NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1

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. \[\text{x}\] 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:

\[a=-2\]

\[b=0\]

\[c=5\]

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables (Ex 4.1) Exercise 4.1

Opting for the NCERT solutions for Ex 4.1 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.1 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

CoolGyan in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 4 Exercise 4.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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