Linear Equations in Two Variables Questions

Linear equations in two variables questions presented here cover a variety of questions asked regarding linear equations in two variables with solutions and proper explanations. By practising these questions students will develop problem-solving skills.

Linear equations in two variables are linear polynomials with two unknowns. They are of the general form ax + by + c = 0, where x and y are the two variables, a and b are non-zero real numbers and c is a constant. The graphical representation of a linear equation in two variables is a straight line.

Linear Equations in Two Variables Questions with Solutions

Below are some practice questions on linear equations in two variables with detailed solutions.

Question 1: Solve for x and y:

Solution:

Put 1/x = u and 1/y = v. The given equations become

u/2 – v = –1 ⇒ u – 2v = – 2 ….(i)

u + v/2 = 8 ⇒ 2u + v = 16 ….(ii)

Multiplying equation (ii) by 2 on both sides and adding (i) and (ii), we get

(u + 4u) + ( –2v + 2v) = –2 + 32

⇒ 5u = 30

⇒ u = 6 ⇒ x = ⅙ and y = ¼

Question 2: Solve the system of linear equations 2x + 3y = 17 and 3x – 2y = 6 by the cross multiplication method.

Solution:

By cross multiplication

We get,

⇒ x/( –52) = y/( –39) = 1/( – 13)

⇒ x = 52/13 = 4 and y = 39/13 = 3

Hence x = 4 and y = 3 is the solution of given equations.

Question 3: Solve the following system of equations by substitution method:

2x + 3y = 0 and 3x + 4y = 5

Solution:

Given equations,

2x + 3y = 0 ….(i)

3x + 4y = 5 …..(ii)

From (i) we get, y = – 2x/3, substituting value of y in (ii), we get

3x + 4(–2x/3) = 5

⇒ 9x – 8x = 15

⇒ x = 15

Then y = (–2 × 15)/3 = – 10

Therefore, x = 15 and y = – 10 is solution of given system of equations.

Video Lesson on Consistent and Inconsistent Equations

Question 4: Find the value of k for which the given system of equations has infinitely many solutions: x + (k + 1)y = 5 and (k + 1)x + 9y + (1 – 8k) = 0.

Solution:

The given equations will have infinitely many solutions if a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, 1/(k + 1) = (k + 1)/9 = – 5/(1 – 8k)

Solving the equations we get k = 2.

Question 5: If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel. Find the value of k.

Solution:

If the lines are parallel, then they are inconsistent system of equations and a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Now, it should be 3/2 = 2k/5

⇒ k = 15/4

Then we have 2k/5 = 30/20 = 3/2, which satisfies the condition of inconsistency.

Question 6: Find the value of k for which the system of equations has a non-zero solution

5x + 3y = 0 and 10x + ky = 0

Solution:

The given equations are homogenous equations, they will have a non-zero solution if a₁/a₂ = b₁/b₂

Then, 5/10 = 3/k

⇒ 1/2 = 3/k

∴ k = 6

Question7: The monthly incomes of A and B are in the ratio 8:7 and their expenditures are in the ratio 19:16. If each saves ₹ 5000 per month, find the monthly income of each.

Solution:

Let the monthly incomes of A and B be 8x and 7x rupees respectively, and let their monthly expenditure be 19y and 16y rupees respectively.

Monthly savings of A = 8x – 19y = 5000 ….(i)

Monthly savings of B = 7x – 16y = 5000 ….(ii)

Multiplying (i) 16 and (ii) by 19 and subtracting (ii) from (i) we get

(16 × 8x – 19 × 7x) = 5000 (16 – 19)

⇒ 5x = 15000 ⇒ x = 3000

Monthly income of A is (8 × 3000) = ₹24,000

Monthly income of B is (7 × 3000) = ₹21,000

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Question 8: The sum of a two-digit number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the original number.

Solution:

Let the original number be (10x + y)

According to the question,

(10x + y) + (10y + x) = 99

⇒ 11(x + y) = 99

⇒ x + y = 9 ….(i)

And x – y = 3 ….(ii)

Adding equations (i) and (ii), we get,

⇒ 2x = 12

⇒ x = 6

and

⇒ y = 3

Hence the required number is 63.

Question 9: A man’s age is three times the sum of the ages of his two sons. After 5 years, his age will be twice the sum of his two son’s age. Find the age of the man.

Solution:

Let the age of the man be x and the sum of the ages of his two sons be y.

According to the question,

x = 3y ⇒ x – 3y = 0 ….(i)

And (x + 5) = 2(y + 5 + 5)

⇒ x – 2y = 15 ….(ii)

Subtracting equation (i) from (ii) we get

Y = 15 and from (i) x = 45.

The present age of the man is 45 years.

Question 10: A man can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours, Find his speed of rowing in still water. Also, find the speed of the stream.

Solution:

Let the speed of the man in still water be x km/hr and let the speed of the current be y kn/hr.

Speed in downstream = (x + y) km/hr

Speed in upstream = (x – y) km/hr

But speed in downstream = 20/2 km/hr = 10 km/hr

And speed in upstream = 4/2 km/hr = 2 km/hr

∴ x + y = 10 and x – y = 2

Solving both the equations we get;

x = 6 and y = 4.

Hence, the speed of the man in still water is 6 km/hr and the speed of the current is 4 km/hr.

Question 11: Find the four angles of a cyclic Quadrilateral ABCD in which ∠A = (2x – 1)^o, ∠B = (y + 5)^o, ∠C = (2y + 15)^o, and ∠D = (4x – 7)^o.

Solution:

We know that sum of opposite angles of a cyclic quadrilateral is 180^o

∴ ∠A + ∠C = 180^o and ∠B + ∠C = 180^o

∠A + ∠C = 180^o ⇒ (2x – 1) + (2y + 15) = 180^o

⇒ x + y = 83 ….(i)

∠B + ∠C = 180^o ⇒ (y + 5) + (4x – 7) = 180^o

⇒ 4x + y = 182 …..(ii)

Subtracting (i) from (ii) we get

3x = 182 – 83 ⇒ x = 33

Substituting in (i), we get y = 50

∴ ∠A = 2 × 33 – 1 = 65^o

∠B = 50 + 5 = 55^o

∠C = 2 × 50 + 15 = 115^o

∠D = 4 × 33 – 7 = 125^o.

Question 12: 8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find time taken by a man and a boy alone to finish the same work.

Solution:

Let 1 man can finish the work in x days and let 1 boy can finish the work in y days.

1 man’s one day work = 1/x

1 boy’s one day work = 1/y

According to the question,

8 men’s 1 day’s work + 12 boy’s one day’s work = ⅕

⇒ 8/x + 12/y = ⅕

⇒ 8u + 12v = ⅕ ….(i) where u = 1/x and v = 1/y

Similarly, 6u + 8v = 1/7 ….(ii)

On solving (i) and (ii) we get, x = 70 and y = 140

∴ One man alone can finish the work in 70 days and one boy alone can finish the work in 140 days.

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Practice Questions:

1. Five years ago Anna was three times older than Mira and ten years later Anna will be two times older than Mira. What are the present ages of Anna and Mira?

2. The difference of two numbers is 4 and the difference of their reciprocals is 4/21. Find the numbers.

3. Find the value of k for which the system of equations 5x – 3y = 0, and 2x + ky = 0 has a non-zero solution.

4. Find the value of a and b for which each of the following systems of linear equations

(a – 1)x + 3y = 2 and 6x + (1 – 2b)y = 6 has infinite number of solutions.