## Math 8 Chapter 3 Lesson 6: Solve math problems by making equations

## 1. Theoretical Summary

### 1.1. Steps to solve the problem by making equations

**Step 1:** Equation

- Select the unknown and set the appropriate conditions for the unknown.
- Express unknown quantities in terms of unknowns and known quantities
- Write an equation that shows the relationship between the quantities.

**Step 2:** Solve the equation

**Step 3: **Reply

Check which of the solutions of the equation satisfy the condition of the unknowns, which do not, and then conclude.

### 1.2. Notes on hiding selection and its proper conditions

– Usually, the problem asks about what quantity, then choose the hidden quantity as that quantity.

– About the suitable condition of hiding

- If \(x\) represents a digit then \(0 ≤ x ≤ 9\).
- If \(x\) represents age, product, or person, then \(x\) is a positive integer.
- If \(x\) represents the velocity of the motion, then \(x > 0.\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Let \(x\) be a two-digit natural number (eg \(x=12\)). Make an expression for the natural number obtained by:

a) Add a digit \(5\) to the left of the number \(x\) (for example, \(12\to512\), ie \(500+12\));

b) Add the digit \(5\) to the right of the number \(x\) (eg \(12\to 125\), i.e. \(12\times 10+5\)).

**Solution guide**

a) The expression for the new natural number when the digit \(5\) is added to the left of the number \(x\) is: \(5. 100 + x\)

b) The expression to represent the new natural number when adding the digit \(5\) to the right of the number \(x\) is: \(10x + 5\)

### 2.2. Exercise 2

The denominator of a fraction that is greater than its numerator is \(3\) by units. If both the numerator and denominator are increased by \(2\) by one unit, the new fraction is equal to \(\dfrac{1}{2}\) . Find the original fraction.

**Solution guide**

Let \(x\) be the numerator of the fraction ( \(x \in Z,x \ne – 3)\)

Since the denominator of a fraction is \(3\) greater than its numerator, the denominator of the fraction is \(x + 3\).

If we increase both the numerator and its denominator by \(2\) units, then we get the following fraction \(\dfrac{{x + 2}}{{x + 3 + 2}} = \dfrac{{x) + 2}}{{x + 5}}\) \((x \ne – 5)\)

Since the new fraction is equal to \(\dfrac{1}{2}\) we have the equation:

\(\eqalign{

& {1 \over 8}x + 3 = {{20} \over {100}}x \cr

& \Leftrightarrow {1 \over 8}x + 3 = {1 \over 5}x \cr

& \Leftrightarrow {{5x} \over {40}} + {{3.40} \over {40}} = {{8x} \over {40}} \cr

& \Leftrightarrow 5x + 120 = 8x \cr

& \Leftrightarrow 5x – 8x = – 120 \cr

& \Leftrightarrow – 3x = – 120 \cr

& \Leftrightarrow x = \left( { – 120} \right):\left( { – 3} \right) \cr

& \Leftrightarrow x = 40 \text{ (satisfied)}\cr} \)

So the original fraction is:\(\dfrac{1}{4}\)

### 2.3. Exercise 3

An odd two-digit natural number that is divisible by 5. The difference of that number and its tens digit is 68. Find the number.

**Solution guide**

Let x be the tens digit of the number to find, the condition that x is an integer and \(0 < x \le 9.\) Because:

* The number to find is odd and divisible by 5, so its unit digit must be equal to 5, deduce the number to find the form: \(\overline {x.5} = 10x + 5\)

* The difference of that number and its tens digit is 68, so:

\((10x + 5) – x = 68 \Leftrightarrow 9x = 63 \Leftrightarrow x = 7,\) satisfy the condition.

So the required number is 75.

## 3. Practice

### 3.1. Essay exercises

**Question 1: **The sum of two numbers is \(80\), their difference is \(14\). Find two of them.

**Verse 2: **The sum of two numbers is \(90\), one is twice the other. Find two of them.

**Question 3:** The difference of two numbers is \(22\) , one is twice the other. Find those two numbers, knowing that :

a) The two numbers mentioned in the lesson are two positive numbers.

b) The two numbers given in the problem are arbitrary.

**Question 4: **Two positive integers whose ratio between the first and the second is \(\displaystyle{3 \over 5}\). If the first number is divided by \(9\) and the second number is divided by \(6\), the quotient of the division of the first number by \(9\) is less than the quotient of the division of the second number by \(6). \) is \(3\) units. Find those two numbers, knowing that the above division is divisible.

### 3.2. Multiple choice exercises

**Question 1:** Instead of multiplying a number by 6, he divides that number by 6. Next instead of having to work by 14, Nam subtracts by 14. The result is 16. If you don’t make the wrong calculation twice, the result of the number is :

A. Less than 400

B. Between 400 and 600

C. Between 800 and 1100

D. Between 600 and 800

**Verse 2:** An goes to work by train or bus. If he goes by train in the afternoon, he will go home by bus; and if he goes home by train in the afternoon, he goes to work by bus in the morning. Out of the total x days of travel, he takes the bus in the morning 8 times, goes home by bus 15 times and takes the train all 9 times both going back and forth. Look for x

A. 19

B. 18

C. 17

D. 16

**Question 3:** An and Binh ran 10km. They started from the same point, ran 5km up the hill and then returned to the starting point according to the route. An ran 10 minutes ahead with a speed of 15km/h when going up the hill and 20km/h when going down the hill. Binh runs 16km/h when going uphill and 22km/h when going downhill. When passing in opposite directions, how far are they from the top of the hill?

A. \(\frac{5}{4}km\)

B. \(\frac{35}{27}km\)

C. \(\frac{27}{20}km\)

D. \(\frac{7}{3}km\)

**Question 4: **Last year, the total population of two provinces A and B was 4 million. The population of province A this year increased by 1.2% and province B by 1.1%. The total population of the two provinces this year is 4045,000 people.

A. The population of province A last year was 1000000 people and this year is 1012,000 people

B. The population of province B last year was 3000000 people and this year is 3033000 people

C. Both a and b are correct

D. Both a and b are wrong

**Question 5: **A right triangle has a perimeter of 24m. The side of the first right angle is 2m more than the side of the second right angle. The length of the first right angle is:

A.42m

B.38m

C.4m

D.8m

## 4. Conclusion

Through this lesson, you will learn some of the main topics as follows:

- Understand how to solve problems by making equations
- How to choose a hide and its suitable condition

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