Linear Equation in Two Variables ⇒⇒ 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 :

The equation y = 3x + 5 is a linear equation in two variables (x and y)in the form of ax + by + c = 0

A solution to a linear equation in two variables is a pair of values (x,y) that makes the equation true.

Substitute 0 for y in given equation,

$ 0 = 3x + 5 $

$ x = - {5 \over 3 }$

Therefore, ($ {-5 \over 3 }$ , 0) is one solution .

⇒ Substitute 1 for y in given equation,

$ 1 = 3x + 5 $

$ 3x = 1 - 5 $

$ x = {-4 \over 3 }$

Therefore, ($ {-4 \over 3 }$ , 1) is another solution .

⇒ Substitute 1 for x in given equation,

$ y = 3(1) + 5 $

$ y = 8 $

Therefore,$ (1 , 8)$ is another solution .

Clearly, This equation is of a line and a line has infinite points on it and each point is a solution .

This means that we can put any value of x and in return, we will get some other value of y,

Since every point (x,y) on the line is a solution to the equation, and there are infinitely many points on the line, the equation must have infinitely many solutions.

Question 2 (i)

Write four solutions for each of the following equations:

(i) 2x + y = 7

Solution :

Given , 2x + y = 7

To find four solutions, we can choose any four values for x (or y) and solve for the corresponding value of the other variable. It's easiest to first solve for one variable, such as y:

y = -2x + 7

For finding out solutions,just put one value in any of the variable in the equation and obtain the other value.

1. Solution where x = 1

y = -2(1) + 7

y = -2 + 7

y = 5

Therefore, (1 , 5) is one solution .

2. Solution where x = 2

y = -2(2) + 7

y = -4 + 7

y = 3

Therefore, (2 , 3) is another solution .

3. Solution where x = -2

y = -2(-2) + 7

y = 4 + 7

y = 11

Therefore, (-2 , 11) is another solution .

4. Solution where x = 3

y = -2(3) + 7

y = -6 + 7

y = 1

Therefore, (3 , 1) is another solution .

The four solutions are : (1 , 5) , (2 , 3), (-2 , 11), (3 , 1)

Question 2 (ii)

Write four solutions for each of the following equations:

(ii) πx + y = 9

Solution :

Given , πx + y = 9

To find four solutions, we can choose any four values for x (or y) and solve for the corresponding value of the other variable. It's easiest to first solve for one variable, such as y:

y = 9 - πx

For finding out solutions,just put one value in any of the variable in the equation and obtain the other value.

1. Solution where x = 1

y = 9 - π(1)

y = 9 - π

Therefore, (1 , 9 - π) is one solution .

2. Solution where x = 2

y = 9 - π(2)

y = 9 - 2π

Therefore, (2 , 9 - 2π) is another solution .

3. Solution where x = 3

y = 9 - π(3)

y = 9 - π

Therefore, (3 , 9 - 3π) is another solution .

4. Solution where x = 4

y = 9 - π(4)

y = 9 - 4π

Therefore, (4 , 9 - 4π) is another solution .

The four solutions are : (1 , 9 - π) , (2 , 9 - 2π), (3 , 9 - 3π), (4 , 9 - 4π)

Question 2 (iii)

Write four solutions for each of the following equations:

(iii) x = 4y

Solution :

Given , x = 4y

For finding out solutions,just put one value in any of the variable in the equation and obtain the other value.

1. Solution where y = 1

x = 4(1)

x = 4

Therefore, (4 , 1) is one solution .

2. Solution where y = 2

x = 4(2)

x = 8

Therefore, (8 , 2) is another solution .

3. Solution where y = 3

x = 4(3)

x = 12

Therefore, (12 , 3) is another solution .

4. Solution where y = 0

x = 4(0)

x = 0

Therefore, (0 , 0) is another solution .

The four solutions are : (4 , 1) , (8 , 2), (12 , 3), (0 , 0)

Question 3 (i)

Check which of the following are solutions of the equation x - 2y = 4 and which are not:
(i) (0, 2)

Solution :

Given , x - 2y = 4

To check if an ordered pair (x,y) is a solution to the equation x− 2y = 4

We substitute the values of x and y into the left-hand side (LHS) of the equation and see if the result equals the right-hand side (RHS), which is 4.

If the equation is satisfied, it is a solution or else it is not.

Here, Given point (0, 2) means x = 0 and y = 2

Let Substitute x = 0 and y = 2 in given equation,

⇒ 0 - 2(2) = 4

⇒ 0 - 4 = 4

⇒ -4 $ \ne $ 4

⇒ L.H.S. $ \ne $ R.H.S.

Therefore, (0, 2) is not the solution of given equation.

Question 3 (ii)

Check which of the following are solutions of the equation x - 2y = 4 and which are not:
(ii) (2, 0)

Solution :

Given , x - 2y = 4

To check if an ordered pair (x,y) is a solution to the equation x− 2y = 4

We substitute the values of x and y into the left-hand side (LHS) of the equation and see if the result equals the right-hand side (RHS), which is 4.

If the equation is satisfied, it is a solution or else it is not.

Here, Given point (2, 0) means x = 2 and y = 0

Let Substitute x = 2 and y = 0 in given equation,

⇒ 2 - 2(0) = 4

⇒ 2 - 0 = 4

⇒ 2 $ \ne $ 4

⇒ L.H.S. $ \ne $ R.H.S.

Therefore, (2,0) is not the solution of given equation.

Question 3 (iii)

Check which of the following are solutions of the equation x - 2y = 4 and which are not:
(iii) ( 4, 0)

Solution :

Given , x - 2y = 4

To check if an ordered pair (x,y) is a solution to the equation x− 2y = 4

We substitute the values of x and y into the left-hand side (LHS) of the equation and see if the result equals the right-hand side (RHS), which is 4.

If the equation is satisfied, it is a solution or else it is not.

Here, Given point (4, 0) means x = 4 and y = 0

Let Substitute x = 4 and y = 0 in given equation,

⇒ 4 - 2(0) = 4

⇒ 4 - 0 = 4

⇒ 4 = 4

⇒ L.H.S. = R.H.S.

Therefore, (4,0) is the solution of given equation.

Question 3 (iv)

Check which of the following are solutions of the equation x - 2y = 4 and which are not:
(iv) ($ \sqrt 2$ , 4 $ \sqrt 2$)

Solution :

Given , x - 2y = 4

To check if an ordered pair (x,y) is a solution to the equation x− 2y = 4

We substitute the values of x and y into the left-hand side (LHS) of the equation and see if the result equals the right-hand side (RHS), which is 4.

If the equation is satisfied, it is a solution or else it is not.

Here, Given point ($ \sqrt 2$ , 4 $ \sqrt 2$) means x = $ \sqrt 2$ and y = 4$ \sqrt 2$

Let Substitute x = $ \sqrt 2$ and y = 4$ \sqrt 2$ in given equation,

⇒ $ \sqrt 2$ - 2(4$ \sqrt 2$) = 4

⇒ $ \sqrt 2$ - 8$ \sqrt 2$ = 4

⇒ - 7$ \sqrt 2$ $ \ne $ 4

⇒ L.H.S. $ \ne $ R.H.S.

Therefore, ($ \sqrt 2$ , 4 $ \sqrt 2$) is not the solution of given equation.

Question 3 (v)

Check which of the following are solutions of the equation x - 2y = 4 and which are not:
(v) (1, 1)

Solution :

Given , x - 2y = 4

To check if an ordered pair (x,y) is a solution to the equation x− 2y = 4

We substitute the values of x and y into the left-hand side (LHS) of the equation and see if the result equals the right-hand side (RHS), which is 4.

If the equation is satisfied, it is a solution or else it is not.

Here, Given point (1, 1) means x = 1 and y = 1

Let Substitute x = 1 and y = 1 in given equation,

⇒ 1 - 2(1) = 4

⇒ 1 - 2 = 4

⇒ -1 $ \ne $ 4

⇒ L.H.S. $ \ne $ R.H.S.

Therefore, (1,1) is not the solution of given equation.

Question 4

Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Solution :

Given , 2x + 3y = k

For finding out value of k , We can substitute the Point Values in the given equation

Here, Given point (2, 1)

Let Substitute x = 2 and y = 1 in given equation,

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

⇒ 7 = k

Therefore, value of k is 7.