CBSE Class 9 Maths Chapter 4 Linear Equations in Two Variables – Solutions 2025–26

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. \[\text{x}\] and that of a pen to be Rs. \[\text{y}\] ) 

Ans. Let the cost of a notebook be Rs. \[x\] and the cost of a pen be Rs. \[y\] .

We know that the cost of a notebook is twice the cost of a pen.

Therefore, we can write the required linear equation in the form:

\[Cost\text{ }of\text{ }notebook\text{ }=\text{ }2\text{ }\times \text{ }Cost\text{ }of\text{ }pen~~\]

\[x=2y~\]

\[x-2y~=0\] 

2. Express the following linear equations in the form \[\text{ax+by+c= 0}\] and indicate the values of \[\text{a}\] , \[\text{b}\] , \[\text{c}\] in each case: 

(i) \[\text{2x+3y=9}\text{.3}\overline{\text{5}}\]

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[2x+3y=9.3\overline{5}\]

\[2x+3y-9.3\overline{5}=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=2\]

\[b=3\]

\[c=-9.3\overline{5}\]

(ii) \[\text{x-}\frac{\text{y}}{\text{5}}\text{-10=0}\]

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[x-\frac{y}{5}-10=0\]

\[5x-y-50=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=5\]

\[b=-1\]

\[c=-50\]

(iii) \[\text{-2x+3y=6}\] 

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[-2x+3y=6\]

\[-2x+3y-6=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=-2\]

\[b=3\]

\[c=-6\]

(iv) \[\text{x=3y}\]

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[x=3y\]

\[x-3y=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=1\]

\[b=-3\]

\[c=0\]

(v) \[2x=-5y\] 

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[2x=-5y\]

\[2x+5y=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=2\]

\[b=5\]

\[c=0\]

(vi) \[\text{3x+2=0}\]

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[3x+2=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=3\]

\[b=0\]

\[c=2\]

(vii) \[\text{y-2=0}\] 

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[y-2=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=0\]

\[b=1\]

\[c=-2\]

(viii) \[\text{5=2x}\]

Ans. Writing the given equation in standard form \[\text{ax+by+c= 0}\] :

\[5=2x\]

\[-2x+5=0\]

Comparing this equation with standard form of the linear equation, \[\text{ax+by+c= 0}\] we have:

\[a=-2\]

\[b=0\]

\[c=5\]

Conclusion

NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables, Exercise 4.1, helps students master the basics of linear equations. Understanding the relationship between variables and practising these problems is crucial for building a strong foundation. Practising Exercise 4.1 improves speed, accuracy, and conceptual clarity, helping students excel in their exams. Solving this exercise with Vedantu's detailed solutions provides clear explanations, making learning easier and more effective.

Class 9 Maths Chapter 4: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 9 Maths

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