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

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

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

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

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

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

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

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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