Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 4 - Linear Equation In Two Variables

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Page No 34:

Question 1:

Write the correct answer in each of the following :
The linear equation 2x – 5y = 7 has
(A) A unique solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solution

Answer:

$2 x - 5 y = 7$ is a linear equation in 2 variables.

Number of variables is two and number of equation is one and moreover, it is the equation of line which will pass through infinite points and all points will satisfy the given equation.

Hence, the correct answer is option C.

Page No 34:

Question 2:

Write the correct answer in each of the following :
The equation 2x + 5y = 7 has a unique solution, if x, y are :
(A) Natural numbers
(B) Positive real numbers
(C) Real numbers
(D) Rational numbers

Answer:

In the case of natural numbers, only one pair is there which satisfies the equation. The pair is (1, 1).
Therefore, if x and y are natural numbers, the given equation has a unique solution.

Hence, the correct answer is option A.

Page No 34:

Question 3:

Write the correct answer in each of the following :
If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is
(A) 4
(B) 6
(C) 5
(D) 2

Answer:

Since (2, 0) is a solution of the linear equation $2 x + 3 y = k$ ,
we can put $x = 2 and y = 0$
⇒ $2 (2) + 3 (0) = k$
⇒ $k = 4$

Hence, the correct answer is option A.

Page No 34:

Question 4:

Write the correct answer in each of the following :
Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form

(A) $(- \frac{9}{2}, m)$

(B) $(n, - \frac{9}{2})$

(C) $(0, - \frac{9}{2})$

(D) (– 9, 0)

Answer:

The given equation is $2 x + 0 y + 9 = 0$
$\Rightarrow 2 x = - 9 \Rightarrow x = \frac{- 9}{2}$

Thus, the solution of the equation is in the form of $(- \frac{9}{2}, m)$ .

Hence, the correct answer is option A.

Page No 34:

Question 5:

Write the correct answer in each of the following :
The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point
(A) (2, 0)
(B) (0, 3)
(C) (3, 0)
(D) (0, 2)

Answer:

The given equation is $2 x + 3 y = 6$ .
Since the graph cuts the y-axis , at that point value of $x = 0$ .
Put x = 0 into the equation $2 x + 3 y = 6$ ,
$\Rightarrow 2 (0) + 3 y = 6 \Rightarrow 3 y = 6 \Rightarrow y = 2$
The required point is (0, 2).

Hence, the correct answer is option D.

Page No 34:

Question 6:

Write the correct answer in each of the following :
The equation x = 7, in two variables, can be written as
(A) 1 · x + 1 · y = 7
(B) 1 · x + 0 · y = 7
(C) 0 · x + 1 · y = 7
(D) 0 · x + 0 · y = 7

Answer:

The given equation is $x = 7$ .
Compare the given equation with generalized linear equation in two variables $a x + b y = c$ .
a = 1 and b = 0
Now, the given equation can be written in the form of $1 \cdot x + 0 \cdot y = 7$ .

Hence, the correct answer is option B.

Page No 34:

Question 7:

Write the correct answer in each of the following :
Any point on the x-axis is of the form
(A) (x, y)
(B) (0, y)
(C) (x, 0)
(D) (x, x)

Answer:

Any on the x-axis has its y-coordinate equal to zero. i.e., y = 0.
The general form of every point on the x-axis is (x, 0).

Hence, the correct answer is option C.

Page No 34:

Question 8:

Write the correct answer in each of the following :
Any point on the line y = x is of the form
(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, –a)

Answer:

The given equation of line is y = x.

Option A, x = a and y = a
y = x
Equation satisfied.

Option B, x = 0 and y = a
Doesnot satisfy the equation.

Option C, x = a and y = 0
Doesnot satisfy the equation.

Option D, x = a and y = $-$ a
Doesnot satisfy the equation.

Hence, the correct answer is option A.

Page No 35:

Question 9:

Write the correct answer in each of the following:
The equation of x-axis is of the form
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y

Answer:

Since, the x-axis is a parallel to itself at a distance 0 from it.

Therefore, the equation of the x-axis is y = 0.

Hence, the correct answer is option B.

Page No 35:

Question 10:

Write the correct answer in each of the following :
The graph of y = 6 is a line
(A) parallel to x-axis at a distance 6 units from the origin
(B) parallel to y-axis at a distance 6 units from the origin
(C) making an intercept 6 on the x-axis.
(D) making an intercept 6 on both the axes.

Answer:

The given equation is y = 6.
Compare the above equation with the generalized equation of a line y = mx + c
m = 0, which means the line is parallel to the x-axis.
c = 6, which means it is at 6 units from the origin.

Hence, the correct answer is option A.

Page No 35:

Question 11:

Write the correct answer in each of the following :
x = 5, y = 2 is a solution of the linear equation
(A) x + 2y = 7
(B) 5x + 2y = 7
(C) x + y = 7
(D) 5x + y = 7

Answer:

If x = 5, y = 2 is a solution of the linear equation.

In option A, x + 2y
= 5 + 2(2)
= 5 + 4
= 9
≠ 7
Does not satifies the equation.

In option B, 5x + 2y
= 5(5) + 2(2)
= 29
≠ 7
Does not satifies the equation.

In option C, x + y
= 5 + 2
= 7
7 = 7
Satifies the equation.

In option D, 5x + y
= 5(2) + 2
= 12
≠ 7
Does not satifies the equation.

Hence, the correct answer is option C.

Page No 35:

Question 12:

Write the correct answer in each of the following :
If a linear equation has solutions (–2, 2), (0, 0) and (2, –2), then it is of the form
(A) y – x = 0
(B) x + y = 0
(C) –2x + y = 0
(D) –x + 2y = 0

Answer:

Let the linear equation be ax + by + c = 0 .....(1)
Thus as, (–2, 2), (0, 0) and (2, –2) are the solution of the linear equation. Therefore, it satisfies the equation (1).
At (–2, 2), the equation is –2a + 2b + c = 0 .....(2)
At (0, 0), the equation is c = 0 .....(3)
At (2, –2), the equation is 2a – 2b + c = 0 .....(4)

From (2) and (3), we get
c = 0 and a = b
On putting a = b and c = 0 in (1), we have
ax + ay = 0
⇒ x + y = 0

Thus, x + y = 0 is the required form of linear equation.

Hence, the correct answer is option B.

Page No 35:

Question 13:

Write the correct answer in each of the following :
The positive solutions of the equation ax + by + c = 0 always lie in the
(A) 1st quadrant
(B) 2nd quadrant
(C) 3rd quadrant
(D) 4th quadrant

Answer:

By solutions, we mean what values of x and y that will satisfy the given equation.
Now if x and y have to be positive, then they will lie in 1st quadrant.
In 1st quadrant, both x and y are positive.
Hence, the correct option is A.

Page No 35:

Question 14:

Write the correct answer in each of the following :
The graph of the linear equation 2x + 3y = 6 is a line which meets the x-axis at the point
(A) (0, 2)
(B) (2, 0)
(C) (3, 0)
(D) (0, 3)

Answer:

For the graph of the linear equation
2x + 3y = 6 to meet the x-axis, put y = 0.
∴ 2x + 3 × 0 = 6
⇒ 2x = 6
⇒ x = 3
Thus, the coordinates of x-axis are (3, 0).

Hence, the correct answer is option C.

Page No 35:

Question 15:

Write the correct answer in each of the following :
The graph of the linear equation y = x passes through the point

(A) $(\frac{3}{2}, \frac{- 3}{2})$

(B) $(0, \frac{3}{2})$

(C) (1, 1)

(D) $(\frac{- 1}{2}, \frac{1}{2})$

Answer:

The linear equation y = x has the same values for x-coordinates and y-coordinates.

Thus, out of all options only (1, 1) satisfies the equation y = x.

Hence, the correct answer is option C.

Page No 35:

Question 16:

Write the correct answer in each of the following :
If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation :
(A) Changes
(B) Remains the same
(C) Changes in case of multiplication only
(D) Changes in case of division only

Answer:

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equations remains unchanged.

Thus, the solution remains same.

Hence, the correct answer is option B.

Page No 35:

Question 17:

Write the correct answer in each of the following :
How many linear equations in x and y can be satisfied by x = 1 and y = 2?
(A) Only one
(B) Two
(C) Infinitely many
(D) Three

Answer:

Consider the linear equation as ax + by + c = 0
Put x = 1 and y = 2, we get
a + 2b + c = 0, where a, b and c are real numbers.
As different values of a, b and c can satisfy the equation above.

Therefore, infinitely many linear equations can be satisfied by x = 1 and y = 2.

Hence, the correct answer is option C.

Page No 35:

Question 18:

Write the correct answer in each of the following :
The point of the form (a, a) always lies on :
(A) x-axis
(B) y-axis
(C) On the line y = x
(D) On the line x + y = 0

Answer:

Since, the given point (a, a) has same values of x and y-coordinates.

Thus, the point (a, a) must lie on the line y = x.

Hence, the correct answer is option C.

Page No 36:

Question 19:

Write the correct answer in each of the following :
The point of the form (a, – a) always lies on the line
(A) x = a
(B) y = – a
(C) y = x
(D) x + y = 0

Answer:

x + y = 0, put (a, – a) in the equation.
a + (–a) = a – a = 0

Therefore, the point of the form (a, – a) always lies on the lines x + y = 0.

Hence, the correct answer is option D.

Page No 37:

Question 1:

State whether the following statement is True or False. Justify your answers :
The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12.

Answer:

True,
Put x = 0 and y = 3 in the L.H.S of the given equation, we get

L.H.S = 3(0) + 4(3) = 0 + 12 = 12 = R.H.S.

Hence, (0, 3) lies on the lines 3x + 4y = 12.

Page No 37:

Question 2:

State whether the following statement is True or False. Justify your answers :
The graph of the linear equation x + 2y = 7 passes through the point (0, 7).

Answer:

False,
Put x = 0 and y = 7 in L.H.S of the given equation, we get

L.H.S = 0 + 2(7) = 0 + 14 = 14 ≠ 7 = R.H.S

Hence, (0, 7) does not lie on the line x + 2y = 7.

Page No 37:

Question 3:

State whether the following statement is True or False. Justify your answers :
The graph given below represents the linear equation x + y = 0.

Answer:

True,
The given points (–1, 1) and (–3, 3) lie on the linear equation x + y = 0, then both points will satisfy the equation.

At point (–1, 1), we put x = –1 and y = 1 in L.H.S of given equation.
L.H.S = x + y = –1 + 1 = 0 = R.H.S

At point (–3, 3)
L.H.S = x + y = –3 + 3 = 0 = R.H.S

Hence, (–1, 1) and (–3, 3) both satisfy the given linear equation.

Page No 37:

Question 4:

State whether the following statement is True or False. Justify your answers :
The graph given below represents the linear equation x = 3.

Answer:

True,

As, the given graph is a line parallel to y-axis at a distance of 3 units to the right of the origin.

Thus, the given graph represents a linear equation of x = 3.

Page No 37:

Question 5:

State whether the following statement is True or False. Justify your answers :
The coordinates of points in the table:

x	0	1	2	3	4
y	2	3	4	–5	6

represent some of the solutions of the equation x – y + 2 = 0.

Answer:

False,
Given equation is x – y + 2 = 0 .....(1)

At (0, 2), L.H.S of equation (1) = 0 – 2 + 2 = 0 = R.H.S
At (1, 3), L.H.S of equation (1) = 1 – 3 + 2 = 0 = R.H.S
At (2, 4), L.H.S of equation (1) = 2 – 4 + 2 = 0 = R.H.S
At (3, –5), L.H.S of equation (1) = 3 – (–5) + 2 = 10 ≠ R.H.S
At (4, 6), L.H.S of equation (1) = 4 – 6 + 2 = 0 = R.H.S

As (3, –5), does not satisfy the equation.

Hence, it doesn't represent the solution of the given equation.

Page No 37:

Question 6:

State whether the following statement is True or False. Justify your answers :
Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.

Answer:

False,

As every point on the graph of the linear equation represents a solution.

Page No 37:

Question 7:

State whether the following statement is True or False. Justify your answers :
The graph of every linear equation in two variables need not be a line.

Answer:

False,

The graph of the linear equation in two variables always represents a line.

Page No 38:

Question 1:

Draw the graphs of linear equations y = x and y = – x on the same cartesian plane.
What do you observe?

Answer:

Given, y = x .....(1)
y = – x .....(2)

Consider (1), we get

x	1	2	4
y	1	2	4

Consider (2), we get

x	–4	0	3
y	4	0	–3

Putting these points on graph altogether.

It is observed that line y = x and y = – x intersect at point O(0, 0).

Page No 39:

Question 2:

Determine the point on the graph of the linear equation 2x + 5y = 19, whose ordinate is $1 \frac{1}{2}$ times its abscissa.

Answer:

Consider x to be abscissa and y to be the ordinate of the given line 2x + 5y = 19 .....(1)

Given,
Ordinate = $1 \frac{1}{2}$ × abscissa

$\Rightarrow y = 1 \frac{1}{2} x \Rightarrow y = \frac{3}{2} x . . . . . (2) Putting (2) in (1), we have 2 x + 5 \times \frac{3}{2} x = 19 \Rightarrow 2 \times 2 x + 5 \times 3 x = 19 \times 2 \Rightarrow 4 x + 15 x = 38 \Rightarrow 19 x = 38 \Rightarrow x = 2 Substituting the value of x = 2 in (2), we get \Rightarrow y = \frac{3}{2} \times 2 \Rightarrow y = 3$

Thus, the required point is (2, 3).

Page No 39:

Question 3:

Draw the graph of the equation represented by a straight line which is parallel to the x-axis and at a distance 3 units below it.

Answer:

All straight lines parallel to the x-axis and below in negative y-direction is given as y = –a, where a is the distance of the line from the X-axis.
Here, a is 3.
Therefore, the equation of the line is y = −3. To draw the graph of this equation, plot the points (1, −3), (2, −3) and (3, −3) and join them.

Hence, the above depicted is the required graph of y = –3.

Page No 39:

Question 4:

Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units.

Answer:

Let x and y be the two variables such that x + y = 10.
Thus,

x	0	5	7	10
y	10	5	3	0

Now, plotting the points (0, 10), (5, 5), (7, 3) and (10, 0) on the graph paper and joining them by a line, we get graph of the linear equation x + y = 10.

Hence, the line AB is the required graph of x + y = 10.

Page No 39:

Question 5:

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.

Answer:

Let x and y be the abscissa and ordinate.

Then, ordinate = 3 × abscissa
⇒ y = 3x

Thus,

x	1	2	3	4
y	3	6	9	12

Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

Page No 39:

Question 6:

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.

Answer:

As, the point (3, 4) lies on the line 3y = ax + 7 ...(1)
putting x = 3 and y = 4 in (1), we get

$\Rightarrow 3 \times 4 = a \times 3 + 7 \Rightarrow 12 = 3 a + 7 \Rightarrow a = \frac{12 - 7}{3} \Rightarrow a = \frac{5}{3}$

Hence, the value of a is $\frac{5}{3}$ .

Page No 39:

Question 7:

How many solution(s) of the equation 2x + 1 = x – 3 are there on the :
(i) Number line
(ii) Cartesian plane

Answer:

The given equation is 2x + 1 = x – 3
Simplifying the equation we have

$\Rightarrow 2 x - x + 1 + 3 = 0 \Rightarrow x + 4 = 0 \Rightarrow x = - 4$

(i) On the number line, x = –4 gives exactly one solution.

(ii) On Cartesian Plane, we have.

x	–4	–4	–4	–4
y	0	1	2	3

There will be infinitely many solutions to the equation x = –4 on a Cartesian Plane.

Page No 39:

Question 8:

Find the solution of the linear equation x + 2y = 8 which represents a point on
(i) x-axis
(ii) y-axis

Answer:

Given, x + 2y = 8 .....(1)

(i) When points is on x-axis, then put y = 0 in (1)
We have,
⇒ x + 2 × 0 = 8
⇒ x = 8

Hence, the required point is (8, 0).

(ii) When the point is on y-axis, then put x = 0, in (1)
We have,

$\Rightarrow 0 + 2 y = 8 \Rightarrow y = \frac{8}{2} \Rightarrow y = 4$

Hence, the required point is (0, 4).

Page No 39:

Question 9:

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution.

Answer:

Given, 2x + cy = 8 .....(1)

As per question, when x and y are same i.e.,
x = y [Putting this in (1)]
Then;

$\Rightarrow 2 x + c x = 8 \Rightarrow (2 + c) x = 8 \Rightarrow c = \frac{8}{x} - 2 \Rightarrow c = \frac{8 - 2 x}{x}$

Hence, the required value is $c = \frac{8 - 2 x}{x}$ .

Page No 39:

Question 10:

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?

Answer:

Given, y varies directly with x.

i.e., y $\propto$ x
⇒ y = kx .....(1)

When x = 4, y = 12, putting thin in (1)
12 = k × 4
$\Rightarrow k = \frac{12}{4} = 3 \Rightarrow k = 3$

∴ Equation becomes y = 3x .....(2)
Then, when x = 5, putting it in (2)
We have,
⇒ y = 3 × 5
⇒ y = 15

Hence, the value of y = 15.

Page No 41:

Question 1:

Show that the points A (1, 2), B (–1, –16) and C (0, –7) lie on the graph of the linear equation y = 9x – 7.

Answer:

Given, the linear equation is y = 9x – 7.

Then,

x	1	–1	2	0
y	2	–16	11	–7

As the equation satisfies the values of the given coordinates the points lie on the linear equation.

Graphically,

From the above graph, we can say that the points lie on line AB.

Page No 41:

Question 2:

The following observed values of x and y are thought to satisfy a linear equation.
Write the linear equation :

x	6	–6
y	–2	6

Draw the graph using the values of x, y as given in the above table.
At what points the graph of the linear equation
(i) cuts the x-axis
(ii) cuts the y-axis

Answer:

Given,

x	6	–6
y	–2	6

The points are A (6, –2) and B (–6, 6).

Consider the linear equation to be of the form y = mx + c.
Now, this equation should satisfy the points A and B.

∴ - 2 = 6 m + c . . . . . (1) And, 6 = - 6 m + c . . . . . (2) Adding (1) and (2), we get 4 = 2 c \Rightarrow c = 2 Putting this in y = m x + c we have y = m x + 2 Form, put x = 6 and y = - 2 ∴ - 2 = 6 m + 2 \Rightarrow - 4 = 6 m \Rightarrow m = \frac{- 2}{3} Thus, the equation becomes y = \frac{- 2}{3} x + 2 \Rightarrow 3 y + 2 x = 6 . . . . . (3)

(1) When the linear equation 2x + 3y = 6 cuts x-axis.
Put y = 0
⇒ 2x + 3 × 0 =6
⇒ 2x = 6
⇒ x = 3

(2) When the linear equation 2x + 3y = 6 cuts y-axis.
Put x = 0
⇒ 2 × 0 + 3y = 6
⇒ 3y = 6
⇒ y = 2

Therefore, the graph of the linear equation 2x + 3y = 6 cuts the x-axis at point (3, 0) and y-axis at point (0, 2).

Page No 41:

Question 3:

Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts the x-axis and the y-axis.

Answer:

Given, linear equation 3x + 4y = 6 .....(1)
$\Rightarrow 4 y = 6 - 3 x \Rightarrow y = \frac{6 - 3 x}{4} . . . . . (2)$
Thus,

x	0	2	1
y	$\frac{3}{2}$	0	$\frac{3}{4}$

∴ We take two points A

(0, \frac{3}{2})

and B (2, 0).
Graphically,

We observe that the line AB cuts x-axis at (2, 0) and y-axis at

(0, \frac{3}{2})

Page No 42:

Question 4:

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $C = \frac{5 F - 160}{9}$
(i) If the temperature is 86°F, what is the temperature in Celsius?
(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?
(iii) 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?
(iv) What is the numerical value of the temperature which is same in both the scales?

Answer:

Given, $C = \frac{5 F - 160}{9}$ .....(1)

$\Rightarrow 9 C - 5 F = - 160 \Rightarrow 9 C + 160 = 5 F \Rightarrow F = \frac{9 C + 160}{5} . . . . . (2)$

(i) As, F = 86
From (1), we get
$\Rightarrow C = \frac{5 \times 86 - 160}{9} \Rightarrow C = \frac{430 - 160}{9} \Rightarrow C = \frac{270}{9} \Rightarrow C = 30$

(ii) As, C = 35
From (2), we get
$\Rightarrow F = \frac{9 \times 35 + 160}{5} \Rightarrow F = \frac{315 + 160}{5} \Rightarrow F = \frac{475}{5} \Rightarrow F = 95$

(iii) As, C = 0
From (2), we get
$\Rightarrow F = \frac{9 \times 0 + 160}{5} \Rightarrow F = \frac{160}{5} \Rightarrow F = 32$

Also, F = 0
From (1), we have
$\Rightarrow C = \frac{5 \times 0 - 160}{9} \Rightarrow C = \frac{- 160}{9}$

(iv) As, C = F
From (1), we have
$C = \frac{5 C - 160}{9} \Rightarrow 9 C = 5 C - 160 \Rightarrow 4 C = - 160 \Rightarrow C = - 40$

As C = F, therefore, F = C = –40.

Thus the value of temperature is –40.

Page No 42:

Question 5:

If the temperature of a liquid can be measured in Kelvin units as x°K or in Fahrenheit units as y°F, the relation between the two systems of measurement of temperature is given by the linear equation $y = \frac{9}{5} (x - 273) + 32$
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313°K.

(ii) If the temperature is 158° F, then find the temperature in Kelvin.

Answer:

Given, $y = \frac{9}{5} (x - 273) + 32$ .....(1)
(i) As, x = 313
From (1), we get
$y = \frac{9}{5} (313 - 273) + 32 \Rightarrow y = \frac{9}{5} \times 40 + 32 \Rightarrow y = 72 + 32 \Rightarrow y = 104$

(ii) As, y = 158
From (1), we get
$∴ 158 = \frac{9}{5} (x - 273) + 32 \Rightarrow 790 = 9 (x - 273) + 160 \Rightarrow 9 (x - 273) = 630 \Rightarrow x - 273 = \frac{630}{9} = 70 \Rightarrow x = 70 + 273 \Rightarrow x = 343$

Page No 42:

Question 6:

The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is
(i) 5 m/sec²,
(ii) 6 m/sec².

Answer:

Given, the Force(F) is directly proportional to the acceleration(a).
F $\propto$ a
⇒ F = ma; where m is a constant mass
m = 6 kg
∴ F = 6a .....(1)

(i) If a = 5 m/sec²
Then From (1), we get
F = 6 × 5 = 30 N

(ii) If a = 6 m/sec²
Then From (1), we get
F = 6 × 6 = 36 N

Thus, the two points are A(5, 30) and B(6, 36).

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