Rd Sharma 2022 for Class 9 Maths Chapter 7 - Linear Equations In Two Variables

Answer:

(i) We are given

$-2x + 3y - 12 = 0$

Comparing the given equation with ,we get

$\boxed{a=-2; b = 3; c = -12}$

(ii) We are given

Comparing the given equation with ,we get

(iii) We are given

Comparing the given equation with ,we get

(iv) We are given

Comparing the given equation with ,we get

(v) We are given

Comparing the given equation with ,we get

(vi) We are given

Comparing the given equation with ,we get

(vii) We are given

Comparing the given equation with ,we get

(viii) We are given

Comparing the given equation with ,we get

Answer:

(i) We are given,

Now, in two variable forms the given equation will be

(ii) We are given,

Now, in two variable forms the given equation will be

(iii) We are given,

Now, in two variable forms the given equation will be

(iv) We are given,

Now, in two variable forms the given equation will be

Answer:

Let the cost of fountain pen be x and cost of ball pen be y.

According to the given equation, we have

Here x is the cost of one fountain pen and y is that of one ball pen.

Answer:

(i) We are given,

Substituting x = 1 in the given equation, we get

Substituting x = 2 in the given equation, we get

(ii) We are given,

Substituting in the given equation, we get

Substituting in the given equation, we get

(iii) We are given,

Substituting x = 0 in the given equation, we get

Substituting x = 4 in the given equation, we get

(iv) We are given,

Substituting x = 0 in the given equation, we get

Substituting x = 3 in the given equation, we get

Answer:

We are given,

2x – y = 6

(i) In the equation 2x – y = 6,we have

Substituting x = 3 and y = 0 in 2x – y, we get

is the solution of 2x – y = 6.
 

(ii) In the equation 2x – y = 6, we have

Substituting x = 0 and y = 6 in 2x – y,we get

is not the solution of 2x – y = 6.
 

(iii) In the equation 2x – y = 6,we have

Substituting x = 2 and y = –2  in 2x – y, we get

is the solution of 2x – y = 6.
 

(iv) In the equation 2x – y = 6, we have

Substituting and y = 0 in 2x – y, we get

is not the solution of 2x – y = 6.
 

(v) In the equation 2x – y = 6, we have

Substituting and y = –5 in 2x – y, we get

is the solution of 2x – y = 6.

Answer:

We are given,

is the solution of equation .

Substituting and in ,we get

Answer:

We are given,

is the solution of equation .

Substituting and in ,we get

Answer:

We are given,

is the solution of equation .

Substituting and in ,we get

$$\begin{gathered} 2\times \left(2a+1\right)-3\times \left(a-1\right)+5=0 \\ \Rightarrow 4a+2-3a+3+5=0 \\ \Rightarrow a+10=0 \\ \Rightarrow a=-10 \left(\mathrm{answer}\right) \end{gathered}$$

Answer:

We are given,

is the solution of equation .

Substituting and in ,we get

Using quadratic factorization

Answer:

(i) We are given,

Substituting x = 0 in the given equation, we get

Substituting y = 0 in the given equation, we get

(ii) We are given,

Substituting in the given equation, we get
$$\begin{gathered} -4\times 0+3y=12 \\ 3y=12 \\ y=4 \end{gathered}$$
Thus x = 0 and y = 4 is the solution of the −4x + 3y = 12

Substituting y = 0 in the given equation, we get

(iii) We are given,

Substituting x = 0 in the given equation, we get

Substituting y = 0 in the given equation, we get

Answer:

Given, 2x + cy = 8      .....(1)

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

$$\begin{gathered} \Rightarrow 2x+cx=8 \\ \Rightarrow cx=8-2x \\ \Rightarrow c=\frac{8-2x}{x} \end{gathered}$$

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

Answer:

Given, y varies directly with x.

i.e., y $\propto$ x 
⇒ y = kx              .....(1)

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

∴ 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.

Answer:

Let x and y be the abscissa and ordinate respectively.

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

Thus,
 

Answer:

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

Hence, the temperature of the liquid is 104° F, if the temperature of the liquid is 313°K.

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

Hence, if the temperature is 158°F, then find the temperature in Kelvin is 343° K.

Answer:

(i) We are given,

x + y = 4

We get,

y = 4 – x,

Now, substituting x = 0 in y = 4 – x, we get

y = 4

Substituting x = 4 in y = 4 – x, we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given

(ii) We are given,

We get,

Now, substituting x = 0 in y = x – 2, we get

Substituting x = 2 in y = x – 2, we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

(iii) We are given,

We get,

Now, substituting in ,we get

Substituting x = –6 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

(iv) We are given,

Now, substituting x = 1 in ,we get

Substituting x = 3 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

(v) We are given,

We get,

Now, substituting x = 0 in ,we get

Substituting x = 5 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

(vi) We are given,

We get,

Now, substituting x = 0 in ,we get

Substituting x = 4 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

(vii) We are given,

We get,

Now, substituting x = 5 in ,we get

Substituting x = 8 in ,we get

y = 5

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

(viii) We are given,

We get,

Now, substituting x = 1 in ,we get

Substituting x = 5 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Answer:

We observe that x = 3 and y = 12 is the solution of the following equations

So, we get the equations of two lines passing through (3, 12) are, 4x – y = 0 and 3x – y + 3 = 0.

We know that passing through the given point infinitely many lines can be drawn. So, there are infinitely many lines passing through (3,12)

Answer:

Total fare of Rs y for covering the distance of x km is given by
y = 15 + 8(x − 1)
y = 15 + 8x − 8
y = 8x + 7

Where, Rs y is the total fare (x – 1) is taken as the cost of first kilometer is already given Rs 15 and 1 has to subtracted from the total distance travelled to deduct the cost of first kilometer.

Answer:

The required graph is below:-

By plotting the given points (3, 5) and (–1, 3) on a graph paper, we get the line BC.

We have already plotted the point A (1, 4) on the given plane by the intersecting lines.

Therefore, it is proved that the straight line passing through (3, 5) and (–1, 3) also passes through A (1, 4).

Answer:

We are given co-ordinates (1, –1) and (–1, 1) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

(i) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given equationdoes not represent the graph in the figure.

(ii) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions satisfy this equation. Thus, it is the equation whose graph is given.

Answer:

We are given co-ordinates (–1, 3) and (2, 0) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

(i) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions does not satisfy this equation.

(ii) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions does not completely satisfy this equation.

(iii) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions satisfy this equation. Thus, it is the equation whose graph is given.

Answer:

It is given that the point lies on the given equation,

Clearly, the given point is the solution of the given equation.

Now,

Substituting in the given equation, we get

Answer:

We are given,

We get,

Substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

By plotting the given equation on the graph, we get the point B (0, 4) and C (6,0).

(i) Co-ordinates of the point whose y axis is 3 are A

(ii) Co-ordinates of the point whose x -coordinate is –3 are D

Answer:

(i) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

(ii) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

(iii) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

(iv) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

Answer:

Total charges of Rs 27 of which Rs x for first three days and Rs y per day for 4 more days is given by

Here, is taken as the charges for the first three days are already given at Rs x and we have to find the charges for the remaining four days as the book is kept for the total of 7 days.

Answer:

The number given to us is in the form of yx,

where y represents the ten’s place of the number

And x represents the unit’s place of the number

Now, the given number is

number obtained by reversing the digits of the number is

It is given to us that the original number is 27 more than the number obtained by reversing its digits

So,

Answer:

The number given to us is in the form of yx.

where y represents the ten^’s place of the number

And x represents the units place of the number

Now, the given number is

number obtained by reversing the digits of the number is

It is given to us that the sum of these two numbers is 121

So,

Answer:

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 6 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

Answer:

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

The region bounded by the graph is ABC which form a traingle.

AC at y axis is the base of traingle having AC = 4 units on y axis.

BC at x axis is the height of traingle having BC = 3 units on x axis.

Therefore,

Area of traingle ABC,say A is given by

Answer:

We are given,

Substituting, we get

Substituting, we get

Substituting, we get

Substituting, we get

For every value of x, whether positive or negative, we get y as a positive number.

Answer:

We are given,

Substituting, we get

Substituting, we get

Substituting, we get

Substituting, we get

Substituting, we get

For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units.

Answer:

We are given,

We get,

Now, substituting in , we get

Substituting in , we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation .

We are given,

We get,

$y=x-1$

Now, substituting in $y=x-1$,we get

$y=-1$

Substituting in $y=x-1$,we get

$y=-2$

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Plotting D(0,1) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by and on the graph, triangle ABC is formed on y axis.

Therefore,

AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis.

FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

$$\begin{gathered} A=\frac{1}{2}\left(\mathrm{Base}\times \mathrm{Height}\right) \\ =\frac{1}{2}\left(\mathrm{AC}\times \mathrm{FE}\right) \\ =\frac{1}{2}\left(5\times 3\right) \\ =\frac{15}{2} \mathrm{sq}. \mathrm{units} \end{gathered}$$

Answer:

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Plotting E(0, ) and A(-1,0) on the graph and by joining the points , we obtain the graph of equation .

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x	0	5
y		0

Plotting D(0, ) and B(5,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by and on the graph, triangle ABC is formed on x axis.

Therefore,

AB at x axis is the base of triangle ABC having AB = 6 units on x axis.

Draw CF perpendicular from C on x axis.

CF parallel to y axis is the height of triangle ABC having CF = 4 units on y axis.

Therefore,

Area of triangle ABC, say A is given by

Answer:

We are given the path of train A,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Plotting A(4,0) and E(0,3) on the graph and by joining the points , we obtain the graph of equation .

We are given the path of train B,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

Plotting C(0,6) and D(8,0) on the graph and by joining the points , we obtain the graph of equation

Answer:

We are given the present age of Ravish as y years and Aarushi as x years.

Age of Ravish seven years ago

Age of Aarushi seven years ago

It has already been said by Ravish that seven years ago he was seven times old then Aarushi was then

So,

Age of Ravish three years from now

Age of Aarushi three years from now

It has already been said by Ravish that three years from now he will be three times old then Aarushi will be then

So,

(1) and (2) are the algebraic representation of the given statement.

We are given,

We get,

Now, substituting in ,we get

Substituting in, we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

The red -line represents the equation.

The blue-line represents the equation.

Answer:

Aarushi is driving the car with the uniform speed of 60 km/h.

We represent time on X-axis and distance on Y-axis

Now, graphically

We are given that the car is travelling with a uniform speed 60 km/hr. This means car travels 60 km distance each hour. Thus the graph we get is of a straight line.

Also, we know when the car is at rest, the distance travelled is 0 km, speed is 0 km/hr and the time is also 0 hr.
Thus, the given straight line will pass through O(0,0) and M(1,60).

Join the points O and M and extend the line in both directions.

Now, we draw a dotted line parallel to y-axis from x = $\frac{1}{2}$ that meets the straight line graph at L from which we draw a line parallel to x-axis that crosses y-axis at 30. Thus, in $\frac{1}{2}$hr, distance travelled by the car is 30 km.

Now, we draw a dotted line parallel to y-axis from x = $2\frac{1}{2}$ that meets the straight line graph at N from which we draw a line parallel to x-axis that crosses y-axis at 150. Thus, in $2\frac{1}{2}$hr, distance travelled by the car is 150 km.

(i)

Distance travelled in hours is given by

(ii)

Distance travelled in hours is given by

Answer:

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

Answer:

Answer:

(i) We are given,

x = 2

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point (2, 0) is shown below

(ii) We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to x axis passing through the point A(0, –3) is shown below

(iii) We are given,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to x axis passing through the point (0, 3) is shown below

(iv) We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane,it is a line parallel to y axis passing through the point is shown below

(v) We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below

Answer:

We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below

Answer:

We are given,

we get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point (–5, 0) is shown below

Answer:

(i) We are given the co-ordinates of the Cartesian plane at (0,3).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to 3.

We get the equation as

(ii) We are given the co-ordinates of the Cartesian plane at (0,-4).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to -4.

We get the equation as

(iii) We are given the co-ordinates of the Cartesian plane at (2,-5).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to -5.

We get the equation as

(iv) We are given the co-ordinates of the Cartesian plane at (3,4).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to 4.

We get the equation as

Answer:

(i) We are given the co-ordinates of the Cartesian plane at (4,0).

For the equation of the line parallel to y axis ,we assume the equation as a one variable equation independent of y containing x equal to 4.

We get the equation as

(ii) We are given the co-ordinates of the Cartesian plane at (–2,0).

For the equation of the line parallel to y axis ,we assume the equation as a one variable equation independent of y containing x equal to –2.

We get the equation as

(iii) We are given the co-ordinates of the Cartesian plane at (3,5).

For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to 3.

We get the equation as

(iv) We are given the co-ordinates of the Cartesian plane at (−4,−3).

For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to −4.

We get the equation as

Answer:

Any straight line parallel to x-axis is given by y = k, where k is the distance of the line from the x-axis.

Here k = 4. Therefore, the equation of the line is y = 4.

To draw the graph of this equation, plot the points (1, 4) and (2, 4) and join them.

This is the required graph.

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.

Answer:

The equation of line representing x axis is given by

Answer:

The equation of line representing y axis is given by

Answer:

We are given the co-ordinates of the Cartesian plane at (0,4).

For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 4.

We get the equation as

Answer:

We are given the co-ordinates of the Cartesian plane at (3,5).

For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 5.

We get the equation as

Answer:

We are given the co-ordinates of the Cartesian plane at (–3,–7).

For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to –3.

We get the equation as

Answer:

We are given the co-ordinates of the Cartesian plane at (–4,6).

For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 6.

We get the equation as

Answer:

We are given,

we get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

Answer:

We are given,

we get,

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below

Answer:

We are given lies on the graph of linear equation.

So, the given co-ordinates are the solution of the equation.

Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation.

Substituting in equation, we get

Answer:

We are given lies on the graph of linear equation.

So, the given co-ordinates are the solution of the equation.

Therefore, we can calculate the value of k by substituting the value of given co-ordinates in equation.

Substituting in equation, we get