Q: Question 2(ii)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :x - y5\dfrac{y}{5}5y - 10 = 0

Given,⇒ x - y5\dfrac{y}{5}5y - 10 = 0⇒ (1)x + (−15)y\Big(-\dfrac{1}{5}\Big)y(−51)y + (-10) = 0Comparing above equation with ax + by + c = 0, we get :a = 1, b = -15\dfrac{1}{5}51 and c = -10Hence, linear equation : x - y5\dfrac{y}{5}5y - 10 = 0, a = 1, b = −15-\dfrac{1}{5}−51 and c = -10.

Question 1

Question 1The 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 ₹ x and that of a pen to be ₹ y).

Accepted Answer

Let cost of notebook be ₹ x and the cost of pen be ₹ yAccording 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 = 0Hence, the required linear equation is x - 2y = 0.

Question 2

Question 2(i)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :2x + 3y = 9.35‾9.3\overline{5}9.35

Accepted Answer

Given,⇒ 2x + 3y = 9.35‾9.3\overline{5}9.35On re-arranging the equation, we have :⇒ 2x + 3y - 9.35‾9.3\overline{5}9.35 = 0⇒ (2)x + (3)y + (-9.35‾9.3\overline{5}9.35) = 0Comparing above equation with ax + by + c = 0, we get :a = 2, b = 3 and c = -9.35‾9.3\overline{5}9.35Hence, linear equation : 2x + 3y - 9.35‾9.3\overline{5}9.35 = 0, a = 2, b = 3 and c = -9.35‾9.3\overline{5}9.35.

Question 3

Question 2(ii)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :x - y5\dfrac{y}{5}5y​ - 10 = 0

Accepted Answer

Given,⇒ x - y5\dfrac{y}{5}5y​ - 10 = 0⇒ (1)x + (−15)y\Big(-\dfrac{1}{5}\Big)y(−51​)y + (-10) = 0Comparing above equation with ax + by + c = 0, we get :a = 1, b = -15\dfrac{1}{5}51​ and c = -10Hence, linear equation : x - y5\dfrac{y}{5}5y​ - 10 = 0, a = 1, b = −15-\dfrac{1}{5}−51​ and c = -10.

Question 4

Question 2(iii)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :-2x + 3y = 6

Accepted Answer

Given,⇒ -2x + 3y = 6On re-arranging the equation, we have :⇒ -2x + 3y - 6 = 0⇒ (-2)x + (3)y + (-6) = 0Comparing above equation with ax + by + c = 0, we get :a = -2, b = 3 and c = -6Hence, linear equation : -2x + 3y - 6 = 0, a = -2, b = 3 and c = -6.

Question 5

Question 2(iv)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :x = 3y

Accepted Answer

Given,⇒ x = 3yOn re-arranging the equation, we have :⇒ x - 3y = 0⇒ (1)x + (-3)y + (0) = 0Comparing above equation with ax + by + c = 0, we get :a = 1, b = -3 and c = 0Hence, linear equation : 1x - 3y + 0 = 0, a = 1, b = -3 and c = 0.

Question 6

Question 2(v)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :2x = -5y

Accepted Answer

Given,⇒ 2x = -5yOn re-arranging the equation, we have :⇒ 2x + 5y = 0⇒ (2)x + (5)y + (0) = 0Comparing above equation with ax + by + c = 0, we get :a = 2, b = 5 and c = 0Hence, linear equation : 2x + 5y + 0 = 0, a = 2, b = 5 and c = 0.

Question 7

Question 2(vi)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :3x + 2 = 0

Accepted Answer

Given,.⇒ 3x + 2 = 0On re-arranging the equation, we have :⇒ 3x + 2 = 0⇒ (3)x + (0)y + 2 = 0Comparing above equation with ax + by + c = 0, we get :a = 3, b = 0 and c = 2Hence, linear equation : 3x + 0.y + 2 = 0, a = 3, b = 0 and c = 2.

Question 8

Question 2(vii)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :y - 2 = 0

Accepted Answer

Given,⇒ y - 2 = 0⇒ (0)x + (1)y + (-2) = 0Comparing above equation with ax + by + c = 0, we get :a = 0, b = 1 and c = -2Hence, linear equation : 0.x + 1.y - 2 = 0, a = 0, b = 1 and c = -2.

Question 9

Question 2(viii)Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case :5 = 2x

Accepted Answer

Given,⇒ 5 = 2xOn re-arranging the equation, we have :⇒ -2x + 5 = 0⇒ (-2)x + (0)y + (5) = 0Comparing above equation with ax + by + c = 0, we get :a = -2, b = 0 and c = 5Hence, linear equation : -2x + 0.y + 5 = 0, a = -2, b = 0 and c = 5.

Question 10

Question 1Which 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

Accepted Answer

Given, linear equation y = 3x + 5We know that,y = 3x + 5 is a linear equation in two variables in the form of ax + by + c = 0Substituting 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.

Question 11

Question 2(i)Write four solutions for the following equation:2x + y = 7

Accepted Answer

Given, equation :⇒ 2x + y = 7Substituting 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).

Question 12

Question 2(ii)Write four solutions for the following equation:πx + y = 9

Accepted Answer

Given, equation :⇒ πx + y = 9Substituting 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)].

Question 13

Question 2(iii)Write four solutions for the following equations:x = 4y

Accepted Answer

Given, equation :⇒ x = 4ySubstituting x = 0, in the given equation, we get :⇒ 0 = 4y⇒ y = 04\dfrac{0}{4}40​⇒ y = 0∴ Solution is (0, 0).Substituting x = 1, in the given equation, we get :⇒ 1 = 4y⇒ y = 14\dfrac{1}{4}41​.∴ Solution is (1,14)\Big(1, \dfrac{1}{4}\Big)(1,41​).Substituting x = 4, in the given equation, we get :⇒ 4 = 4y⇒ y = 44\dfrac{4}{4}44​⇒ y = 1∴ Solution is (4, 1).Substituting x = -4, in the given equation, we get :⇒ -4 = 4y⇒ y = −44\dfrac{-4}{4}4−4​⇒ y = -1∴ Solution is (-4, -1).Hence, solutions for equation x = 4y are (0, 0), (1,14)(1, \dfrac{1}{4}\Big)(1,41​), (4, 1) and (-4, -1).

Question 14

Question 3Check 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) (2,42)(\sqrt{2}, 4\sqrt{2})(2​,42​)(v) (1, 1)

Accepted Answer

(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 = 2Since, 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 = 4Since, L.H.S. = R.H.S.Hence, (4, 0) is a solution for equation x – 2y = 4.(iv) Substituting (2,42\sqrt{2}, 4\sqrt{2}2​,42​), in L.H.S. of the equation x - 2y = 4, we get :⇒ x - 2y = 2−2(42)=2−82=−72\sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = -7\sqrt{2}2​−2(42​)=2​−82​=−72​.Since, L.H.S. ≠ R.H.S.Hence, (2,42)(\sqrt{2}, 4\sqrt{2})(2​,42​) 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 = -1Since, L.H.S. ≠ R.H.S.Hence, (1, 1) is not a solution for equation x – 2y = 4.

Question 15

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

Accepted Answer

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 = 7Hence, the required value of k is 7.

Linear Equations in Two Variables

Exercise 41

Exercise 42