NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Ex 4.2

Exercise 4.2

1. Which one of the Following Options is True, and Why? \[\text{y=3x+5}\] has  

(i) A unique solution, 

(ii) only two solutions, 

(iii) infinitely many solutions

Ans. We are given that \[y=3x+5\] is a linear equation. 

• For \[x=0\] , \[y=5\] . Therefore, \[(0,5)\] is a solution of the equation. 

• For \[x=1\] , \[y=8\] . Therefore \[(1,8)\] is another solution of the equation. 

• For \[x=2\] , \[y=11\] . Therefore \[(2,11)\] is another solution of the equation. 

Clearly, for different values of \[x\] , we get another distinct value of \[y\] .  

So, there is no end to different solutions of a linear equation in two variables. Therefore, a linear equation in two variables has infinitely many solutions.

Hence (iii) is the correct answer.  

2. Write Four Solutions for Each of the Following Equations:

(i) \[\text{2x+y=7}\]

Ans. Given equation \[2x+y=7\] , can be written as,

\[y=7-2x\]

Let us now take different values of \[x\] and substitute in the above equation-

• For \[x=0\] ,

\[y=7\]

So, \[(0,7)\] is a solution.

• For \[x=1\] ,

\[y=5\]

So, \[(1,5)\] is a solution.

• For \[x=2\] ,

\[y=3\]

So, \[(2,3)\] is a solution.

• For \[x=3\] ,

\[y=1\]

So, \[(3,1)\] is a solution.

Therefore, the four solutions of \[2x+y=7\] are \[(0,7)\] , \[(1,5)\] , \[(2,3)\] , \[(3,1)\] .

(ii) \[\pi \text{x+y=9}\]

Ans. Given equation \[\pi x+y=9\] , can be written as,

\[y=9-\pi x\]

Let us now take different values of \[x\] and substitute in the above equation-

• For \[x=0\] ,

\[y=9\]

So, \[(0,9)\] is a solution.

• For \[x=1\] ,

\[y=9-\pi \]

So, \[(1,9-\pi )\] is a solution.

• For \[x=2\] ,

\[y=9-2\pi \]

So, \[(2,9-2\pi )\] is a solution.

• For \[x=3\] ,

\[y=9-3\pi \]

So, \[(3,9-3\pi )\] is a solution.

Therefore, the four solutions of \[\pi x+y=9\] are \[(0,9)\] , \[(1,9-\pi )\] , \[(2,9-2\pi )\] , \[(3,9-3\pi )\] .

(iii) \[\text{x=4y}\]

Ans. Given equation \[x=4y\] .

Let us now take different values of \[y\] and substitute in the above equation-

• For \[y=0\] ,

\[x=0\]

So, \[(0,0)\] is a solution.

• For \[y=1\] ,

\[x=4\]

So, \[(4,1)\] is a solution.

• For \[y=2\] ,

\[x=8\]

So, \[(8,2)\] is a solution.

• For \[y=3\] ,

\[x=12\]

So, \[(12,3)\] is a solution.

Therefore, the four solutions of \[x=4y\] are \[(0,0)\] , \[(4,1)\] , \[(8,2)\] , \[(12,3)\].

3. Check Which of the Following are Solutions of the Equation \[x-2y=4\] and Which are not:

(i) \[(0,2)\]

Ans. Substituting \[x=0\] and \[y=2\] in the L.H.S. of the given equation \[x-2y=4\] :

\[\Rightarrow 0-2(2)\]

\[\Rightarrow -4\]

Since \[L.H.S.\ne R.H.S.\] , therefore \[(0,2)\] is not a solution of the equation.

(ii) \[(2,0)\]

Ans. Substituting \[x=2\] and \[y=0\] in the L.H.S. of the given equation \[x-2y=4\] :

\[\Rightarrow 2-2(0)\]

\[\Rightarrow 2\]

Since \[L.H.S.\ne R.H.S.\] , therefore \[(2,0)\] is not a solution of the equation.

(iii) \[(4,0)\]

Ans. Substituting \[x=4\] and \[y=0\] in the L.H.S. of the given equation \[x-2y=4\] :

\[\Rightarrow 4-2(0)\]

\[\Rightarrow 4\]

Since \[L.H.S.=R.H.S.\] , therefore \[(4,0)\] is a solution of the equation.

(iv) \[(\sqrt{2},4\sqrt{2})\]

Ans. Substituting \[x=\sqrt{2}\] and \[y=4\sqrt{2}\] in the L.H.S. of the given equation \[x-2y=4\] :

\[\Rightarrow \sqrt{2}-2(4\sqrt{2})\]

\[\Rightarrow -7\sqrt{2}\]

Since \[L.H.S.\ne R.H.S.\] , therefore \[(\sqrt{2},4\sqrt{2})\] is not a solution of the equation.

(v) \[(1,1)\]

Ans. Substituting \[x=1\] and \[y=1\] in the L.H.S. of the given equation \[x-2y=4\] :

\[\Rightarrow 1-2(1)\]

\[\Rightarrow -1\]

Since \[L.H.S.\ne R.H.S.\] , therefore \[(1,1)\] is not a solution of the equation.

4. Find the value of \[k\] , if \[x=2\] , \[y=1\] is a solution of the equation \[2x+3y=k\] .

Ans. We are given the equation \[2x+3y=k\] along with the values \[x=2\] and \[y=1\] .

Substituting the given values in the L.H.S. of the equation:

\[\Rightarrow 2(2)+3(1)=k\]

\[\Rightarrow 4+3=k\]

\[\Rightarrow k=7\]

Therefore, we get \[k=7\] on solving the equation.

Conclusion

NCERT of Class 9 maths exercise 4.2 focuses on Linear Equations in Two Variables, primarily dealing with understanding and solving linear equations by finding pairs of values that satisfy the given equations. It emphasizes graphing solutions and interpreting the relationships between variables. By practicing these problems, students learn to visualize equations and improve their problem-solving skills, preparing them for more complex algebraic concepts in Class 9th maths chapter 4 exercise 4.2. The solutions provided by Vedantu offer step-by-step explanations and can be a valuable resource for students aiming to clear their doubts and perform well in exams.

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