NCERT Solutions Class 9 Maths Chapter 4- Linear Equation in Two variables

Linear Equation in Two variables
Exercise 4.1

Q1. 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 ₹ x and that of a pen to be ₹ y)

Answer:
Let’s say the price of a laptop is = x.
Let’s say the price of a pen is = y.
In response to the question,
A notebook is twice as expensive as a pen.
Therefore, we can write: Cost of a notebook =2 x Cost of a pen
x=2y
=>x-2y=0 The linear equation with two variables x-2y=0 represents the statement 'The price of a notebook is twice the price of a pen.'

Q2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
2x+3y=9.3\underline{5}
Answer:
2x+3y=9.3\underline{5}
2x+3y-9.3\underline{5}=0
2x+3y+\left(-9.3\underline{5}\right)=0
Finally on comparing the above equation with ax+by+c=0\ we will get,
a=2
b=3
c=9.3\underline{5}\

x\ –y5–10 = 0
Answer:
x\ –y5–10 = 0
=> 1x+\left(-\frac{1}{5}\right)y\ +(–10) = 0
Finally on comparing the above equation with ax+by+c=0\ we will get,
a=1
b=-\frac{1}{5}
c=-10

-2x+3y=6
Answer:
-2x+3y=6
=>-2x+3y-6=0
=> \left(-2\right)x+3y+\left(-6\right)=0
Finally on comparing the above equation with ax+by+c=0 we will get,
a=-2
b=-3
c=-6

x=3y
Answer:
x=3y
=> x-3y=0
=> x+(-3)y+(0)c=0
Finally on comparing the above equation with ax+by+c=0\ we will get,
a=1
b=-3
c=0

2x\ =-5y
Answer:
2x\ =-5y
=> 2x+5y=0
=> 2x+5y+0=0
Finally on comparing the above equation with ax+by+c=0\ we will get,
a=2
b=5
c=0

3x+2=0
Answer:
3x+2=0
=> 3x+0y+2=0
Finally on comparing the above equation with ax+by+c=0 we will get,
a=3
b=0
c=2

y-2=0
Answer:
y-2=0
=> 0x+y-2=0
=> 0x+1y-2=0
Finally on comparing the above equation with ax+by+c=0 we will get,
a=0
b=1
c=-2

5=2x
Answer:
5=2x
=> 2x-5=0
=> 2x+0y-5=0
Finally on comparing the above equation with ax+by+c=0 we will get,
a=2
b=0
c=-5

Exercise 4.2

Q1. Which one of the following options is true, and why?
y\ =\ 3x+5 has
A unique solution
Only two solutions
Infinitely many solutions
Answer: In the linear equation y = 3x+5, we will try substituting different values for x.

It is evident from the table that x could have infinite values, and that there exist infinite y values for all infinite x values.
Therefore, option 3. Infinitely many solutions is the correct answer.

Q2. Write four solutions for each of the following equations:
2x+y=\ 7
Answer:
We will be substituting different values for x and y to get the four values satisfying 2x+y\ =7 .
At x=0:
2\left(0\right)+y=7
y=7
(0,7)
At x=1:
2\left(1\right)+y=7
y=\frac{7}{2}
\left(1,\frac{7}{2}\ \right)
At y=0
2x+0=7
x=\frac{7}{2}
\left(\frac{7}{2},0\right)
At y=1:
2x+1=7
2x=6
x=3
\left(3,1\right)
The answers will be: \left(0,7\right),\ \left(1,\frac{7}{2}\right),\left(\frac{7}{2},0\right)\ and\ \left(3,1\right)
\pi x+y=9
Answer:
At x=0:
\pi\left(0\right)+y=9
y=9
\left(0,9\right)
At x=1:
\pi\left(1\right)+y=9
y=\frac{9}{\pi}
\left(1,\frac{9}{\pi}\right)
At y=0:
\pi x+0=9
x=\frac{9}{\pi}
\left(\frac{9}{\pi},0\right)
At y=1:
\pi x+1=9
x=\frac{8}{\pi}
\left(\frac{8}{\pi},1\right)
The answers will be: \left(0,9\right),\ \left(1,\frac{9}{\pi}\right),\ \left(\frac{9}{\pi},0\right)\ and\ \left(\frac{8}{\pi},1\right)

x=4y
Answer:
At x=0:
4y=0
y=0
\left(0,0\right)
At x=1:
1=4y
y=\frac{1}{4}
\left(1,\frac{1}{4}\right)
At y=1:
x=4(1)
x=4
\left(4,1\right)
At y=2:
x=4(2)
x=8
\left(8,2\right)
The answers will be: \left(0,0\right),\ \left(1,\frac{1}{4}\right),\ \left(4,1\right)\ and\ \left(8,2\right)

Q3. Check which of the following are solutions of the equation x–2y = 4 and which are not:
\left(0,2\right)
Answer:
To check whether the given values of x and y are the solutions, we will substitute these values in the equation.
x–2y = 4
\Longrightarrow\ 0\ – (2×2) = 4
But, \ -4\ \neq\ 4
Therefore, (0,2) is not the solution of the equation.

(2,0)
Answer:
To check whether the given values of x and y are the solutions, we will substitute these values in the equation.
x\ -2y\ =\ 4
\Longrightarrow\ 2-(2\times0)\ =\ 4
\Longrightarrow\ 2\ -0\ =\ 4
But, \ 2\ \neq\ 4
Therefore, (2,0) is not the solution of the equation.

(4,0)
Answer:
To check whether the given values of x and y are the solutions, we will substitute these values in the equation.
x–2y = 4
\Longrightarrow\ 4\ – 2×0 = 4
\Longrightarrow\ 4-0\ =\ 4
\Longrightarrow\ 4\ =\ 4
Therefore, (4,0) is the solution of the equation.

\left(\sqrt2,\ 4\sqrt2\right)
Answer:
To check whether the given values of x and y are the solutions, we will substitute these values in the equation.
x\ –2y = 4
\Longrightarrow\ \sqrt2-(2\times4\sqrt2)\ =\ 4
\sqrt2-8\sqrt2\ =\ 4
But, \ -7\sqrt2\ \neq\ 4
Therefore, \left(\sqrt2,\ 4\sqrt2\right) is not the solution of the given equation.

(1,1)
Answer:
To check whether the given values of x and y are the solutions, we will substitute these values in the equation.
x\ –2y = 4
\Longrightarrow\ 1\ -(2\times1)\ =\ 4
\Longrightarrow\ 1-2\ =\ 4
But, \ -1\ \neq\ 4
Therefore, (1,1) is not the solution of the given equation.

Q4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.
Answer:
The given formula is 2x+3y = k
x = 2 and y = 1 according to the question.
Now, in the equation 2x+3y = k, substitute the values of x and y.
We will be getting,
(2\times2)+(3\times1)\ =\ k
\Longrightarrow\ 4+3\ =\ k
\Longrightarrow\ 7\ =\ k
k\ =\ 7
Therefore, the value of k will be 7.

Exercise 4.3

Q1. Draw the graph of each of the following linear equations in two variables:
x+y=4
Answer:
Let us first determine the points to plot on a graph of linear equations in two variables. To determine the points, we must determine the possible values for x and y that meet the equation.
At x=0:
y=4
\left(0,4\right)
At y=0:
x=4
(4,0)

x-y=2
Answer:
Let us first determine the points to plot on a graph of linear equations in two variables. To determine the points, we must determine the possible values for x and y that meet the equation.
At x=0:
y=-2
(0,-2)
At y=0:
x=2
So, the points are: (0,-2) and (2,0)

y=3x
Answer:
Let us first determine the points to plot on a graph of linear equations in two variables. To determine the points, we must determine the possible values for x and y that meet the equation.
At x=0:
y=0
(0,0)
At x=1:
y=3
(1,3)
So, the points are: (0,0) and (1,3).

3=2x+y
Answer:
Let us first determine the points to plot on a graph of linear equations in two variables. To determine the points, we must determine the possible values for x and y that meet the equation.
At x=0:
y=3
(0,3)
At y=0:
x=3/2
(3/2, 0)
So, the points are: (0,3) and (3/2, 0)

Q2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Answer:
We know that a point can have an endless number of lines passing through it.
The point should be satisfied by the equation of two lines going through (2,14).
Let us assume that 7x = y is the equation.
7x–y = 0
When y = 14 and x = 2
(7×2)-14 = 0
14–14 = 0
0 Equals 0
L.H.S = R.H.S
4x = y-6 is another equation to consider.
4x-y+6 = 0
When x equals 2 and y equals 14,
(4×2)–14+6 = 0
8–14+6 = 0
0 Equals 0
L.H.S = R.H.S
Because both equations satisfy the point (2, 14), the equations of two lines passing through it are 7x = y and 4x = y-6.

Q3. If the point (3, 4) lies on the graph of the equation 3y = ax+7, find the value of a.
Answer: The provided equation is:
3y = ax+7
x = 3 and y = 4 are the answers to the question.
Now, in the given equation 3y = ax+7, substitute the values of x and y.
We will get,
(3×4) = (a×3)+7
12 = 3a+7
3a = 12–7
3a = 5
a = 5/3
If the point (3,4) is on the graph of the equation 3y = ax+7, the value of a will be 5/3.

Q4. The taxi fare in a city is as follows: For the first kilometer, the fare is ₹8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
Answer: Given,
x = total distance travelled
y = total fare
The first kilometer costs ₹8 per kilometer.
After the first 1km, the fare is ₹5 per km.
The distance after one kilometer = (x-1) km if x is the total distance.
Fare after the first kilometer = 5 (x-1)
In response to the question,
Total fare = first-kilometer fare + second-kilometer fare
y = 8+5(x-1)
y = 8+5(x-1)
y = 8+5x – 5
y = 5x+3
we will be solving the equation,
When the value of x = 0,
y = 5x+3
y = 5×0+3
y = 3
(0,3)
When the value of y = 0,
y = 5x+3
0 = 5x+3
5x = -3
x = -3/5
(-3/5, 0)

Q5. From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
For Fig. 4. 6
(i) y = x
(ii) x+y = 0
(iii) y = 2x
(iv) 2+3y = 7x Answer: The points in the given figure 4.6 are (0,0), (-1,1), and (-1,2).
The points in the given figure 4.6 are (0,0), (-1,1), and (-1,2).
On substituting the given values in the equations, we will get:
(i) y = x
(0,0) ⟹ 0 = 0
(-1, 1) ⟹ -1 ≠ 1 -equation is not satisfied
(1, -1) ⟹ 1≠ -1 -equation is not satisfied

(ii) x+y = 0
(0,0) ⟹ 0+0 = 0
(-1, 1) ⟹ -1+1 = 0
(1, -1) ⟹ 1+(-1) =0

(iii) y = 2x
(0,0)
⟹ 0 = 2×0
0 = 0
(-1, 1)
⟹ 1 = 2×(-1)
1≠ -2 ————————— equation is not satisfied
(1, -1) ⟹ -1 = 2×1
-1 ≠ 2 ————————— equation is not satisfied

(iv) 2+3y = 7x
(0,0) ⟹ 2+(30) = 7×0
2 ≠ 0 ————————— equation is not satisfied
(-1, 1) ⟹ 2+(3×1) = 7×-1
5 ≠ -7 ————————— equation is not satisfied
(1, -1) ⟹ 2+(3×-1) = 7×1
-1 ≠ 7 ————————— equation is not satisfied
Because only the equation x+y = 0 satisfies all of the points, the equation whose graphs are shown in Fig. 4.6 is the equation.
x+y = 0
For Fig. 4. 7
(i) y = x+2
(ii) y = x–2
(iii) y = –x+2
(iv) x+2y = 6

The points in the given figure 4.7 are: (0,2), (2,0) and (-1,3)
On substituting the given values of x and y in the equations, we will get:
(i)y=x+2
(0,2) ⟹2 = 0+2
2 = 2
(2, 0) ⟹ 0= 2+2
0 ≠ 4 -equation is not satisfied
(-1, 3) ⟹ 3 = -1+2
3 ≠ 1 -equation is not satisfied
(ii) y = x–2
(0,2) ⟹ 2 = 0–2
2 ≠ -2 -equation is not satisfied
(2, 0) ⟹ 0 = 2–2
0= 0
(-1, 3) ⟹ 3= –1–2
3 ≠ –3 – equation is not satisfied
(iii) y = –x+2
(0,2) ⟹ 2 = -0+2
2 = 2
(2, 0) ⟹ 0 = -2+2
0 = 0
(-1, 3) ⟹ 3= -(-1)+2
3 = 3
(iv) x+2y = 6

(0,2) ⟹ 0+(2×2) = 6
4 ≠ 6 -equation is not satisfied
(2, 0) ⟹ 2+(2×0) = 6
2 ≠ 6 ————————— equation is not satisfied
(-1, 3) ⟹ -1+(2×3) = 6
5 ≠ 6 -equation is not satisfied
Because only the equation y = –x+2satisfies all of the points, the equation whose graphs are shown in Fig. 4.7 is the only one that may be used.
y = –x+2

Q6. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 unit
Answer: Let x be the distance travelled by the body and y be the force applied to it.
It is mentioned that the amount of work a body does is related to the distance it travels.
In response to questions, we can write
y\ \propto\ x
y = 5x (constant of proportionality will be 5)
in solving this equation
(i) If x equals 2 units,
y = 5x2 = 10 units, then the point will be
(2, 10)
(ii) If x equals 0 units,
y = 0 units if y = 0 units.
(0, 0)
The plotting points will be (2, 10) and (0, 0)

Q7. Yamini and Fatima, two students of Class IX of a school, together contributed ₹ 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as ₹ x and ₹ y.) Draw the graph of the same.
Answer:
Let x be Yamini's donation and y be Fatima's donation.
In response to the question;
x+y = 100
We're aware of this.
when the value of x = 0 , y = 100
when the value of x = 50, y = 50
when the value of x = 100, y = 0
The points to plot are (0,100), (50,50), and (100,0).

Q8. In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F=\left(\frac{9}{5}\right)C+32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer:
(i) According to the given question
F\ =\ \left(\frac{9}{5}\right)C\ +\ 32
solving the problem
We will be getting,
When the value of C = 0, F = 32
When the value of C = -10 , F = 14
(0, 32) and (-10, 14) are the points to be plotted.

(ii)When the value of C will be 30,
F\ =\ (9/5)C\ +32
F\ =\ (9\times30)/5+32
=\ (9\times6)+32
=\ 54+32
=\ {86}^oF

(iii)When the value of F is 95,
95\ =\ (9/5)C\ +32
(9/5)C\ =\ 95-32
(9/5)C\ =63
C\ =\ (63\times5)/9
={35}^oC
(iv)When the value of C is 0,
F\ =\ (9/5)C\ +32
F\ =\ (9\times0)/5\ +32
=0+32
={32}^oF

(iv)When the value of F is 0,
0\ =\ (9/5)C+32
(9/5)C\ =\ 0-32
(9/5)C\ =\ -32
C\ =\ (-32\times5)/9
=-17.7777
=-{17.8}^oC

(v)When the value of F is C
C\ =\ (9/5)C+32
C\ – (9/5)C = 32
\frac{\left(5-9\right)C}{5}\ =32
\left(-\frac{4}{5}\right)C\ =\ 32
\left(-\frac{4}{5}\right)C\ =\frac{\left(-32\times5\right)}{4}
=\ – 40oC
As a result, -40o is the temperature that has the same numerical value in Fahrenheit and Celsius.

Exercise 4.4

Q1. Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Answer:
(i)

(ii)

Q2. Give the geometric representations of 2x+9 = 0 as an equation
(i) in one variable
(ii) in two variables
Answer:
In one variable
2x+9\ =\ 0
2x\ =\ -9
x\ =\ -9/2
x\ =\ -4.5

(ii)
2x+9\ =\ 0 2x+0y+9\ =\ 0 When the value of \ y\ =\ 0,\ x\ =\ -4.5
When the value of y\ =\ 1,\ x\ =\ -4.5