Linear Equations in Two Variables

Let cost of notebook be ₹ x and the cost of pen be ₹ y

According to the question, we have :

The cost of a notebook is twice the cost of a pen.

i.e (Cost of a notebook) = 2 x (Cost of a pen)

⇒ x = 2y

⇒ x - 2y = 0

Hence, the required linear equation is x - 2y = 0.

Given,

⇒ 2x + 3y = $9.3\overline{5}$

On re-arranging the equation, we have :

⇒ 2x + 3y - $9.3\overline{5}$ = 0

⇒ (2)x + (3)y + (-$9.3\overline{5}$) = 0

Comparing above equation with ax + by + c = 0, we get :

a = 2, b = 3 and c = -$9.3\overline{5}$

Hence, linear equation : 2x + 3y - $9.3\overline{5}$ = 0, a = 2, b = 3 and c = -$9.3\overline{5}$.

Given,

⇒ x - $\dfrac{y}{5}$ - 10 = 0

⇒ (1)x + $\Big(-\dfrac{1}{5}\Big)y$ + (-10) = 0

Comparing above equation with ax + by + c = 0, we get :

a = 1, b = -$\dfrac{1}{5}$ and c = -10

Hence, linear equation : x - $\dfrac{y}{5}$ - 10 = 0, a = 1, b = $-\dfrac{1}{5}$ and c = -10.

Given,

⇒ -2x + 3y = 6

On re-arranging the equation, we have :

⇒ -2x + 3y - 6 = 0

⇒ (-2)x + (3)y + (-6) = 0

Comparing above equation with ax + by + c = 0, we get :

a = -2, b = 3 and c = -6

Hence, linear equation : -2x + 3y - 6 = 0, a = -2, b = 3 and c = -6.

Given,

⇒ x = 3y

On re-arranging the equation, we have :

⇒ x - 3y = 0

⇒ (1)x + (-3)y + (0) = 0

Comparing above equation with ax + by + c = 0, we get :

a = 1, b = -3 and c = 0

Hence, linear equation : 1x - 3y + 0 = 0, a = 1, b = -3 and c = 0.

Given,

⇒ 2x = -5y

On re-arranging the equation, we have :

⇒ 2x + 5y = 0

⇒ (2)x + (5)y + (0) = 0

Comparing above equation with ax + by + c = 0, we get :

a = 2, b = 5 and c = 0

Hence, linear equation : 2x + 5y + 0 = 0, a = 2, b = 5 and c = 0.

Given,.

⇒ 3x + 2 = 0

On re-arranging the equation, we have :

⇒ 3x + 2 = 0

⇒ (3)x + (0)y + 2 = 0

Comparing above equation with ax + by + c = 0, we get :

a = 3, b = 0 and c = 2

Hence, linear equation : 3x + 0.y + 2 = 0, a = 3, b = 0 and c = 2.

Given,

⇒ y - 2 = 0

⇒ (0)x + (1)y + (-2) = 0

Comparing above equation with ax + by + c = 0, we get :

a = 0, b = 1 and c = -2

Hence, linear equation : 0.x + 1.y - 2 = 0, a = 0, b = 1 and c = -2.

Given,

⇒ 5 = 2x

On re-arranging the equation, we have :

⇒ -2x + 5 = 0

⇒ (-2)x + (0)y + (5) = 0

Comparing above equation with ax + by + c = 0, we get :

a = -2, b = 0 and c = 5

Hence, linear equation : -2x + 0.y + 5 = 0, a = -2, b = 0 and c = 5.

Given, linear equation y = 3x + 5

We know that,

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

Substituting x = 0, in y = 3x + 5, we get :

⇒ y = 3(0) + 5 = 0 + 5 = 5.

∴ (0, 5) is one solution.

Substituting x = 1, in y = 3x + 5, we get :

⇒ y = 3(1) + 5 = 3 + 5 = 8.

∴ (1, 8) is another solution.

Substituting x = 2, in y = 3x + 5, we get :

⇒ y = 3(2) + 5 = 6 + 5 = 11.

∴ (2, 11) is another solution.

Clearly, for different values of x, we get different values of y.

Thus, y = 3x + 5 has infinitely many solutions.

Hence, Option (iii) is the correct answer.

Given, equation :

⇒ 2x + y = 7

Substituting x = 0 in the given equation, we get :

⇒ 2(0) + y = 7

⇒ 0 + y = 7

⇒ y = 7

∴ Solution is (0, 7).

Substituting x = 1 in the given equation, we get :

⇒ 2(1) + y = 7

⇒ 2 + y = 7

⇒ y = 7 - 2

⇒ y = 5

∴ Solution is (1, 5).

Substituting x = 2 in the given equation, we get :

⇒ 2(2) + y = 7

⇒ 4 + y = 7

⇒ y = 7 - 4

⇒ y = 3

∴ Solution is (2, 3).

Substituting x = 3 in the given equation, we get :

⇒ 2(3) + y = 7

⇒ 6 + y = 7

⇒ y = 7 - 6

⇒ y = 1

∴ Solution is (3, 1).

Hence, solutions for equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (3, 1).

Given, equation :

⇒ πx + y = 9

Substituting x = 0, in the given equation, we get :

⇒ π(0) + y = 9

⇒ 0 + y = 9

⇒ y = 9

∴ Solution is (0, 9).

Substituting x = 1, in the given equation, we get :

⇒ π(1) + y = 9

⇒ π + y = 9

⇒ y = 9 - π

∴ Solution is [1, (9 - π)].

Substituting x = 2, in the given equation, we get :

⇒ π(2) + y = 9

⇒ 2π + y = 9

⇒ y = 9 - 2π

∴ Solution is [2, (9 - 2π)].

Substituting x = 3, in the given equation, we get :

⇒ π(3) + y = 9

⇒ 3π + y = 9

⇒ y = 9 - 3π

∴ Solution is [3, (9 - 3π)].

Hence, solutions for equation πx + y = 9 are (0, 9), [1, (9 - π)], [2, (9 - 2π)] and [3, (9 - 3π)].

Given, equation :

⇒ x = 4y

Substituting x = 0, in the given equation, we get :

⇒ 0 = 4y

⇒ y = $\dfrac{0}{4}$

⇒ y = 0

∴ Solution is (0, 0).

Substituting x = 1, in the given equation, we get :

⇒ 1 = 4y

⇒ y = $\dfrac{1}{4}$.

∴ Solution is $\Big(1, \dfrac{1}{4}\Big)$.

Substituting x = 4, in the given equation, we get :

⇒ 4 = 4y

⇒ y = $\dfrac{4}{4}$

⇒ y = 1

∴ Solution is (4, 1).

Substituting x = -4, in the given equation, we get :

⇒ -4 = 4y

⇒ y = $\dfrac{-4}{4}$

⇒ y = -1

∴ Solution is (-4, -1).

Hence, solutions for equation x = 4y are (0, 0), $(1, \dfrac{1}{4}\Big)$, (4, 1) and (-4, -1).

(i) Substituting (0, 2), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 0 - 2(2) = 0 - 4 = -4.

Since, L.H.S. ≠ R.H.S.

Hence, (0, 2) is not a solution for equation x – 2y = 4.

(ii) Substituting (2, 0), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 2 - 2(0) = 2 - 0 = 2

Since, L.H.S. ≠ R.H.S.

Hence, (2, 0) is not a solution for equation x – 2y = 4.

(iii) Substituting (4, 0), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 4 - 2(0) = 4 - 0 = 4

Since, L.H.S. = R.H.S.

Hence, (4, 0) is a solution for equation x – 2y = 4.

(iv) Substituting ($\sqrt{2}, 4\sqrt{2}$), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = $\sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = -7\sqrt{2}$.

Since, L.H.S. ≠ R.H.S.

Hence, $(\sqrt{2}, 4\sqrt{2})$ is not a solution for equation x – 2y = 4.

(v) Substituting, (1, 1), in L.H.S. of the equation x - 2y = 4, we get :

⇒ x - 2y = 1 - 2(1) = 1 - 2 = -1

Since, L.H.S. ≠ R.H.S.

Hence, (1, 1) is not a solution for equation x – 2y = 4.

Given,

x = 2 and y = 1 is the solution of the equation 2x + 3y = k.

On substituting the values of x = 2 and y = 1 in the given equation, we get :

⇒ 2 x 2 + 3 x 1 = k

⇒ 4 + 3 = k

⇒ k = 7

Hence, the required value of k is 7.