# Excercise 2.6 Linear Equations in One Variable- NCERT Solutions Class 8

Linear Equations in One Variable

## Question 1

Solve: \(\begin{align}\frac{{8x - 3}}{{3x}} = 2\end{align}\)

### Solution

**Video Solution**

**What is known?**

Equation

**What is unknown?**

Value of the variable

**Reasoning:**

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

**Steps:**

\(\begin{align}\frac{{8x - 3}}{{3x}} = 2\end{align}\)

On multiplying both sides by \(3x\), we obtain

\[\begin{align}8x - 3 = 6x \\8x - 6x = 3 \\\,\,\,\,\,\,\,\,\,\,2x = 3 \\\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{3}{2} \\\end{align} \]

## Question 2

Solve: \(\begin{align}\frac{{9x}}{{7 - 6x}} = 15\end{align}\)

### Solution

**Video Solution**

**What is known?**

Equation

**What is unknown?**

Value of the variable

**Reasoning: **

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

**Steps:**

On multiplying both sides by \(7{\text{ }} - {\text{ }}6x\) we obtain

\[\begin{align}9x &= 15\left( {7 - 6x} \right) \\

9x &= 105 - 90x \\

9x + 90x &= 105 \\

\,\,\,\,\,\,\,\,\,\,99x &= 105 \\

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &= \frac{{105}}{{99}} \\

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{35}}{{33}} \\

\end{align} \]

## Question 3

** **Solve: \(\begin{align}\frac{z}{{z + 15}} = \frac{4}{9}\end{align}\)

### Solution

**Video Solution**

**What is known?**

Equation

**What is unknown?**

Value of the variable

**Reasoning:**

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

**Steps:**

On multiplying both sides by\(\begin{align} {\text{ 9(}}z{\text{ + 15), }} \hfill \\ \end{align} \)we obtain

\[\begin{align}9z& = 4\left( {{\rm{z + 15}}} \right)\\9z &= 4z + 60\\9z - 4z &= 60\\\,\,\,\,\,\,\,5z &= 60\\\,\,\,\,\,\,\,\,\,z &= 12\end{align}\]

## Question 4

Solve:\(\begin{align}\frac{{3y + 4}}{{2 - 6y}} = \frac{{ - 2}}{5}\end{align}\)

### Solution

**Video Solution**

**What is known?**

Equation

**What is unknown?**

Value of the variable

**Reasoning:**

**Steps:**

On multiplying both sides by \(5(2-6 y),\) we obtain

\[\begin{align} 5\left( 3\text{ }y+4 \right)&=-2\left( 2-6\text{ }y \right) \\15y+20&=-4+12\text{ }y \\15y-12y&=-4-20 \\3\,y&=-24 \\y &=-8\end{align}\]

## Question 5

Solve:\(\begin{align} \frac{{7y + 4}}{{y + 2}} = \frac{{ - 4}}{3}\end{align}\)

### Solution

**Video Solution**

**What is known?**

Equation

**What is unknown?**

Value of the variable

**Reasoning:**

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

**Steps:**

On multiplying both sides by \(3(y + 2)\),we obtain

\[\begin{align}3\left( {7{\rm{ }}y + 4} \right) &= - 4\left( {y + 2} \right)\\21y + 12 &= - 4y - 8\\21y + 4y &= - 8 - 12\\\,\,\,\,\,\,\,\,\,\,25y &= - 20\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y &= - \frac{4}{5}\end{align}\]

## Question 6

The ages of Hari and Harry are in the ratio \(5:7\). Four years from now the ratio of their ages will be \(3:4\) Find their present ages.

### Solution

**Video Solution**

**What is known?**

i) Ages of Hari and Harry are in the ratio \(5:7\)

ii) Four years from now the ratio of their ages will be \(3:4\)

**What is unknown?**

Present ages of Hari and Harry.

**Reasoning:**

Use the ratio condition and express ages of Hari and Harry in the form of variable. Use second condition to form the equation.

**Steps:**

Let the common ratio between their ages be *\(x\)*.

Therefore, Hari’s age and Harry’s age will be \(5x\) years and \(7x\) years respectively and four years later, their ages will be \((5x + 4)\) years and \((7x + 4) \)years respectively.

According to the situation given in the question,

\begin{align}\frac{{5x + 4}}{{7x + 4}} &= \frac{3}{4} \\4\left( {5x + 4} \right) &= 3\left( {7x + 4} \right) \\

\,20x + 16 &= 21x + 12 \\\,\,\,\,16 - 12 &= 21x - 20x \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 &= x \\\end{align}

Hari’s age \(=5x \) years \(=\rm{}\, (5 × 4)\) years \(= \rm{}\,20 \) years

Harry’s age \(= 7x\) years \(=\rm{}\, (7 × 4)\) years \(= \rm{}\,28 \) years

Therefore, Hari’s age and Harry’s age are \(\rm{}\,20\) years and \(\rm{}\,28\) years respectively.

## Question 7

The denominator of a rational number is greater than its numerator by \(8\). If the numerator is increased by \(17 \) and the denominator is decreased by \(1\), the number obtained is \(\begin{align}\frac{3}{2}\end{align}\).Find the rational number.

### Solution

**Video Solution**

**What is known?**

i) The denominator of a rational number is greater than its numerator by \(8\)

ii) If the numerator is increased by \(17\) and the denominator is decreased by 1, the number obtained is \(\begin{align}\frac{3}{2}\end{align}\)

**What is unknown?**

the rational number

**Reasoning:**

Assume numerator of the fraction as variable. Use first condition to express denominator in the form of variable and use second condition to form the equation.

**Steps:**

Let the numerator of the rational number be* *\(x\) Therefore, its denominator will be \(x + 8\).

The rational number will be\(\begin{align}\frac{x}{{x + 8}}\end{align}\). According to the question,

\[\begin{align}\,\,\frac{{x + 17}}{{x + 8 - 1}} &= \frac{3}{2} \\\,\,\,\,\,\,\frac{{x + 17}}{{x + 7}} &= \frac{3}{2} \\2\left( {x + 17} \right) &= 3\left( {x + 7} \right) \\\,\,\,2x + 34 &= 3x + 21 \\\,\,\,34 - 21 &= 3x - 2x \\\,\,\,\,\,\,\,\,\,\,\,\,\,13 &= x \\\end{align}\]

Numerator of the rational number \(= x = 13\)

Denominator of the rational number \(= x + 8 = 13 + 8 = 21\)

Rational number \(=\begin{align} \frac{{13}}{{21}}\end{align}\)

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