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Explore comprehensive, step-by-step solutions for Class 9 Maths Chapter 4, Exercise 4.1, focusing on Linear Equations in Two Variables. This exercise is designed to strengthen students’ understanding of formulating and interpreting linear equations involving two variables. By engaging with these problems, learners will enhance their algebraic reasoning skills and prepare for more complex mathematical concepts in higher classes. Exercise 4.1 introduces students to the foundational aspects of linear equations in two variables, emphasizing the importance of expressing real-life situations mathematically. By defining variables for the costs and establishing a relationship between them, students learn to translate verbal statements into mathematical equations.
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Exercise 4.1
1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be Rs \(x\) and that of a pen to be Rs. \(y\))
(Take the cost of a notebook to be Rs \(x\) and that of a pen to be Rs. \(y\))
Answer
Let the cost of pen be \( y \) and the cost of notebook be \( x \)As per the question,
Cost of a notebook \( = \) Twice the cost of pen \( =2 \mathrm{y} \) So,
\(2 y=x\)
\(x-2 y=0\)
This is a linear equation in two variables to represent this statement.
2. Express the following linear equations in the form \(a x + by +c=0 \) and indicate the values of \( \mathrm{a}, \mathrm{b} \) and c in each case:
(i) \( 2 x+3 y=9 . \overline{35} \)
Answer
\( 2 x+3 y=9.3 \overline{5} \)\(2 x+3 y-9.3 \overline{5}=0\)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\(a=2\)
\(b=3\)
and \( {c}=-9.3 \overline{5} \)
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(ii) \(x-\frac{y}{5}-10=0 \)
Answer
\( {x}-\frac{y}{5}-10=0 \)Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\({a}=1,\)
\( b=\frac{-1}{5} \)
and \( {c}=-10 \)
(iii) \( -2 x+3 y=6 \)
Answer
\( -2 x+3 y=6 \)\( -2 {x}+3 {y}-6=0 \)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\( a=-2 \),
\( {b}=3 \)
and \( c=-6 \)
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(iv) \( x=3 y \)
Answer
\( x=3 y \)\( x-3 y=0 \)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\( {a}=1 \),
\( b=-3 \)
and \( {c}=0 \)
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(v) \( 2 x=-5 y \)
Answer
\( 2 x=-5 y \)\( 2 x+5 y=0 \)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\(a=2 \text {, }\)
\( b=5 \)
and \( {c}=0 \)
(vi) \( 3 x+2=0 \)
Answer
\( 3 x+2=0 \)\( 3 {x}+0 {y}+2=0 \)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\( a=3 \),
\( b=0 \)
And \( c=2 \)
(vii) \( y-2=0 \)
Answer
\( {y}-2=0 \)\( 0 {x}+{y}-2=0 \)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\( a=0 \),
\( {b}=1 \)
And \( c=-2 \)
(viii) \( 5=2 x \)
Answer
\( 5=2 x \)\(-2 x+0 y+5=0\)
Now,
On comparing this equation with \( {ax}+{by}+{c}=0 \)
We get,
\(a=-2\)
\(b=0\)
And \( c=5 \)
