Class 9 Linear Equation in Two Variables Ex 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 1:

Given:

y = 3 x + 5 is a linear Equation.

For x = 0, y = 0 + 5 = 5

Therefore, (0, 5) is another solution.

For x = 1, y = 3 × 1 + 5 = 8

Therefore (1, 8) is another solution.

For y = 0 , 3x + 5 = 0 is the one solution

Clearly, Here in a linear equation in two variables has infinitely many solutions

Hence (iii) is the correct answer

Question 2:

Write four solutions for each of the following equations:

(i) 2x + y = 7

Solution:

Linear Equation, 2x + y = 7

y = 7- 2x ----------Equation (1)

Let  x = 0,

2(0) + y = 7

y = 7

Hence, we get (x, y) = (0, 7)

For x = 1,

2(1) + y = 7

⇒ y = 5

Hence, we get (x, y) = (1, 5)

For x = 2,

2(2) + y = 7

⇒ y = 3

Hence, we get (x, y) = (2, 3)

For x = 3,

2(3) + y = 7

⇒ y = 1

Hence we get (x, y) = (3, 1)

Therefore the four solutions of the given equation are (0,7) , (1,5), (2, 3), (3,1)

(ii) πx + y = 9

Solution 

πx + y = 9

y = 9 - πx ----------Equation (1)

Let us now take different values of x and substituting in the Equation (1), we get

For x = 0,

y = 9 - π(0)

⇒y = 9

Hence we get (x, y) = (0, 9)

For x = 1,

y = 9- π (1)

= 9- π

Hence we get (x, y) = (1, 9- π )

For x = 2,

y = 9- π (2)

Hence we get (x, y) = (2, 9 −2π)

For x = 3,

y = 9- π (3)

Hence we get (x, y) = (3, 9 - 3π)

Therefore the four solutions of the given equation are (0, 9), (1, 9 - π), (2, 9 −2π ), (3, 9 - 3π)

(iii) x = 4y

Solution 

x = 4y

\(y = {x\over 4}\) ----------Equation (1)

Let x = 0,

\(y = {0\over 4} = 0\)

Hence we get (x, y) = (0, 0)

For x = 1,

\(y = {1\over 4}\)

Hence we get (x, y) = (1, {1\over 4})\)

For x = 2,

\(y = {2\over 4} = {1\over 2}\)

Hence we get \((x, y) = (2, {1\over 2})\)

For x = 3,

\(y = {3\over 4}\)

Hence we get \((x, y) = (3, {3\over 4})\)

Therefore the four solutions of the given equation are \( (0, 0), (1, {1\over 4}), (2, {1\over 2}), (3, {3\over 4})\)

Question 3:

Check which of the following solutions of the equation are x − 2y = 4 and which are not:

(i)(0, 2)

(ii)(2, 0)

(iii)(4, 0)

(iv) \((\sqrt{2},4\sqrt{2})\)

(v)(1, 1)

Solution 3:

Given : x − 2y = 4 is a Linear Equation----------Equation(1)

(i)(0, 2)

B Substituting x = 0 and y = 2 in the L.H.S of the given Equation (1)

x − 2y

\(= (0) – (2)^2\)

= − 4 ≠ 4 ≠ RHS

L.H.S ≠ R.H.S

Therefore, (0, 2) is not a solution of this equation.

(ii)(2, 0)

By Substituting, x = 2 and y = 0 in the L.H.S of the given Equation (1),

x − 2y = 2 – 2(0) = 2 ≠ 4 ≠ RHS

L.H.S ≠ R.H.S

Therefore, (2, 0) is not a solution of this equation.

(iii)(4, 0)

By Substituting, x = 4 and y = 0 in the L.H.S of the given Equation (1)

x − 2y

= 4 − 2(0)

= 4

L.H.S = R.H.S

Therefore, (4, 0) is a solution of this equation.

(iv) \((\sqrt{2},4\sqrt{2})\)

By Substituting,

\(x=\sqrt{2}\) and \(y=4\sqrt{2}\)  in the L.H.S of the given Equation (1)

\(x-2y =\sqrt{2}-8\sqrt{2}\)

\(= -7\sqrt{2} ≠ 4 ≠\) R.H.S

L.H.S ≠ R.H.S

Therefore, is not a solution of this equation.

(v)(1, 1)

By Substituting, x = 1 and y = 1 in the L.H.S of the given Equation (1)

x − 2y

= 1 − 2(1)

= 1 − 2

= − 1 ≠ 4 ≠ RHS

L.H.S ≠ R.H.S

Therefore, (1, 1) is not a solution of this equation.

Question 4:

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

Solution 4:

Given : 2x + 3y = k is the Linear Equation ---------------- Equation (1)

x = 2

y = 1

k =?

By substituting the values of x and in the Equation (1),

2x + 3y = k

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

\(\therefore, k = 7\)