## Math 8 Review chapter 4: Unequal first degree one unknown

## 1. Theoretical Summary

### 1.1. Inequality

An inequality is a relation of the form \(a > b\) or \(a < b,a \ge b,a \le b\)

**a. The property of adding the same number to both sides of the inequality**

When we add the same number to both sides of an inequality, we get a new inequality in the same direction as the two given inequalities \( a > b \Rightarrow a + c > b + c\)

**b. The property of multiplying the same number on both sides of the inequality**

+) When we multiply both sides of an inequality by the same positive number, we get a new inequality in the same direction as the given inequality.

With \(a > b\) and \(c > 0 \Rightarrow ac > bc\)

+) When we multiply both sides of an inequality by the same negative number, we get a new inequality in the same direction as the given inequality.

With \(a > b\) and \( c < 0 \Rightarrow ac < bc\)

**c. Bridging properties**

If a > b and b > c then a > c

**Attention: **Cosic inequality

The average of two non-negative numbers is greater than or equal to their multiplicative mean.

\(\dfrac{{a + b}}{2} \ge \sqrt {ab} \) with \(a \ge 0;b \ge 0\)

### 1.2. Solution set of the inequality

The number x = a is called the solution of an inequality. If we substitute x = a into the inequality, we get a true inequality.

The set of all solutions of an inequality is called the set of solutions to the inequality. Solving an inequality is to find the set of solutions to that inequality.

Equivalent Inequality: Two equivalent inequalities are two inequalities that have the same set of solutions.

### 1.3. One-hidden first-order inequality

Inequality of the form \(ax + b > 0\) (or \(ax + b < 0,ax + b \ge 0,ax + b \le 0\)) in there \(a\) and \(b\) are two given numbers, \(a \ne 0\), is called a first-order inequality with one unknown.

Transition rule: When moving a term of an inequality from one side to the other, we must change the sign of that term.

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

+ Keep the same dimension if the number is positive;

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

### 1.4. Equations containing absolute sign

Repeat:.

\(\left| a \right| = \left\{ \begin{array}{l}a\;\;when\;\;a \ge 0\\ – a\;\;when\;\;a < 0\end{array} \right..\)

To solve an equation containing the absolute value sign (GDT) of the form \(\left| {A\left( x \right)} \right| = B\left( x \right)\), we remove the GTTD sign by considering 2 cases:

– Case 1: \( \left\{ \begin{array}{l}A\left( x \right) \ge 0\\A\left( x \right) = B\left( x \right)\end{array} \ right.\)

– Case 2: \(\left\{ \begin{array}{l}A\left( x \right) < 0\\ - A\left( x \right) = B\left( x \right)\end{array} \ right.\)

+) With an equation of the form \(\left| {A\left( x \right)} \right| = m\) with \(m > 0\), We have:

\(\left| {A\left( x \right)} \right| = m \Leftrightarrow A\left( x \right) = m\) or \(A\left( x \right) = – m\)

+) With an equation of the form \(\left| {A\left( x \right)} \right| = \left| {B\left( x \right)} \right|\) We have:

\( \left| {A\left( x \right)} \right| = \left| {B\left( x \right)} \right| \Leftrightarrow A\left( x \right) = B\left( x \right)\) or \(A\left( x \right) = – B\left( x \right)\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Let \(m > n\), prove

a) \(m + 2 > n +2\)

b) \(-2m < -2n\)

**Solution guide**

a) We have \(m > n\)

Adding both sides of the inequality \(m > n\) with \(2\) we get:

\( m + 2 > n + 2\) (which must be proved).

b) We have \(m > n\)

Multiplying both sides of the inequality \(m > n\) by \((-2)\) we get:

\(- 2m < - 2n\) (something to be proved)

### 2.2. Exercise 2

Solve the inequalities and represent the solution set on the number line:

a) \(x – 1 < 3\)

b) \(x + 2 > 1\)

**Solution guide**

a) \(x – 1 < 3 x < 1 + 3 x < 4\)

So the solution set \(S = \left\{ {x \,|\,x < 4} \right\}\)

Representation on the number line:

b) \(x +2 > 1 ⇔ x > 1 – 2 ⇔ x > -1\)

So the solution set \(S = \left\{ {x \,|\,x > -1} \right\}\)

Representation on the number line:

### 2.3. Exercise 3

Solve the inequalities:

a) \(3 \leqslant \dfrac{{2x + 3}}{5}\)

b) \(\dfrac{{4x – 5}}{3} > \dfrac{{7 – x}}{5}\)

**Solution guide**

a) \(3 \leqslant \dfrac{{2x + 3}}{5}\)

\(\Leftrightarrow 5.3 \leqslant 5.\dfrac{{2x + 3}}{5}\)

\(\Leftrightarrow 15 \le 2x + 3\)

\(⇔15 – 3 \le 2x \)

\(\Leftrightarrow 12 \le 2x\)

\(\Leftrightarrow 6 \le x\)

\(\Leftrightarrow x \ge 6\)

So the solution of the inequality is: \(x \ge 6\)

b) \(\dfrac{{4x – 5}}{3} > \dfrac{{7 – x}}{5}\)

\(\Leftrightarrow 15.\dfrac{{4x – 5}}{3} > 15.\dfrac{{7 – x}}{5}\)

\(\Leftrightarrow 5\left( {4x – 5} \right) > 3\left( {7 – x} \right)\)

\(⇔20x – 25 > 21 – 3x\)

\(⇔20x + 3x > 21 + 25\)

\(⇔23x > 46\)

\(⇔x > 46 : 23\)

\(⇔x > 2\)

So the solution of the inequality is: \(x > 2\)

### 2.4. Exercise 4

Find x such that:

a) The value of the expression \(x + 3\) is less than the value of the expression \(4x – 5\);

b) The value of the expression \(2x +1\) is not less than the value of the expression \(x + 3\);

**Solution guide**

a) We have the inequality: \(x + 3 < 4x - 5\)

\(⇔x – 4x < -5 - 3\)

\(⇔-3x < -8\)

\(⇔x > \dfrac{8}{3}\)

So for \(x + 3\) to be less than \(4x – 5\) then \(x >\dfrac{8}{3}\) .

b) We have the inequality: \(2x +1 x + 3\)

\(⇔ 2x – x 3 – 1\)

\(⇔ x ≥ 2\)

So, so that \(2x +1\) is not less than the value of the expression \(x + 3\), then \(x ≥ 2\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let \(a > b\), prove that

a) \(3a + 5 > 3b + 2\) ;

b) \(2 – 4a < 3 - 4b\).

**Verse 2:**

a) Show that \(2,99\) is a solution of the inequality \(3 > x\). Name three numbers greater than \(2.99\) that are also solutions to the inequality.

b) Show that \(4,01\) is a solution of the inequality \(4 < x\). Name three numbers less than \(4,01\) that are also solutions to the inequality.

**Question 3: **Solve the inequalities and represent their solutions on the number line:

a) \(2\left( {3x – 1} \right) – 2x < 2x + 1\;;\)

b) \(4x – 8 \ge 3\left( {3x – 2} \right) + 4 – 2x.\)

**Question 4: **Solving the inequalities:

a) \(\displaystyle2x + 1.4 < {{3x - 7} \over 5}\)

b) \(\displaystyle1 + {{1 + 2x} \over 3} > {{2x – 1} \over 6} – 2\)

### 3.2. Multiple choice exercises

**Question 1:** Compare \(m^{3}\) and \(m^{2}\) with 0 < m < 1.

A. \(m^{2} > m^{3}\)

B. \(m^{2} < m^{3}\)

C. \(m^{2} = m^{3}\)

D. incomparable

**Verse 2: **Equation |x − 1| + |x − 3| = 2x − 1 has a number of solutions

A. 2

B. 1

C. 3

D. 0.

**Question 3:** Let a > b, c > d. Which of the following assertion true?

A. a + d > b + c

B. a + c > b + d

C. b + d > a + c

D. a + b > c + d

**Question 4: **Minimum solution of the equation |2 + 3x| = |4x − 3| to be

A. \(\frac{1}{7}\)

B. 5

C. -\(\frac{1}{7}\)

D. -5

**Question 5: **Let a > b. Which of the following assertion true?

A. – 3a – 1 > – 3b – 1

B. – 3(a – 1) < - 3(b - 1)

C. – 3(a – 1) > – 3(b – 1)

D. 3(a – 1) < 3(b - 1)

## 4. Conclusion

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

- Understand the relationship between order and multiplication, the transitive property of order applies to solving problems
- Practice presenting assignments.
- Applying it in real life

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