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

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Chapter 2 Ex.2.6 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

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

Chapter 2 Ex.2.6 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

\[\begin{align}9x &= 15\left( {7 - 6x} \right) \\
9x &= 105 - 90x \\
9x + 90x &= 105 \\
\,\,\,\,\,\,\,\,\,\,99x &= 105 \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &= \frac{{105}}{{99}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{35}}{{33}} \\
\end{align} \]

Chapter 2 Ex.2.6 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

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

Chapter 2 Ex.2.6 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:

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 \(5(2-6 y),\) we obtain

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

Chapter 2 Ex.2.6 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

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

Chapter 2 Ex.2.6 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.

Chapter 2 Ex.2.6 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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