Linear Equation In Two Variables

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.) Table of solutions for the given equation-

GRAPH:

(i) Yes, all these are equations of the form y = mx, where m is a real number and m = 1,2,-2,3,-3 respectively in the above equations.

(ii) Yes, all these are graphs passing through the origin, i.e., pt. A in every graph

(iii) ∴ we can conclude that every graph of type y = mx passes through origin, where m is a real number.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

From the graph (pt. E) we can see that for x = 1, the y = 3.

(Note: Also we can put x = 1 in the given equation and can find the value of y-

We have,

At x = 1,

⇒ y = 3

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

i)the value of y when x = 4 is y = 6 (pt. E)

ii) the value of x when y = -3 is y = -5 (pt. F)

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

(i) From the graph, we can see that for y = 3 is pt. E and the

(ii) From the graph, we can see that for x = -3 is pt. F and the corresponding y = 6 for that.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

⇒ ∴ the pts. Where graph cuts the co-ordinate axis(i.e., where x = 0 and where y = 0) are pt. A = (2,0) and pt. B = (0,-4)

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

⇒ ∴ the pts. Where graph cuts the co-ordinate axis(i.e., where x = 0 and where y = 0) are pt. A = (-8,0) and pt. B = (0,2).

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

⇒ ∴ the pts. Where graph cuts the co-ordinate axis(i.e., where x = 0 and where y = 0) are pt. A = (-2,0) and pt. B = (0,-3).

Given that together Rajiya and Preethi collected Rs.1000.

Now, Let the amount collected by Rajiya be Rs. x and by Preethi be Rs. y.

∴ the linear equation will be-

⇒ x + y = 1000

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

Given that total area = 5000 sq. m.

Let the area of field with wheat = x sq. m and area of field with paddy = y sq.m

∴ the linear equation will be-

⇒ x + y = 5000

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

Given that mass (m) = 5 kg

Also, from Newton’s send law of motion, we have-

⇒ f = ma (where, f is force and a is acceleration)

⇒f = 6a is the equation to express this observation .

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

(Note: a = -1 means, a is negative and negative acceleration is called retardation.)

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.) Table of solutions for the given equation-

GRAPH:

From the graph (pt. E) we can see that for t = 4, the V = 39.2.

Let the number of voters who cast their votes be ‘x’ and the total number of voters be ‘y’

⇒ x = 60% of y

⇒ 5x = 3y is the equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

From the graph we can see that for x = 1200, the value of y = 2000

And for y = 800, the value of x = 480

(where, no. of voters who cast their vote = x, and Total no. of voters = y)

Give that, When Rupa was born, his father was 25 years old.

Now, let the age of Rupa’s father be ‘x’ and Rupa’s be ‘y.’

⇒ x - y = 25 is the equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

(i) Given that y = 25.

∴ from the graph x = 50 (pt. E)

(ii) Given that x = 40

∴ from the graph y = 40 (pt. F) (; 15

(where, Father’s age = x and Rupa’s age = y)

Given that, An auto charges ₹ 15 for first kilometer and ₹ 8 each for each subsequent kilometer. For a distance of ‘x’ km. an amount of ₹ ‘y’ is paid.

⇒ the extra charge for first kilometer apart from 8-

= 15-8 = Rs.7

⇒ y = 8x + 7 is the linear equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

Now, from the graph, distance travelled if the fare paid is Rs.55

is 6 km (pt. E)

And the money to be paid for 7 kilometers is Rs. 63. (pt. F)

John paid Rs. 27 for a book kept for seven days. If the fixed charges be Rs. x and subsequent per day charges be

Rs. y

⇒ x + 4y = 27 is the equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

Now, from the graph if the fixed charge is Rs. 7 the subsequent per day charge is Rs. 5 (pt. E)

and if the per day charge is Rs. 4, then the fixed charge is Rs. 11. (pt. F)

Give that, the parking charges of a car for first two hours is Rs. 50 and Rs. 10 for each subsequent hour.

⇒ the extra charges for first two hours apart from Rs. 10 per hour = 50-(10× 2)

= 30

Now, let ‘x’ be no. of hours and ‘y’ be the parking charges.

⇒ y = 10x + 30 is the equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

(Note: Negative values are taken just to make graph but these values are practically impossible)

(i)Given, x = 3 hr.

From the graph, the corresponding y = 60 hr. (pt. E)

(ii)Given, x = 6 hr.

From the graph, the corresponding y = 90 hr. (pt. F)

(iii)Given, y = Rs. 80

From the graph, the corresponding x = 5 hr. (pt. G)

(where, No. of hours = x ; Parking charges = y)

Given that, speed = 60 kmph.

Also, distance = speed× time

⇒ d = 60 t (where, d = distance, t = time);

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

(i) Given, t = 1.5 hr.

From the graph, we have corresponding y = 90 km. (pt. E)

(ii) Given, t = 2 hr.

From the graph, we have corresponding y = 120 km. (pt. F)

(iii) Given, t = 3.5 hr.

From the graph, we have corresponding y = 210 km. (pt. G)

Given that, ratio of molecular weight of Hydrogen and Oxygen in water is 1:8.

Now, Let the quantity of hydrogen and oxygen or ‘x’ and ‘y’ respectively.

⇒ 8x = y

⇒ y = 8x is the equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

From the graph, the quantity of Hydrogen if Oxygen is 12 grams is grams. (pt. E)

And quantity of oxygen if hydrogen is grams is 12 grams (pt. E).

Given that, the ratio of milk and water is 5:2.

∴ Ratio between mixture and milk = 5 + 2 : 5 = 7 : 5

Now, let the quantity of mixture be ‘x’ and milk be ‘y’.

⇒ 7y = 5x is the equation.

For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

Now, given mixture(x) = 28

From the graph, (pt. E)- the corresponding quantity of milk(y) = 20

(i) For graph, we’ll first make the table of solutions by putting some random values of x and thereafter we’ll find corresponding values of y and then we’ll plot these points on graph, join them and extend them in straight line to find the graph.

(Note: ∵ equation is linear graph will always be straight line.)

Table of solutions for the given equation-

GRAPH:

(ii) If the temperature is 30°C, then from the graph, the temperature in Fahrenheit is 86° F. (pt. E)

(iii) If the temperature is 30°C, what is the temperature in Fahrenheit is 35° C. (pt. F)

(iv) Yes.

The temperature that has numerically the same value in both Fahrenheit and Celsius is -40 (pt. G)

We are given, x = 3

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

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

We are given, y + 3 = 0

We get, y = -3

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

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

We are given, y = 4

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

The representation of the solution on the Cartesian plane, it is parallel to y axis passing through-the point (0,4) is shown below

We are given, 2x – 9 = 0

We get, 2x = 9 ⇒ x =

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

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

We are given, 3x + 5 = 0

We get, 3x = -5 ⇒ x =

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

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

We are given, 2x -11 = 0

We get,

2x = 11

x =

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

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

We are given that, 3x + 2 = 8x - 8

We get,

3x – 8x = -8 – 2

⇒ -5x = -10

⇒ x =

⇒ 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 (5,0) is shown below

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,

y = -3

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,

y = 4

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,

y = -5

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,

y = 4

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,

x = -4

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,

x = 2

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,

x = 3

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,

x = -4

i). Equation of lines Parallel to x-axis

1: y = 4

2: y = -3

3: y = 8

ii). Equation of lines parallel to y-axis

1: x = 6

2: x = -2

3: x = 1