## Math 8 Chapter 3 Lesson 5: Equations hidden in denominators

## 1. Theoretical Summary

### 1.1. Determination condition of an equation

The defining condition of the equation is the set of values of the unknowns that make all the samples in the equation non-zero. The defining condition of the equation is abbreviated as CONCLUSION.

### 1.2. Solve the equation contained in the sample

- Step 1: Find the exact condition of the equation
- Step 2: Reduce the denominator on both sides of the equation and then desample.
- Step 3: Solve the equation just received.
- Step 4: Conclusion.

Among the values of the unknowns found in step 3, the values that satisfy the determination condition are the solutions of the given equation.

## 2. Illustrated exercise

### 2.1. Exercise 1

Find the conditions for each of the following equations:

\(\eqalign{& a)\,\,{x \over {x – 1}} = {{x + 4} \over {x + 1}} \cr & b)\,\,{3 \over {x – 2}} = {{2x – 1} \over {x – 2}} – x \cr} \)

**Solution guide**

a) \(x – 1 0\) when \(x ≠ 1\)

\(x + 1 0\) when \(x ≠ – 1\)

So, the SCC of the equation \(\dfrac{x}{{x – 1}} = \dfrac{{x + 4}}{{x + 1}}\) is \(x ≠ 1\) and \(x) – 1\)

b) \(x – 2 0\) when \(x ≠ 2\)

So, the SCC of the equation \(\dfrac{3}{{x – 2}} = \dfrac{{2x – 1}}{{x – 2}} – x\) is \(x ≠ 2\)

### 2.2. Exercise 2

Solve the following equation

\(\eqalign{& a)\,\,{x \over {x – 1}} = {{x + 4} \over {x + 1}} \cr & b)\,\,{3 \over {x – 2}} = {{2x – 1} \over {x – 2}} – x \cr} \)

**Solution guide**

a) \(\dfrac{x}{{x – 1}} = \dfrac{{x + 4}}{{x + 1}}\)

Term: \(x\ne 1\) and \(x\ne -1\)

\( \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \ dfrac{{\left( {x – 1} \right)\left( {x + 4} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right )}}\)

\(\eqalign{

& \Rightarrow x\left( {x + 1} \right) = \left( {x – 1} \right)\left( {x + 4} \right) \cr

& \Leftrightarrow {x^2} + x = {x^2} + 4x – x – 4 \cr

& \Leftrightarrow {x^2} + x = {x^2} + 3x – 4 \cr

& \Leftrightarrow {x^2} + x – {x^2} – 3x = – 4 \cr

& \Leftrightarrow – 2x = – 4 \cr

& \Leftrightarrow x = \left( { – 4} \right):\left( { – 2} \right) \cr

& \Leftrightarrow x = 2 \text{(satisfy the CONDITIONS)}\cr} \)

So the solution set of the equation is: \(S = \{2\}\)

b) \(\dfrac{3}{{x – 2}} = \dfrac{{2x – 1}}{{x – 2}} – x\)

TCM: \(x\ne2\)

\(\eqalign{

& \Leftrightarrow {3 \over {x – 2}} = {{2x – 1} \over {x – 2}} – {{x\left( {x – 2} \right)} \over {x – 2 }} \cr

& \Rightarrow 3 = 2x – 1 – x\left( {x – 2} \right) \cr

& \Leftrightarrow 3 = 2x – 1 – {x^2} + 2x \cr

& \Leftrightarrow 3 = – {x^2} + 4x – 1 \cr

& \Leftrightarrow {x^2} – 4x + 3 + 1 = 0 \cr

& \Leftrightarrow {x^2} – 4x + 4 = 0 \cr

& \Leftrightarrow {x^2} – 2.x.2 + {2^2} = 0 \cr

& \Leftrightarrow {\left( {x – 2} \right)^2} = 0 \cr

& \Leftrightarrow x = 2\,\text{ (type)} \cr} \)

So the solution set of the equation is: \(S = \phi \)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** When solving the equation \(\displaystyle{{2 – 3x} \over { – 2x – 3}} = {{3x + 2} \over {2x + 1}}\) , Ha’s friend did the following:

By the definition of two equal fractions, we have:

\(\displaystyle {{2 – 3x} \over { – 2x – 3}} = {{3x + 2} \over {2x + 1}} \)

\(\displaystyle \Leftrightarrow \left( {2 – 3x} \right)\left( {2x + 1} \right) \) \(\displaystyle= \left( {3x + 2} \right)\left( { – 2x – 3} \right) \)

\(\displaystyle \Leftrightarrow – 6{x^2} + x + 2 = – 6{x^2} – 13x – 6 \)

\(\displaystyle \Leftrightarrow 14x = – 8 \Leftrightarrow x = – {4 \over 7} \)

So the equation has a solution \(\displaystyle x = – {4 \over 7}\)

Please let me know your opinion about Ha.

**Verse 2: **The following statements are true or false:

a) The equation \(\displaystyle{{4x – 8 + \left( {4 – 2x} \right)} \over {{x^2} + 1}} = 0\) has a solution of \(x = 2 \).

b) The equation \(\displaystyle{{\left( {x + 2} \right)\left( {2x – 1} \right) – x – 2} \over {{x^2} – x + 1} } = 0\) has a solution set of \(S = \{ -2; 1 \}\).

c) The equation \(\displaystyle{{{x^2} + 2x + 1} \over {x + 1}} = 0\) has a solution of \(x = -1\).

d) The equation \(\displaystyle{{{x^2}\left( {x – 3} \right)} \over x} = 0\) has a solution set of \(S = \{ 0; 3 \} \).

**Question 3: **Solve the following equation:

a) \(\displaystyle{{1 – x} \over {x + 1}} + 3 = {{2x + 3} \over {x + 1}}\)

b) \(\displaystyle{{{{\left( {x + 2} \right)}^2}} \over {2x – 3}} – 1 = {{{x^2} + 10} \over { 2x – 3}}\)

c) \(\displaystyle{{5x – 2} \over {2 – 2x}} + {{2x – 1} \over 2} = 1 – {{{x^2} + x – 3} \over {1 – x}}\)

**Verse 4**

a) Find \(x\) such that the value of the expression \(\displaystyle{{2{x^2} – 3x – 2} \over {{x^2} – 4}}\) equals \(2 .\)

b) Find \(x\) such that the values of the two expressions \(\displaystyle{{6x – 1} \over {3x + 2}}\) and \(\displaystyle{{2x + 5} \over { x – 3}}\) are equal.

c) Find \(y\) such that the value of the two expressions \(\displaystyle{{y + 5} \over {y – 1}} – {{y + 1} \over {y – 3}}\ ) and \(\displaystyle{{ – 8} \over {\left( {y – 1} \right)\left( {y – 3} \right)}}\) are equal.

### 3.2. Multiple choice exercises

**Question 1: **The defining condition of the equation \(\frac{x}{{x – 2}} – \frac{{2x}}{{{x^2} – 1}} = 0\) is

A. \(x \ne 1;x \ne – 2\)

B. \(x \ne 0\)

C. \(x \ne 2\) and \(x \ne \pm 1\)

D. \(x \ne – 2;x \ne 1\)

**Verse 2: **The number of solutions to the equation \(\frac{{{\rm{x – 5}}}}{{x – 1}} + \frac{2}{{x – 3}} = 1\) is:

A. 3

B. 2

C. 0

D. 1

**Question 3: **The defining condition of the equation \(\frac{1}{{x – 2}} + 3 = \frac{{3 – x}}{{x – 2}}\) is

A. \(x \ne 3\)

B. \(x \ne 2\)

C. \(x \ne -2\)

D. \(x \ne -3\)

**Question 4: **The equation \(\frac{{6{\rm{x}}}}{{9 – {x^2}}} = \frac{x}{{x + 3}} – \frac{3}{{ 3 – x}}\) has a solution of

A. x = -3

B. x = -2

C. inexperienced

D. countless solutions

**Question 5: **The equation \(\left( {{x^2} – 1} \right)\left( {{x^2} – 4} \right)\left( {{x^2} – 9} \right) = 0\) how many solutions

A. 3

B. 4

C. 5

D. 6

## 4. Conclusion

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

- The concept of the definite condition of an equation, how to find the definite condition of the equation.
- How to solve the equation containing the hidden in the sample

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