🏠 首页 CHAPTER 5 · SIMPLE EQUATIONS IN ONE VARIABLE 5.4

5.4

Formulate Linear Equations to Solve Word Problems

💡 5.1–5.3 学的是怎么解方程（已经给好的方程）。5.4 是真正的实战——把文字描述翻译成方程。
关键：选一个未知量设成 $x$，把其他量用 $x$ 表示，然后根据题目里的"等量关系"列方程。

We can use linear equations to represent some real-life situations. By solving these equations, we can provide solutions to these real-life problems.

In the following activity, we shall learn how to formulate linear equations to solve word problems.

🧪 Activity 2 · Formulate linear equations

Objective: To formulate linear equations to solve word problems.

1. There are 54 students in a choir. If there are 6 more girls than boys, how many girls are there?

(a) Let $x$ be the number of girls in the choir. As there are 6 more girls than boys, express the number of boys in terms of $x$.
Boys = 检查

Girl
$x$
⎫
Boy
$x - 6$
⎬ 54
⎭

(b) As the total number of girls and boys is 54, formulate an equation in $x$.
Equation: 检查

(c) Solve the equation and hence write down the number of girls.
Number of girls = 检查

💡 思路

$x + (x - 6) = 54$；$2x - 6 = 54$；$x = 30$. Girls = 30, boys = 24.

2. A bakery made three times as many buns as cakes. After selling 200 buns and 40 cakes, it had twice as many buns as cakes left. Find the number of cakes made.

(a) Let $x$ be the number of cakes made. Express the following quantities in terms of $x$.

Before selling Buns
$3x$
Before selling Cakes
$x$
After selling Buns
$3x - 200$
After selling Cakes
$x-40$
  1. (i) Number of buns made: 检查
  2. (ii) Number of buns left after selling: 检查
  3. (iii) Number of cakes left after selling: 检查

(b) Using the condition that the bakery had twice as many buns as cakes left, formulate an equation in $x$.
Equation: 检查

(c) Solve the equation and hence write down the number of cakes made.
Number of cakes = 检查

💡 思路

$3x - 200 = 2(x - 40)$；$3x - 200 = 2x - 80$；$x = 120$. 蛋糕 120 个，包子 360 个。

From Activity 2, we can summarise the steps involved in problem solving with linear equations as follows:

📋 Six-Step Strategy for Word Problems

  1. Read the question carefully and identify the unknown quantity.
  2. Use an appropriate notation (like $x$ or $y$) to represent one unknown quantity.
  3. Express other quantities in terms of the unknown quantity.
  4. Form an equation based on the given information.
  5. Solve the equation.
  6. Write down the answer statement.

It is a good practice to check whether the solution satisfies the conditions in the original problem. For example, if we are asked to find the number of students in the question, then the solution should be a positive integer. In this case, a negative or fractional value of $x$, if obtained, should be rejected.

📖 Worked Example 13 · Consecutive integers

The sum of three consecutive integers is 111. Form an equation and solve it to find the integers.

SOLUTION

STEP 1  The three consecutive integers are unknown. When we set one of the integers to be $x$, the other two integers can be expressed in terms of $x$.

STEP 2  Let the smallest integer be $x$.

STEP 3  Then the middle integer is $x + 1$ and the largest integer is $x + 2$.

Smallest
$x$
Middle
$x + 1$
Largest
$x + 2$

STEP 4  Sum of the three integers is 111. Hence $x + (x + 1) + (x + 2) = 111$.

STEP 5  Solve:

  • $3x + 3 = 111$
  • $3x = 111 - 3 = 108$
  • $x = \dfrac{108}{3} = 36$
  • ∴ $x + 1 = 37$ and $x + 2 = 38$.

STEP 6  The three integers are 36, 37 and 38.

CHECK: 36, 37 and 38 are consecutive integers and $36 + 37 + 38 = 111$ ✓. Hence the solution $x = 36$ is correct.

📌 SPOTLIGHT: It is important to set just one unknown $x$ and express the other unknowns in terms of $x$ based on the given information. Alternatively, we can let the largest integer be $y$. Hence $y + (y-1) + (y-2) = 111 \Rightarrow 3y - 3 = 111 \Rightarrow y = 38$. Same answer, different unknown.

✍️ Try It Yourself 13

Form an equation for each of the following and solve it to find the integers.

  1. (a) The sum of three consecutive integers is 144. → integers are 检查
    💡 思路

    $x + (x+1) + (x+2) = 144 \Rightarrow 3x + 3 = 144 \Rightarrow x = 47$. So 47, 48, 49.

  2. (b) The sum of three consecutive even integers is 300. → integers are 检查
    💡 思路

    $x + (x+2) + (x+4) = 300 \Rightarrow 3x + 6 = 300 \Rightarrow x = 98$. So 98, 100, 102.

📖 Worked Example 14 · Age problem

Mrs Li is 3 times as old as her daughter now. In 5 years' time, the sum of their ages will be 62 years. How old is her daughter now?

SOLUTION

Let her daughter's present age be $x$ years. Mrs Li's present age is $3x$ years.

Present age:

Daughter
$x$
Mrs Li
$x \cdot 3 = 3x$

Age in 5 years' time:

Daughter
$x$
+5
Mrs Li
$3x$
+5
⎬ 62

In 5 years' time, her daughter's age = $(x + 5)$ years; Mrs Li's age = $(3x + 5)$ years.
∴ the equation is $(x + 5) + (3x + 5) = 62$, i.e., $4x + 10 = 62$.

  • $4x = 52$
  • ∴ $x = 13$

Her daughter is 13 years old now.

📌 SPOTLIGHT: The unknown $x$ has no unit. Hence we need to state the unit of the unknown quantity. In Worked Example 14, we let the unknown age of her daughter be $x$ years.

✍️ Try It Yourself 14
  1. (a) Mr Rashid is 4 times as old as his son. Four years ago, the sum of their ages was 37 years. How old is his son now? → years 检查
    💡 思路

    Let son now = $x$; Rashid = $4x$. 4 years ago: son = $x-4$, Rashid = $4x-4$. Sum = $5x - 8 = 37 \Rightarrow x = 9$.

  2. (b) Ten years ago, Bryan was twice as old as his brother, Ken. Now, the sum of their ages is 44 years. Find the present age of Ken. → years 检查
    💡 思路

    Let Ken now = $x$. 10 yrs ago: Ken = $x - 10$, Bryan = $2(x-10)$. So Bryan now = $2(x-10) + 10 = 2x - 10$. Sum now: $x + (2x-10) = 44 \Rightarrow 3x = 54 \Rightarrow x = 18$.

Which unknown variable in Try It Yourself 14(b) would you set to $x$ years so that it is easier to formulate the equation?
💡 思考

设 Ken 现在为 $x$ 更直接（最少的变换）。如果设 Bryan 现在为 $y$，需要从 $y$ 推 Ken 当年再推 Ken 现在，步骤更多。
一般规律：设"被比较的基础量"为 $x$，因为其他量是它的倍数 / 偏移，表达更简洁。

📖 Worked Example 15 · Pricing problem

The price of a chair is $50 more than $\dfrac{1}{8}$ of the price of a table. A set of one table and 4 chairs is priced at $1400. Find the price of a table and the price of a chair.

SOLUTION

Let the price of a table be $x$.
Price of a chair = $\left(50 + \dfrac{1}{8}x\right)$.
Total price of a table and 4 chairs = $1400.

  • $x + 4\left(50 + \dfrac{1}{8}x\right) = 1400$
  • $x + 200 + \dfrac{1}{2}x = 1400$
  • $\dfrac{3}{2}x + 200 = 1400$
  • $\dfrac{3}{2}x = 1200$
  • $x = 1200 \times \dfrac{2}{3}$
  • ∴ $x = 800$

∴ price of a chair = $50 + \dfrac{1}{8}x = 50 + \dfrac{1}{8} \times 800 = 50 + 100 = 150$.

The price of a table is $800 and the price of a chair is $150.

✍️ Try It Yourself 15

The price of a book is $11 less than half of the price of a calculator. The total price of 3 books and 2 calculators is $229.50. Find the total price of 4 books and 3 calculators.

Price of 1 calculator = $  Price of 1 book = $ 检查

Total for 4 books + 3 calculators = $ 检查

💡 思路

Let calculator price = $x$; book price = $\dfrac{x}{2} - 11$.
Total: $3\left(\dfrac{x}{2} - 11\right) + 2x = 229.5$
$\dfrac{3x}{2} - 33 + 2x = 229.5 \Rightarrow \dfrac{7x}{2} = 262.5 \Rightarrow x = 75$.
Book = $\dfrac{75}{2} - 11 = 26.5$. Check: $3 \cdot 26.5 + 2 \cdot 75 = 79.5 + 150 = 229.5$ ✓.
Total for 4 books + 3 calculators = $4 \cdot 26.5 + 3 \cdot 75 = 106 + 225 = \mathbf{331}$.

PPractice Exercise 5.4

BASIC MASTERY
  1. Q1. There are 7 more Mathematics books than Science books on a shelf. Let $x$ be the number of Science books on the shelf.
    (a) Express the number of Mathematics books in terms of $x$. → 检查
    (b) If there is a total of 39 Mathematics and Science books on the shelf, find the number of Science books. → 检查
    💡 思路

    $x + (x+7) = 39 \Rightarrow 2x = 32 \Rightarrow x = 16$. 数学书 23 本，科学书 16 本。

  2. Q2. Every week Rahim works 3 times as many hours as Jinyan. Let the number of hours Jinyan works in a week be $t$ hours.
    (a) Express the number of hours Rahim works in a week in terms of $t$. → hours 检查
    (b) If the total number of hours they work in a week is 56 hours, find the number of hours Jinyan works in a week. → hours 检查
  3. Q3. The sum of three consecutive odd integers is 141. Form an equation and solve it to find the integers.
    Integers: 检查
    💡 思路

    Let smallest = $x$ (odd). Then $x + (x+2) + (x+4) = 141 \Rightarrow 3x + 6 = 141 \Rightarrow x = 45$. So 45, 47, 49.

INTERMEDIATE
  1. Q4. A shop displays a total of 50 type A and type B mice. The cost of a type A mouse is $6 and that of a type B mouse is $13. If the total cost of all 50 mice is $433, find the number of type A mice displayed.
    Number of type A mice = 检查
    💡 思路

    Let A = $x$; B = $50-x$. $6x + 13(50-x) = 433 \Rightarrow -7x + 650 = 433 \Rightarrow -7x = -217 \Rightarrow x = 31$.

  2. Q5. Tom is twice as old as Arul. In 4 years' time, the sum of their ages will be 32 years. Find Arul's age now.
    Arul's age = years 检查
    💡 思路

    Let Arul = $x$, Tom = $2x$. In 4 yrs: Arul = $x+4$, Tom = $2x+4$. Sum: $3x + 8 = 32 \Rightarrow x = 8$.

  3. Q6. Halim has 4 more $50 notes than $10 notes in his wallet. If the total value of the notes is $380, how many $10 notes does Halim have?
    Number of $10 notes = 检查
    💡 思路

    Let # of $10 notes = $x$; # of $50 notes = $x + 4$. Total: $10x + 50(x+4) = 380 \Rightarrow 60x + 200 = 380 \Rightarrow x = 3$.

  4. Q7. In a Mathematics competition, the winner won $100 more than twice the amount won by the runner-up. How much money did the runner-up win if both of them won $2350 altogether?
    Runner-up won $ 检查
    💡 思路

    Let runner-up = $r$; winner = $2r + 100$. $r + 2r + 100 = 2350 \Rightarrow 3r = 2250 \Rightarrow r = 750$.

  5. Q8. A group of boys and girls planted a total of 148 trees. Each boy planted 7 trees and each girl planted 5 trees. There were 4 more boys than girls in the group. How many boys were there in the group?
    Number of boys = 检查
    💡 思路

    Let girls = $g$; boys = $g + 4$. $7(g+4) + 5g = 148 \Rightarrow 12g + 28 = 148 \Rightarrow g = 10$. Boys = 14.

  6. Q9. Aaron, Ben and Chetan shared 110 marbles. Ben received twice as many marbles as Aaron. Chetan received 10 more marbles than Aaron. How many marbles did each boy receive?
    Aaron: ; Ben: ; Chetan: 检查
    💡 思路

    Let Aaron = $a$; Ben = $2a$; Chetan = $a + 10$. Sum: $4a + 10 = 110 \Rightarrow a = 25$. So 25, 50, 35.

ADVANCED
  1. Q10. A triathlon includes swimming, cycling and running in one event. The cycling distance is 4 times the running distance. The swimming distance is 8.5 km less than the running distance and 38.5 km less than the cycling distance. Find the total distance of the race.
    Total distance = km 检查
    💡 思路

    Let running = $r$. Then cycling = $4r$. Swimming = $r - 8.5$ AND $4r - 38.5$ — set them equal:
    $r - 8.5 = 4r - 38.5 \Rightarrow -3r = -30 \Rightarrow r = 10$.
    So run = 10, cycle = 40, swim = 1.5. Total = $\mathbf{51.5}$ km.

  2. Q11. Before a chemical reaction, the mass of iron is 3 times the mass of sulphur. After the reaction, the masses of iron and sulphur are reduced by 10 g and 6 g respectively, and the mass of iron is 4 times the mass of sulphur. Find the mass of iron before the reaction.
    Mass of iron before = g 检查
    💡 思路

    Let sulphur before = $s$. Iron before = $3s$. After: sulphur = $s - 6$, iron = $3s - 10$.
    $3s - 10 = 4(s - 6) \Rightarrow 3s - 10 = 4s - 24 \Rightarrow s = 14$. Iron before = $3 \times 14 = 42$ g.

  3. Q12. Mrs Tan has some money to buy fruits. She can buy $n$ mangoes at $1.60 each and have $0.80 left. Alternatively, she can buy $(n + 10)$ apples at $0.70 each and have $0.10 left.
    (a) Find the value of $n$. → 检查
    (b) How much money does Mrs Tan have for buying fruits? → $ 检查
    (c) If Mrs Tan buys 3 mangoes and uses the rest of the money to buy apples,
    (i) how many apples can she buy? → 检查
    (ii) how much money will she have left? → $ 检查
    💡 思路

    (a) Money is fixed both ways: $1.60n + 0.80 = 0.70(n+10) + 0.10$；$1.6n + 0.8 = 0.7n + 7.1$；$0.9n = 6.3$；$n = 7$.
    (b) Money = $1.60 \times 7 + 0.80 = \$12$.
    (c)(i) After 3 mangoes: $12 - 3 \times 1.60 = \$7.20$. Apples: $\lfloor 7.20 / 0.70 \rfloor = 10$.
    (c)(ii) $7.20 - 10 \times 0.70 = \$0.20$.

  4. Q13. (OPEN) Write an application problem such that the equation to be formed for solving the problem is $5x + 4(x - 10) = 140$.
    💡 参考

    开放题。先解方程：$5x + 4x - 40 = 140 \Rightarrow 9x = 180 \Rightarrow x = 20$. 然后编一个 $x = 20$ 合理的情境：
    示例：玛丽买了 $x$ 支铅笔，每支 5 元；同时买了 $(x - 10)$ 本练习本，每本 4 元；总共花了 140 元。求她买了多少支铅笔。

  5. Q14. The denominator of a fraction is 3 more than its numerator. If 2 is added to both the numerator and the denominator, the new fraction is equivalent to $\dfrac{2}{3}$. Find the original fraction.
    Original fraction = 检查
    💡 思路

    Let numer = $x$; denom = $x + 3$. $\dfrac{x+2}{x+5} = \dfrac{2}{3} \Rightarrow 3(x+2) = 2(x+5) \Rightarrow 3x + 6 = 2x + 10 \Rightarrow x = 4$. Original: $\dfrac{4}{7}$.