🏠 首页 CHAPTER 4 · BASIC ALGEBRA AND ALGEBRAIC MANIPULATIONS 4.2

4.2

Simplification of Linear Expressions

💡 4.1 学了怎么写和求值代数表达式；4.2 学怎么把表达式 化简（simplify）——把同类项合在一起，让表达式变短。算盘技巧：代数 disc 配 zero pair。

AIdentifying Like Terms and Unlike Terms

Consider the algebraic expression $4xy - x + 6x - 2y + 7$. The various parts, $4xy, -x, 6x, -2y$ and $7$, are called terms of the expression. In general, a term can be a constant, a variable or a product of a constant and one or more variables. A term that has no variable such as the term $7$ is called a constant term.

The term $6x$ denotes the product of 6 and $x$. The number, including the sign, of a term is called the coefficient of the term. The coefficient of the term $6x$ is 6 and the coefficient of the constant term 7 is 7.

The terms, $-x$ and $6x$, have an identical variable component, $x$. We call these two terms like terms. All constant terms are like terms. On the other hand, the terms $6x$ and $-2y$ are called unlike terms as the variable components, $x$ and $y$, are not the same. Similarly, the terms $6x$ and $4xy$ are also unlike terms as the variable components, $x$ and $xy$, are different.

📖 Worked Example 8

Given the expression $4p + 3q - pq - 7$, state

  1. (a) the number of terms in the expression,
  2. (b) the constant term,
  3. (c) the coefficient of each term.

SOLUTION

(a) The terms are $4p$, $3q$, $-pq$ and $-7$. ∴ there are 4 terms in the expression.

(b) The constant term is −7.

(c) The coefficient of $4p$ is 4, the coefficient of $3q$ is 3, the coefficient of $-pq$ is −1 and the coefficient of $-7$ is −7.

✍️ Try It Yourself 8

Given the expression $-9mn + n - 6$, state

  1. (a) the number of terms in the expression. → 检查
  2. (b) the constant term. → 检查
  3. (c) the coefficient of $-9mn$ → 检查; the coefficient of $n$ → 检查; the coefficient of $-6$ → 检查.
📖 Worked Example 9

Identify the like terms in each of the following expressions.

  1. (a) $7x + 4a - \tfrac{1}{2}x$
  2. (b) $ab - 0.3ab - 7a + 3ab$
  3. (c) $\tfrac{1}{3} + y + y^2 - 2$

SOLUTION

(a) The like terms are $7x$ and $-\tfrac{1}{2}x$ as they have an identical variable component, $x$.

(b) The like terms are $ab$, $-0.3ab$ and $3ab$ as they have an identical variable component, $ab$.

(c) The like terms are $\tfrac{1}{3}$ and $-2$ as they are both constants.

✍️ Try It Yourself 9

Identify the like terms in each of the following expressions. (Just list them separated by commas.)

  1. (a) $a^2 + 2a - \tfrac{1}{2}a^2$. → 检查
  2. (b) $mn - \tfrac{1}{6}nm + 3m^2 - 2n$. → 检查
    💡 注意

    $mn$ 和 $nm$ 完全一样（乘法可交换）。

  3. (c) $7 + yx + xy^2 - 2y + 9$. → 检查
    💡 提示

    $yx$ 和 $xy^2$ 不是同类项（$x \neq xy$ 不同变量组合）。两个常数 $7, 9$ 才是同类项。

BSimplifying Linear Expressions

In Secondary One, we will focus mainly on linear expressions, where the power (or index) of each term containing a variable is 1. For example, $2x + y - 5$ is a linear expression, where the terms containing a variable, such as $2x$, have a power of 1.

In contrast, the expression $2x^3 - x + 5y - 7xy$ is not a linear expression. The term $2x^3$ has a power of 3. The term $-7xy$ has a power of 2, (the sum of the power of $x$ and the power of $y$ is 2). Can you write down two examples of linear expressions and two examples of non-linear expressions?

In Chapter 2, we used algebra discs to help us make sense of the operations involving negative integers. Similarly, we can use algebra discs to help us make sense of and interpret linear expressions with coefficients that are integers. Each disc has two sides as shown below.

1 −1 front back x −x front back y −y front back
positive (front) negative (back, "flipped")

Each disc has two sides: a positive face (blue) and a negative face (orange). Flipping a disc changes its sign.

Recall that 1 and −1 form a zero pair as their sum = 0. Similarly, $x$ and $-x$ is a zero pair, and $y$ and $-y$ is another zero pair. We write $x + (-x) = 0$ and $y + (-y) = 0$. Here are other examples.

x x −x −x 2 zero pairs $2x + (-2x) = 2x - 2x$ = 0 −y −y −y y y y 3 zero pairs $(-3y) + 3y = -3y + 3y$ = 0

We can use algebra discs to represent and simplify linear expressions with coefficients that are integers as illustrated in the following activity.

🧪 Activity 2 · Collecting like terms with discs

Objective: To simplify algebraic expressions by collecting like terms.

Representing algebraic expressions — examples:

  • (i) $2x + 3$ or $3 + 2x$ → 2 blue $x$ discs + 3 blue 1 discs
  • (ii) $y - 4$ or $-4 + y$ → 1 blue $y$ disc + 4 orange −1 discs
  • (iii) $-x + 3y - 2$, $-2 + 3y - x$, $-2 - x + 3y$ or $3y - x - 2$ → 1 orange −$x$ disc + 3 blue $y$ discs + 2 orange −1 discs

1. Represent the following expressions using algebra discs (describe what discs you'd lay down):

  1. (a) $3x + 2$ → 检查
  2. (b) $x + 4y$ → 检查
  3. (c) $-2x - 3y + 4$ → 检查

Simplifying algebraic expressions — example: Simplify $2x + 3y - 5x + y$.

Start: 2x + 3y − 5x + y x x y y y −x −x −x −x −x y → Find 2 zero pairs (x, −x) x x y y y −x −x −x −x −x y → After cancel: −3x + 4y −x −x −x y y y y

Step: Find 2 zero pairs ($x$ + $(-x)$ = 0). Remove them. Result: $-3x + 4y$.

Algebraically:

  • $2x + 3y - 5x + y$
  • $= 2x \underline{-5x} + 3y + y \qquad$ (Collect like terms)
  • $= (2-5)x + (3+1)y \qquad$ (Combine coefficients)
  • $= -3x + 4y$

2. Simplify the following expressions.

  1. (a) $x + 5x = $ 检查
  2. (b) $-2x + 5x = $ 检查
  3. (c) $-2x - 5x = $ 检查
  4. (d) $3y + 1 - 4y - 3 = $ 检查
  5. (e) $-3y - 1 + 4y + 3 = $ 检查
  6. (f) $-3x + 2y - 4 - x + y = $ 检查

From Activity 2, we see that we can simplify linear expressions by collecting like terms.

📖 Worked Example 10

Simplify the following expressions.

  1. (a) $7x - 3x + 2x$
  2. (b) $\tfrac{1}{2}a + \tfrac{2}{3}a$
  3. (c) $-6c + c - \tfrac{5}{4}c$

SOLUTION

(a) $7x - 3x + 2x = (7 - 3 + 2)x = \mathbf{6x}$   (add/subtract like terms in same way as real numbers)

(b) $\tfrac{1}{2}a + \tfrac{2}{3}a = \left(\tfrac{1}{2} + \tfrac{2}{3}\right)a = \left(\tfrac{3}{6} + \tfrac{4}{6}\right)a = \mathbf{\tfrac{7}{6}a}$   (LCM of 2 and 3 is 6)

(c) $-6c + c - \tfrac{5}{4}c = \left(-6 + 1 - \tfrac{5}{4}\right)c = \left(-\tfrac{24}{4} + \tfrac{4}{4} - \tfrac{5}{4}\right)c = \mathbf{-\tfrac{25}{4}c}$

📌 SPOTLIGHT: If the coefficient of a term is an improper fraction, we do not express it as a mixed number. Hence in Worked Example 10(b) and (c), we give the answer as $\tfrac{7}{6}a$ and $-\tfrac{25}{4}c$.

✍️ Try It Yourself 10
  1. (a) $5t + 6t - 7t = $ 检查
  2. (b) $3m - \tfrac{4}{5}m - 10m = $ 检查
    💡 思路

    $3 - \tfrac{4}{5} - 10 = \tfrac{15-4-50}{5} = -\tfrac{39}{5}$. So answer = $-\tfrac{39}{5}m$.

  3. (c) $-4z - \tfrac{1}{2}z + \tfrac{5}{3}z = $ 检查
    💡 思路

    LCM of 1, 2, 3 is 6. $-4 - \tfrac{1}{2} + \tfrac{5}{3} = -\tfrac{24}{6} - \tfrac{3}{6} + \tfrac{10}{6} = -\tfrac{17}{6}$.

📖 Worked Example 11

Simplify the following expressions.

  1. (a) $3a + 4b - 2a + 5b$
  2. (b) $-3x - 8y + \tfrac{3}{7} + x + 5y - \tfrac{1}{3}$

SOLUTION

(a) $3a + 4b - 2a + 5b = 3a - 2a + 4b + 5b$  (Collect like terms.)
$= (3-2)a + (4+5)b$  (Simplify like terms.)
$= \mathbf{a + 9b}$  (Cannot simplify further — unlike terms.)

(b) $-3x - 8y + \tfrac{3}{7} + x + 5y - \tfrac{1}{3}$
$= -3x + x + (-8y + 5y) + \tfrac{3}{7} - \tfrac{1}{3}$
$= (-3 + 1)x + (-8 + 5)y + \tfrac{9}{21} - \tfrac{7}{21}$  (LCM of 3 and 7 is 21)
$= \mathbf{-2x - 3y + \tfrac{2}{21}}$

📌 SPOTLIGHT: An expression is the sum of all the terms. When collecting like terms, the signs in front of the terms should be kept. For example, the sum of $-8y$ and $5y$ is $(-8y + 5y) = -3y$. Its sum is neither $-(8y + 5y)$ nor $(8y + 5y)$.

✍️ Try It Yourself 11
  1. (a) $5c + 4d - 3c - d = $ 检查
  2. (b) $5x + 9y + (-11x) - (-4y) = $ 检查
    💡 提示

    $+(-11x) = -11x$；$-(-4y) = +4y$。所以 = $5x + 9y - 11x + 4y = -6x + 13y$.

  3. (c) $2t - 7x + \tfrac{2}{3} - 8t - 2x + \tfrac{1}{12} = $ 检查
    💡 思路

    $t$ 项：$2 - 8 = -6$，得 $-6t$。$x$ 项：$-7 - 2 = -9$，得 $-9x$。常数：$\tfrac{2}{3} + \tfrac{1}{12} = \tfrac{8}{12} + \tfrac{1}{12} = \tfrac{9}{12} = \tfrac{3}{4}$。合起来 $-6t - 9x + \tfrac{3}{4}$。

📖 Worked Example 12 · Word Problem

Amy spent $x. Jason spent $23 less than twice the amount Amy spent. Louis spent $53 more than half of what Amy spent.

  1. (a) Express the total amount Amy, Jason and Louis spent in terms of $x$.
  2. (b) If Amy spent $200, how much did they spend in total?

SOLUTION

(a) Amy spent $x. Jason spent $$(2x - 23)$. Louis spent $$\left(\tfrac{1}{2}x + 53\right)$.

Total: $x + 2x - 23 + \tfrac{1}{2}x + 53 = x + 2x + \tfrac{1}{2}x + (-23 + 53) = \left(1 + 2 + \tfrac{1}{2}\right)x + 30 = \tfrac{7}{2}x + 30$.

The total amount they spent was $$\left(\tfrac{7}{2}x + 30\right)$.

(b) If $x = 200$, then $\tfrac{7}{2}(200) + 30 = 700 + 30 = 730$. They spent $730 in total.

📌 EQUIVALENCE: During each step of the simplification, the terms in the expression may have changed. However, writing it differently does not change the value of the expression at each step when the same value for $x$ is substituted into the expressions.

✍️ Try It Yourself 12

Bag $A$ has $m$ coins. Bag $B$ has twice as many coins as bag $A$. Bag $C$ has 42 fewer coins than bag $B$.

  1. (a) Find the total number of coins in the 3 bags in terms of $m$. → 检查
    💡 思路

    A=$m$; B=$2m$; C=$2m - 42$. 合计 = $m + 2m + (2m-42) = 5m - 42$.

  2. (b) If there are 123 coins in bag $A$, how many coins are there in total? → 检查
    💡 计算

    $5(123) - 42 = 615 - 42 = 573$.

PPractice Exercise 4.2

BASIC MASTERY
  1. Q1. State the number of terms and the constant term in each of the following expressions.
    (a) $2a - 3b - 1$ → number of terms: ; constant: 检查
    (b) $7x + 6 - 4y + 5z$ → number of terms: ; constant: 检查
  2. Q2. Write down the coefficients of $x$ and $y$ in each of the following expressions.
    (a) $3x - 4y + 6xy$ → coeff of $x$: ; coeff of $y$: 检查
    (b) $x^2 - x + y + 8$ → coeff of $x$: ; coeff of $y$: 检查
  3. Q3. In each of the following pairs, identify the pairs which are like terms (Yes/No).
    (a) $5x^2y$ and $3x^2y$ → 检查
    (b) $0.4ab^3$ and $-9b^3 a$ → 检查
    (c) $8ab$ and $8abc$ → 检查
    (d) $mn$ and $-nm$ → 检查
    (e) $3(5^2 a)$ and $5(3^2 a)$ → 检查
    (f) $π$ and $2.5^2$ → 检查
    💡 思路

    (b) $ab^3 = b^3 a$（乘法可交换）→ same variable component → 同类项。
    (c) $ab \neq abc$（多了一个 $c$）→ unlike.
    (d) $mn = nm$ → same variable component → 同类项。
    (e) $5^2 a = 25a$, $3^2 a = 9a$ — 都是 $a$ 的常数倍，所以同类项（系数不同但变量部分相同）。
    (f) $π$ 和 $2.5^2 = 6.25$ 都是常数，所以同类项。

  4. Q4. Simplify the following.
    (a) $7a + 2a = $ 检查
    (b) $5b - 8b = $ 检查
    (c) $-4x + 6x = $ 检查
    (d) $-2y - 3y = $ 检查
    (e) $c - 8c + 2c = $ 检查
    (f) $d + 2d - 9d = $ 检查
    (g) $-3p + p + 4p = $ 检查
    (h) $-2q - 5q - q = $ 检查
    (i) $4y - 9y + 5y = $ 检查
    (j) $-4m - 2m + 5m - m = $ 检查
  5. Q5. Simplify the following.
    (a) $3n + 10 - 4n - 11 = $ 检查
    (b) $-6 + 3k - 4k + 7 = $ 检查
    (c) $3m - (-4n) - 2n + 5m = $ 检查
    (d) $-7x - 3y + (-2x) - (-3y) = $ 检查
    (e) $\tfrac{2}{3}p - \tfrac{1}{4}q + \tfrac{1}{6}p - \tfrac{1}{2}q = $ 检查
    (f) $7t + (-4av) - \tfrac{5}{3}t - \left(-\tfrac{1}{2}av\right) = $ 检查
    (g) $\tfrac{10}{3}rs - rs - 6sv - \tfrac{3}{2}sr = $ 检查
    (h) $-\tfrac{11}{4}uw + \tfrac{2}{7}w + 2uw - \tfrac{9}{7}w = $ 检查
    💡 思路 (g)

    $rs$ 项 ($rs = sr$)：$\tfrac{10}{3} - 1 - \tfrac{3}{2} = \tfrac{20}{6} - \tfrac{6}{6} - \tfrac{9}{6} = \tfrac{5}{6}$，得 $\tfrac{5}{6}rs$。$sv$ 项：$-6sv$。合 $= \tfrac{5}{6}rs - 6sv$。

    💡 思路 (h)

    $uw$ 项：$-\tfrac{11}{4} + 2 = -\tfrac{11}{4} + \tfrac{8}{4} = -\tfrac{3}{4}$，得 $-\tfrac{3}{4}uw$。$w$ 项：$\tfrac{2}{7} - \tfrac{9}{7} = -1$，得 $-w$。合 $= -\tfrac{3}{4}uw - w$。

INTERMEDIATE
  1. Q6. (a) Simplify $-4 - 2x + 5 + x$. → 检查
    (b) Find the value when $x = -3$. → 检查
  2. Q7. (a) Simplify $2a - (-5b) - ab + (-7a) - 10b$. → 检查
    (b) Find the value when $a = -1$ and $b = 5$. → 检查
    💡 验证 (b)

    $-5(-1) - 5(5) - (-1)(5) = 5 - 25 + 5 = -15$.

  3. Q8. (a) Simplify $\tfrac{3}{4}x - \tfrac{2}{5}y + \tfrac{1}{3} - \tfrac{1}{8}x$. → 检查
    (b) Find the value when $x = -2$ and $y = -6$. → 检查
  4. Q9. The total number of atoms in the compound CnH2n+2 is $n + 2n + 2$.
    (a) Simplify $n + 2n + 2$. → 检查
    (b) Find the total number of atoms if $n = 10$. → 检查
  5. Q10. A concert hall has 3 sections, A, B and C. Each ticket for section A costs $x. Each ticket for section B costs $50 more than a ticket for section A. Each ticket for section C costs $30 less than twice the cost of a ticket for section A.
    (a) Find, in terms of $x$, the total cost of three tickets, one from each section. → $ 检查
    (b) Find the total cost if each ticket for section A costs $90. → $ 检查
    💡 思路

    A=$x$; B=$x+50$; C=$2x-30$. 合 $= x + (x+50) + (2x-30) = 4x+20$. $x=90$: $4(90)+20 = 380$.

ADVANCED
  1. Q11. The sides of a triangle are $2x$ cm, $3x$ cm and $4y$ cm.
    (a) Express the perimeter of the triangle in terms of $x$ and $y$. → cm 检查
    (b) Find its perimeter when $x = 12$ and $y = 15$. → cm 检查
  2. Q12. Alice works $2t$ hours each day from Monday to Friday and $(2t - y)$ hours on Saturdays. She does not work on Sundays.
    (a) Express the number of hours she works in a week in terms of $t$ and $y$. → hours 检查
    (b) How many hours does she work each week when $t = 4\tfrac{1}{2}$ and $y = 3$? → hours 检查
    💡 思路

    (a) 5 × (2t) + (2t−y) = 10t + 2t − y = 12t − y. (b) 12(4.5) − 3 = 54 − 3 = 51.

  3. Q13. (OPEN) Write a linear expression that has 3 terms involving the variables $p$ and $q$.
    💡 提示

    开放题。需要确保3 项，且各项变量幂为 1 (linear)。
    示例：$2p + 3q - 5$、$\tfrac{1}{2}p - q + 8$、$-p + 4q + \tfrac{1}{3}$ 等。
    注意：$7p + q - 4q$ 应当简化成 $7p - 3q$，只剩 2 项，不算 3 项。

  4. Q14. (OPEN) When the algebraic expression $-4x + 7x + mx$ is simplified, the coefficient of $x$ in the result is negative. Write down two possible values of $m$ given that $m$ is a constant.
    $m_1 = $ (任填一个), $m_2 = $ 检查
    💡 思路

    简化后系数 $= -4 + 7 + m = 3 + m$。要 $3 + m < 0$，所以 $m < -3$。
    任意取 $m = -4, -5, -10, -100$ 等都可以。

  5. Q15. Ryan and Danny worked on the following question:
    Simplify the given expression and find its value when $x = 12$ and $y = 16$.
    $6x - \tfrac{1}{2}y - 19x - \tfrac{1}{2}y + 66 + 13x + y$

    Ryan substituted the incorrect values into the expression but his answer was the same as Danny's. Both boys were puzzled. Explain why their answers were the same.

    💡 关键发现

    化简：
    $x$ 项：$6 - 19 + 13 = 0$ → 消失了！
    $y$ 项：$-\tfrac{1}{2} - \tfrac{1}{2} + 1 = 0$ → 也消失了！
    常数：$66$。
    所以原式 $= 66$，与 $x$ 和 $y$ 的值无关。Ryan 代错值结果也是 66，Danny 代对的值结果也是 66——所以一样。