## Math 8 Chapter 4 Lesson 4: Unequal first degree one unknown

## 1. Theoretical Summary

### 1.1. Define

Inequality of the form \(ax + b < 0\) (or \(ax + b > 0\), \(ax + b ≤ 0\), \(ax + b ≥ 0\)) where \(a \) and \(b\) are two given numbers, \(a\ne 0\), which is called a one-hidden first-order inequality.

### 1.2. Two rules for transforming inequalities

**a) Transition rule**

When we move a term of an inequality from one side to the other, we change the sign of that term.

**b) The rule for multiplying by a number**

When multiplying both sides of the inequality by the same non-zero number, we must:

– Keep the direction of the equation unchanged if the number is positive.

– Change the direction of the equation if the number is negative.

### 1.3. Apply

Applying the above two transformation rules, we solve the first-order inequality in one unknown as follows:

Format: \(ax + b > 0 \Leftrightarrow ax > -b\)

\( \Leftrightarrow x > \dfrac{-b}{a}\) if \(a > 0\) or \(x < \dfrac{-b}{a}\) if \(a < 0\).

So the solution of the inequality \(ax + b > 0\) is:

\({S_1} = \left\{ {x|x > \dfrac{ – b}{ a},a > 0} \right\}\) or \({S_2} = \left\{ {x|x) < \dfrac{{ - b}}{a},a < 0} \right\}\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Solve the following inequalities:

a) \(x + 12 > 21\);

b) \(-2x > -3x – 5\).

**Solution guide**

a) \(x + 12 > 21 x > 21 – 12 x > 9\)

So the solution set of the inequality \(x + 12 > 21\) is \(S=\{x|x > 9\}\)

b) \(-2x > -3x – 5 ⇔ -2x + 3x > -5 \)\(\,⇔ x > -5\)

So the solution set of the inequality \(-2x > -3x – 5\) is \(S=\{x|x > -5\}\)

### 2.2. Exercise 2

Solve the following inequalities (using the multiplication rule):

a) \(2x < 24\)

b) \(-3x < 27\)

**Solution guide**

a) \(2x < 24\)

\( \Leftrightarrow 2x.\dfrac{1}{2} < 24.\dfrac{1}{2}\)

\( \Leftrightarrow x < 12\)

So the solution set of the inequality \(2x < 24\) is \(S=\{x|x < 12\}\).

b) \(-3x < 27 \)

\( \Leftrightarrow – 3x.\dfrac{{ – 1}}{3} > 27.\dfrac{{ – 1}}{3}\)

\(⇔ x > -9\)

So the solution set of the inequality \(-3x < 27\) is \(S=\{x|x > -9\}\).

### 2.3. Exercise 3

Solve the inequality: \(-0.2x-0.2>0.4x-2\)

**Solution guide**

\(\eqalign{

& – 0.2x – 0.2 > 0.4x – 2 \cr

& \Leftrightarrow – 0.2x – 0.4x > – 2 + 0.2 \cr

& \Leftrightarrow – 0.6x > – 1.8 \cr

& \Leftrightarrow -0.6x:(-0.6) < \left( { - 1.8} \right):\left( { - 0.6} \right) \cr

& \Leftrightarrow x < 3 \cr} \)

So the solution set of the inequality is: \(\{x|x<3\}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Applying the transformation rule, solve the following inequalities:

a) \(x – 2 > 4\)

b) \(x + 5 < 7\)

c) \(x – 4 < -8\)

**Verse 2:** Applying the multiplication rule, solve the following inequalities:

a) \(\displaystyle{1 \over 2}x > 3\)

b) \(\displaystyle – {1 \over 3}x < - 2\)

c) \(\displaystyle{2 \over 3}x > – 4\)

d) \(\displaystyle – {3 \over 5}x > 6\)

**Question 3: **Equivalence explanation:

a) \(2x < 3 \Leftrightarrow 3x < 4.5\)

b) \(x – 5 < 12 \Leftrightarrow x + 5 < 22\)

c) \( – 3x < 9 \Leftrightarrow 6x > – 18\)

**Question 4:** Solve the inequalities and represent their solutions on the number line:

a) \(2x – 4 < 0\)

b) \(3x + 9 > 0\)

c) \( – x + 3 < 0\)

### 3.2. Multiple choice exercises

**Question 1: **The positive integers satisfying the two inequalities 2x>-9 and 7-3x>0 are:

A. 1;2

B. 3;4

C. 5,6

**D. Another answer**

**Verse 2: **The negative integers satisfying both the inequalities 16-2x>0 and 4x-3>0 are:

A. -1;-2

B. -3;-4

C. -5;-6

D. Another answer

**Question 3: **The solution set of the inequality \(\frac{5x-2}{4}>\frac{1-2x}{12}\) is:

A. S={x|x>\(\frac{7}{17}\)}

B. S={x|x>\(\frac{7}{16}\)}

C. S={x|x>\(\frac{7}{15}\)}

D. S={x|x>\(\frac{7}{12}\)}

**Question 4:** The solution set of the inequality \(\frac{12x+5}{8}<\frac{3x-1}{12}\) is:

A. S={x|x<\(-\frac{17}{26}\)}

B. S={x|x<\(-\frac{17}{28}\)}

C. S={x|x<\(-\frac{17}{30}\)}

D. Another answer

**Question 5: **The solution set of the inequality \(\frac{2}{3}x-1<-\frac{1}{2}\) is:

A. S={x|x<\(-\frac{3}{4}\)}

B. S={x|x<\(\frac{17}{26}\)}

C. S={x|x>3}

D. S={x|x<\(-\frac{4}{3}\)}

## 4. Conclusion

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

- Identify first order inequalities 1 hidden
- Know how to apply each inequality transformation rule to solve simple inequalities
- Know how to use inequalities transformation rules to explain the equivalence of BPT

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