ .
Subtracting Brackets: To subtract an expression in brackets, subtract every term in it: $-(3x - 4) = -3x + 4$ .
The Unit Rule: When an algebraic expression has more than one term and carries a unit, always enclose the entire expression in brackets, such as $(6x + 6)\text{ cm}$ or $(3m + 2n + 3)$ items.