# Linear Equations in Two Variables

### Linear Equation in One Variable

The equation with one variable in it is known as a Linear Equation in One Variable.

The general form is

px + q = s, where p, q, and s are real numbers and p ≠ 0.

Example

x + 5 = 10

y – 3 = 19

These are called Linear Equations in One Variable because the highest degree of the variable is one.

### Graph of the Linear Equation in One Variable

We can mark the point of the linear equation in one variable on the number line.

x = 2 can be marked on the number line as follows –

### Linear Equation in Two Variables

An equation with two variables is known as a Linear Equation in Two Variables. The general form of the linear equation in two variables is

ax + by + c = 0

where a and b are coefficients and c is the constant. a ≠ 0 and b ≠ 0.

Example

6x + 2y + 5 = 0, etc.

### Slope Intercept form

Generally, the linear equation in two variables is written in the slope-intercept form as this is the easiest way to find the slope of the straight line while drawing the graph of it. The slope-intercept form is

• Where m represents the slope of the line.
• and b tells the point of intersection of the line with the y-axis. tells the point of intersection of the line with the y-axis

### Solution of a Linear Equation

• There is only one solution in the linear equation in one variable but there are infinitely many solutions in the linear equation in two variables.
• As there are two variables, the solution will be in the form of an ordered pair, i.e. (x, y).
• The pair which satisfies the equation is the solution of that particular equation.

Example:

Find the solution for the equation 2x + y = 7.

Solution:

To calculate the solution of the given equation we will take x = 0

2(0) + y = 7

y = 7

Hence, one solution is (0, 7).

To find another solution we will take y = 0

2x + 0 = 7

x = 3.5

So another solution is (3.5, 0).

### Graph of a Linear Equation in Two Variables

To draw the graph of linear equation in two variables, we need to draw a table to write the solutions of the given equation, and then plot them on the Cartesian plane.

By joining these coordinates, we get the line of that equation.

• The coordinates which satisfy the given Equation lies on the line of the equation.As there are two variables, the solution will be in the form of an ordered pair, i.e. (x, y).
• Every point (x, y) on the line is the solution x = a, y = b of the given Equation.
• •Any point, which does not lie on the line AB, is not a solution of Equation.

Example:

Draw the graph of the equation 3x + 4y = 12.

Solution:

To draw the graph of the equation 3x + 4y = 12, we need to find the solutions of the equation.

Let x = 0

3(0) + 4y = 12

y = 3

Let y = 0

3x + 4(0) = 12

x = 4

Now draw a table to write the solutions.

Now we can draw the graph easily by plotting these points on the Cartesian plane.

### Equations of Lines Parallel to the x-axis and y-axis

When we draw the graph of the linear equation in one variable then it will be a point on the number line.

x – 5 = 0

x = 5

This shows that it has only one solution i.e. x = 5, so it can be plotted on the number line.

But if we treat this equation as the linear equation in two variables then it will have infinitely many solutions and the graph will be a straight line.

x – 5 = 0

or

x + (0) y – 5 = 0

This shows that this is the linear equation in two variables where the value of y is always zero. So the line will not touch the y-axis at any point.

x = 5, x = number, then the graph will be the vertical line parallel to the y-axis.

All the points on the line will be the solution of the given equation. Similarly if y = – 3, y = number then the graph will be the horizontal line parallel to the x-axis.

Exercise 4.1

1. 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 Rs. x and that of a pen to be Rs.y).

Solution:

Let the cost of a notebook = Rs. x
and the cost of a pen = Rs. y
According to the condition, we have
[Cost of a notebook] =2 x [Cost of a pen]
i. e„ (x) = 2 x (y) or, x = 2y
or, x – 2y = 0
Thus, the required linear equation is x – 2y = 0.

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

(i) 2x + 3y = 9.35¯¯¯
(ii) x−y/5−10=0
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = -5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x

Solution:
(i) We have 2x + 3y = 9.35¯¯¯
or (2)x + (3)y + (−9.35¯¯¯ ) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –9.35¯¯¯ .

(ii) We have x−y/5−10=0
or x + (- 1/5) y + (10) = 0
Comparing it with ax + by + c = 0, we get
a =1, b =- 1/5 and c= -10

(iii) We have -2x + 3y = 6 or (-2)x + (3)y + (-6) = 0
Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.

(iv) We have x = 3y or (1)x + (-3)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 1, b = -3 and c = 0.

(v) We have 2x = -5y or (2)x + (5)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0 or (3)x + (0)y + (2) = 0 Comparing it with ax + by + c = 0, we get a = 3, b = 0 and c = 2.

(vii) We have y – 2 = 0 or (0)x + (1)y + (-2) = 0 Comparing it with ax + by + c = 0, we get a = 0, b = 1 and c = -2.

(viii) We have 5 = 2x ⇒ 5 – 2x = 0
or -2x + 0y + 5 = 0
or (-2)x + (0)y + (5) = 0
Comparing it with ax + by + c = 0, we get a = -2, b = 0 and 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

Solution:
Option (iii) is true because, for every value of x, we get a corresponding value of y and vice-versa in the given equation.
Hence, a given linear equation has infinitely many solutions.

Question 2
Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y

Solution:
(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)

When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)

When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

(iii) x = 4y
When x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
When x = 1, 4y = 1 ⇒ y = 14
∴ Solution is (1,1/4 )
When x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
When x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)

Question 3
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0,2)
(ii) (2,0)
(iii) (4, 0)
(iv) (√2, 4√2)
(v) (1, 1)

Solution:
(i) (0,2) means x = 0 and y = 2
Puffing x = 0 and y = 2 in x – 2y = 4, we get
L.H.S. = 0 – 2(2) = -4.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x =0, y =2 is not a solution.

(ii) (2, 0) means x = 2 and y = 0
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H:S. 2 – 2(0) = 2 – 0 = 2.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (2,0) is not a solution.

(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = o in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.

(iv) (√2, 4√2) means x = √2 and y = 4√2
Putting x = √2 and y = 4√2 in x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (√2 , 4√2) is not a solution.

(v) (1, 1)means x =1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4
∴ LH.S. ≠ R.H.S.
∴ (1, 1) is not a solution.

Question 4
Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.

Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.

Exercise 4.3

Question 1
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x  (iv) 3 = 2x + y

Solution:
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4

(ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of x – y = 2

(iii) y = 3x
If we have x = 0,
then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) = 3
x= -1, then y = 3(-1) = -3

∴ We get the following table:

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.

Thus, the line LM is the required graph of y = 3x

(iv) 3 = 2x + y ⇒ y = 3 – 2x
If we have x = 0, then y = 3 – 2(0) = 3
x = 1,then y = 3 – 2(1) = 3 – 2 = 1
x = 2,then y = 3 – 2(2) = 3 – 4 = -1
∴ We get the following table:

Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.

Thus, the line CD is the required graph of 3 = 2x + y.

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Solution:
(2, 14) means x = 2 and y = 14

Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0

There are infinite number of lines which passes through the point (2, 14) because infinite number of lines can be drawn through a point.

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

Solution:
The equation of the given line is 3y = ax + 7
∵ (3, 4) lies on the given line.
∴ It must satisfy the equation 3y = ax + 7
We have, (3, 4) ⇒ x = 3 and y = 4.
Putting these values in given equation, we get
3 x 4 = a x 3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7 = 5 ⇒ a = 5/3
Thus, the required value of a is 5/3

4. The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.

Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,

which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table:

Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.

5. From the choices given below, choose the equation whose graphs are given ¡n Fig. (1) and Fig. (2).
For Fig. (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

For Fig. (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6

Solution:
For Fig. (1), the correct linear equation is x + y = 0
[As (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]

For Fig.(2), the correct linear equation is y = -x + 2
[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)
⇒ 2 = -(0) + 2 ⇒ 2 = 2]

6. 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

Solution:
Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force x Distance
⇒ y = 5 x x ⇒ y = 5x
For drawing the graph, we have y = 5x
When x = 0, then y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = -1, then y = 5(-1) = -5
∴ We get the following table:

Plotting the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.

From the graph, we get
(i) Distance travelled =2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.

(ii) Distance travelled = 0 unit i.e., x = 0
∴ If x = 0 ⇒ y = 5(0) – 0
⇒ Work done = 0 unit.

7. Yamani and Fatima, two students of Class IX of a school, together contributed Rs. 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 Rs.xand Rs.y.) Draw the graph of the same.

Solution:

Let the contribution of Yamini = Rs. x
and the contribution of Fatima Rs. y
∴ We have x + y = 100 ⇒ y = 100 – x
Now, when x = 0, y = 100 – 0 = 100
x = 50, y = 100 – 50 = 50
x = 100, y = 100 – 100 = 0
∴ We get the following table:

Plotting the ordered pairs (0,100), (50,50) and (100, 0) on a graph paper using proper scale and joining these points, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation x + y = 100.

8. 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 = (9/5 )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.

Solution:
(i) We have
F = (9/5 )C + 32
When C = 0 , F = (9/5 ) x 0 + 32 = 32
When C = 15, F = (9/5)(-15) + 32= -27 + 32 = 5
When C = -10, F = 9/5 (-10)+32 = -18 + 32 = 14
We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(ii) From the graph, we have 86°F corresponds to 30°C.

(iii) From the graph, we have 95°F corresponds 35°C.

(iv) We have, C = 0
From (1), we get
F = (9/5)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (9/5)C + 32 ⇒ -32 x 5/9 = C ⇒ C = -17.8

(V) When F = C (numerically)
From (1), we get
F = 9/5 F + 32 ⇒ F – 9/5F = 32
⇒ -4/5 F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.

Exercise 4.4

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

Solution:
(i) y = 3
∵ y = 3 is an equation in one variable, i.e., y only.
∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3
We can write y = 3 in two variables as 0.x + y = 3
Now, when x = 1, y = 3
x = 2, y = 3
x = -1, y = 3
∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get a line AB as solution of 0. x + y = 3,
i.e. y = 3.

2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Solution:
(i) 2x + 9 = 0
We have, 2x + 9 = 0 ⇒ 2x = – 9 ⇒ x = -9/2
which is a linear equation in one variable i.e., x only.
Therefore, x = -9/2 is a unique solution on the number line as shown below:

(ii) 2x +9=0
We can write 2x + 9 = 0 in two variables as 2x + 0, y + 9 = 0
or x=-9-0.y/2
∴ When y = 1, x = x=-9-(1)/2 = -9/2

Thus, we get the following table:

Now, plotting the ordered pairs on a graph paper and joining them, we get a line PQ as solution of 2x + 9 = 0.