Class 9 Maths Exercise 4.1 Solutions NCERT

📘 Concept & Theory Concept Used ›

A linear equation in two variables is an equation of the form:

\[ ax+by+c=0 \]

where \(x\) and \(y\) are variables and \(a\), \(b\), and \(c\) are constants.

In daily life situations, quantities like cost, distance, age, marks etc. can be represented using variables and converted into linear equations.

In this question:

Cost of notebook depends on cost of pen
“Twice” means multiplication by \(2\)
We first assign variables to unknown quantities

🗺️ Solution Roadmap Step-by-step Plan ›

Assume variables for notebook cost and pen cost
Translate the statement “notebook costs twice the pen”
Convert the relation into standard linear equation form

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Let the cost of a pen be \(\small y\) rupees.
Let the cost of a notebook be \(\small x\) rupees.
According to the question:
“The cost of a notebook is twice the cost of a pen.”
Therefore,
\[\small x=2y\]
Bringing all terms to one side:
\[\small \begin{align} x&=2y\\ x-2y&=0 \end{align} \]
Hence, the required linear equation in two variables is:
\[\small x-2y=0 \]

🎯 Exam Significance Exam Significance ›

Helps students learn how real-life statements are converted into algebraic equations.
Important for CBSE Board examinations because statement-to-equation conversion questions are frequently asked.
Builds foundation for Coordinate Geometry and Graph plotting in higher classes.
Useful for competitive examinations such as NTSE, Olympiads and Polytechnic entrance tests where algebraic modelling is tested.

NUMERIC3 marks

Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) \(2x + 3y = 9.3\bar{5}\)
(ii) \(x-\frac{y}{5}-10=0\)
(iii) \(–2x + 3y = 6\)
(iv) \(x = 3y\)
(v) \(2x = –5y\)
(vi) \(3x + 2 = 0\)
(vii) \(y – 2 = 0\)
(viii) \(5 = 2x\)

📘 Concept & Theory Concept Used ›

A linear equation in two variables can be written in the standard or general form:

\[ ax+by+c=0 \]

where:

\(a\) = coefficient of \(x\)
\(b\) = coefficient of \(y\)
\(c\) = constant term

To convert any equation into general form:

Bring all terms to one side.
Arrange terms in the order \(x\), \(y\), constant.
Compare with \[ ax+by+c=0 \] to identify \(a\), \(b\), and \(c\).

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given equation clearly.
Move all terms to one side so that RHS becomes \(0\).
Compare the obtained equation with \[ ax+by+c=0 \]
Identify the values of \(a\), \(b\), and \(c\).

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

(i) \[2x+3y=9.3\bar5\]
General form:
\[ax+by+c=0\]
Bringing all terms to the left side:
\[ \begin{align} 2x+3y &= 9.3\bar5\\ 2x+3y-9.3\bar5 &=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=2\\ b&=3\\ c&=-9.3\bar5 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=2,\quad b=3,\quad c=9.3\bar5\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(ii) \[x-\frac{y}{5}-10=0\]
To remove the fraction, multiply the complete equation by \(5\).
\[ \begin{align} 5\left(x-\frac{y}{5}-10\right)&=5\times0\\ 5x-y-50&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=5\\ b&=-1\\ c&=-50 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=5,\quad b=-1, \quad c=-50\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(iii) \[-2x+3y=6\]
Bringing all terms to one side:
\[ \begin{align} -2x+3y&=6\\ -2x+3y-6&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=-2\\ b&=3\\ c&=-6 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=-2,\quad b=3,\quad c=-6\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(iv) \[x=3y\]
Bringing all terms to one side:
\[ \begin{align} x&=3y\\ x-3y&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=1\\ b&=-3\\ c&=0 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=1,\quad b=-3,\quad c=0\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(v) \[2x=-5y\]
Bringing all terms to one side:
\[ \begin{align} 2x&=-5y\\ 2x+5y&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=2\\ b&=5\\ c&=0 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=2,\ b=5,\quad c=0\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(vi) \[3x+2=0\]
Writing coefficient of \(y\) explicitly:
\[ \begin{align} 3x+2&=0\\ 3x+0\cdot y+2&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=3\\ b&=0\\ c&=2 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=3,\ b=0,\quad c=2\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(vii) \[y-2=0\]
Writing coefficient of \(x\) explicitly:
\[ \begin{align} y-2&=0\\ 0\cdot x+y-2&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=0\\ b&=1\\ c&=-2 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=0,\ b=1,\quad c=-2\)

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(viii) \[5=2x\]
Bringing all terms to one side:
\[ \begin{align} 5&=2x\\ 5-2x&=0\\ -2x+5&=0\\ -2x+0\cdot y+5&=0 \end{align} \]
Comparing with \[ ax+by+c=0 \]
\[ \begin{aligned} a&=-2\\ b&=0\\ c&=5 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(a=-2,\ b=0,\quad c=5\)

🎯 Exam Significance Exam Significance ›

Questions based on identifying coefficients are frequently asked in CBSE Board examinations.
Understanding general form is essential for graph plotting in Coordinate Geometry.
Helps in solving higher-level algebra questions in NTSE, Olympiads and Polytechnic entrance examinations.
Builds strong conceptual understanding for equations of straight lines in higher mathematics.

LINEAR EQUATIONS IN TWO VARIABLES

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