NCERT Solutions for Maths Class 9 Chapter 4

Exercise 4.1

Question 1: The cost of a notebook is twise cost of a pen. Write a linear equation in two variable to represent this statement. (Take the cost of a notebook to be Rs. \(x\) and that of a pen to be Rs. \(y\).)

Answer: 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 iss twise the cost of a pen.

Therefore, we can write the required linear equationon theform:

Cost of notebook \(=2\times\) cost of a pen

\(x=2y\)

\(x-2y=0\)

Question 2: Express the following linear equations in the form \(ax+by+c=0\) and indicate the valus of \(a, b, c\) in each case.

(i) \(2x+3y=9.\bar{35}\)

Answer: Writing the given equation in standard form \(ax+by+x=0\)

\(2x+3y=9.\bar{35}\)

\(2x+3y-9.\bar{35}=0\)

Comparing this equation with standard form of the linear equation, \(ax+by+c=0\) we have:

\(a=2\)

\(b=3\)

\(c=-9\bar{35}\)

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

Answer: Writing the given equation in standard form \(ax-by+c=0\)

\(x-\frac{y}{5}-1=0\)

\(5x-y-50=0\)

Comparing this equation with standard form of the equation, \(ax+by+x=0\) we have:

\(a=5\)

\(b=-1\)

\(c=-50\)

(iii) \(-2+3y=6\)

Answer: Writing the givn equation in standard form \(ax+by+x=0\)

\(-2x+3y=6\)

\(-2x+3y-6=0\)

Comparing this equation with standard form of the linear equation, \(ax+by+c=0\) we have

\(a=-2\)

\(b=3\)

\(c=-6\)

(iv) \(x=3y\)

Answer: Writing th given quation in standard form \(ax+by+c=0\):

\(x=3y\)

\(x-3y=0\)

Comparing the equation with standard form of the linear equation, \(ax+by+c=0\)  we have

\(a=1\)

\(b=-3\)

\(c=0\)

(v) \(2x=-5y\)

Answer: Writing the given equation in standard form \(ax+by+c=0\)

\(2x=-5y\)

\(2x+5y=0\)

Comparing this equation with standard form of the linear equation, \(ax+by+c=0\) we have:

\(a=2\)

\(b=5\)

\(c=0\)

(vi) \(3x+2=0\)

Answer Writing the given equation in standard form \(ax+by+c=0\)

\(3x+2=0\)

Comparing tis equation with standard form of the linear equation, \(ax+by+c=0\) we have:

\(a=3\)

\(b=0\)

\(c=2\)

(vii) \(y-2=0\)

Answe: Writing the given equation in standard form \(ax+by+c0\):

\(y-2=0\)

Comparing this equation with standard form of the linear equation, \(ax+by+c=0\) we have:

\(a=0\)

\(b=1\)

\9c=-2\)

(viii) \(5=2x\)

Answer: Writing the given equation in standad form \(ax+by+c=0\)

\(5=2x\)

\(2x+5=0\)

Comparing this equaton tith standard form of the linear equation, \(ax+by+c=0\) we have:

\(a=-2\)

\(b=0\)

\(c=5\)

Exercise 4.2

Question 1: Which one of the following options is true, and why? \(y=3x+5\) has
(i) A unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
Answer: We are given that \(y=3x+5\) is a linear equation.

● For \(x=0,\space y=5\). Therefore, (0,5) is a solution of the equation.
● For \(x=1,\space y=8\). Therefore (1,8) is another solution of the equation.
● For \(x=2,\space y=11\) . 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.

Answer: Write four solutions for each of the following equatons

(i) \(2x+y=7\)

Answer: Given equation \(2x+y=7\), can be written as,

\(y=7-2x\)

Let us now take different value 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 \92x+y=7\) are \((0,7) , (1,5) , (2,3) , (3,1)\)

ii. \(\pi x+y=9\)

Answer: Given equation \(\pi x+y=9\), can written as, \(y=9-\pi x\)

Let us now take different value 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, \((1,9-π\)  is a solution.
● For \(x=2\),  \(y=9-2\pi\)
So, \((2,9-2π\)  is a solution.
● For \(x=3\),  \(y=9-3\pi\)

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

Therefore, the four solution of \(\pi x+y=9\) are \((0.9)\), \((1.9-\pi)\), \((2.0-2\pi)\),  \(93.9-3\pi)\)

(iii) \(x=4y\)

Answer: Let us now take different values of \(y\) and substitute in the above equation-

For \(y=0\),  \9x=0\)
So, \((0,0)\) is a solution.
● For \(y=1\),  \9x=4\)
So, \((4,1)\) is a solution.
● For \9y=2\),  \(x=8\)
So, \((8,2)\) is a solution.
● For \(y=3\),  \9x=1\)
So, \((12,3)\) is a solution.

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

Qustion 3:  Check which of the following are solutions of the equation \(x-2y=4\) and which are not:

(i) \((0,2)\)

Answer: Substituting \(x=0\) and \(y=2\) in the L.H.S of the given equation \(x-2\y=4\):

\(⇒0-2(2)\)

\(⇒-4\)

Since L.H.S.R.H.S.≠ , therefore (0,2) is not a solution of the equation.

(iii) \((4,0)\)

Answer: Substituting \(x=4\) and \(y=0\) in th L.H.S of th equation \(x-2y=4\):

\(⇒4-2(0)\)

\(⇒4\)

Since L.H.S=R.H.S, therefore \((4.0)\) is a solution of the equation.

(iv) \((\sqrt2,\space4\sqrt2)\)

Answer: Substituting \(x=\sqrt2\) and \(y=4\sqrt2 \) in the L.H.S of the equation \(x-2y=4\)

\(⇒\sqrt2-2(4\sqrt2)\)

\(⇒-7\sqrt2\)

Since L.H.S.≠R.H.S, therefore \((\sqrt2,\space 4\sqrt2)\) is not a solution of the equation.

(v) \((1,1)\)

Answer: Substituting \(x=1\) and \(y=1\) in the L.H.S of the given equation \(x-2y=4\):

\(⇒1-2(1)\)

\(⇒-1\)

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

Question 4: Find the value of \(K\), if \(x=2,\space y=1\) is a solution of the equation \92x+3y=k\).

Answer: We are given the equation \(2x+3y=k\) along with the value \(x=2\) and \(y=1\) substituting the given values if the L.H.S of the equation:

\(⇒292)+3(1)=k\)

\(⇒4+3=k\)

\(⇒k=7\)

Terefore, we get \(k=7\) on solving equation.