Tough questions on linear equations in two variables class 10

Tough and Hots questions on linear equations in two variables class 10.

Are you planning to practice Tough and Hots questions in two variables to score maximum marks in the board examination? Do you search Hots questions of linear equations in two variables? Do you find all important previous year's questions related to the equations in two variables? If yes, then you are at the right place.

I understand your problem of not being able to collect all important and previous year's questions related to equations in two variables because these questions are often asked with higher weightage, I have collected all the important and previous year's questions from last year's papers, reference book, sample papers. As for your knowledge, the solutions to all questions are digestible and there are no comparisons of ease at all.

In this article, I shall guide you on how to solve the word problems of linear equations in two variables. Our team made it easy to understand very soon. I hope solutions to all important questions provide unique ideas and clear all doubts regarding the equations. If you have any relevant doubts regarding these topics. You should immediately consult with our expert teams. Our team reach you soon.

You should know about the main concepts to understand the graph behaviour and the skills required to solve the word problems related to linear equations in two variables. These concepts are essential for a better understanding of two variable equations. No doubt, without concepts, any solutions to equations may be tough for you. 

How do you prepare Chapter 3 "linear equations in two variables"?

• Understand all the concepts related to it carefully.
• You can use your textbook or our platform for the best concepts in a descriptive manner.
• Collect all important and previous years' questions of linear equations in two variables, and try to solve them regularly.
• Prefer textbook for revisions and question practice for a reference book, sample papers or our online platforms.

The advantage of Solving all important questions of a Pair of linear equations in two variables.

• Now, case-based study questions are asked, which helps to prepare for that.
• All important and previous years' questions given the real experience of board-level and aware of board-level questions papers.
• It helps to understand the trend of equations in two variables questions in the board paper.
• It helps to understand which types of questions have to give more attention.
• It helps to be aware of the difficulty levels of questions.

Why should you solve all the important and previous year's questions of pair of linear equations in two variables?

There are several reasons to solve all important questions. some reasons can be expressed in words, and some are feelings while getting higher marks. Some points are given below.

• To maintain higher accuracy. 
• To develop the conceptual understanding.
• To develop critical thinking regarding word problems.
• To achieve higher marks in board examinations.
•  To build confidence which helps in the examination hall.

What is the right time to solve all important and the previous year's questions pair of linear equations in two variables?

• After completing a one-time revision of this chapter from the textbook.
• After September, solve all-important questions on a weekly or monthly basis.
• During the last month of your board examinations, prepare all previous year's questions.
• During the last minute of your board examination, prepare questions from the sample papers.

Where and How do find all important and previous year's questions of pair of linear equations in two variables for class 10th?

If you are looking for all the important and previous years' questions of pair of linear equations in two variables at one place. Then follow the simple steps which are given below.

• Open your browser and search aimathic.com.
• Navigate to Math Jano or look at the image below.

• various categories are shown on the sidebar, select the Linear equations,

So, learning step-by-step helps to reduce your stress regarding tough equations in two variables.

What do you mean by a Pair of linear equations in two variables?

A pair of linear equations in the form of ax+by+c=0 where a & b are not equal to zero. An equation in which two variables are present with the highest degree '1' and having a straight line in the graphical representation is known as a Linear equation in two variables.

For examples:

1. 2x+3y-2=0
2. 3x-4y=5
3. 2y-2=7x
4. 2a+3b=2
5. 3u-4v=7

Types of Pair of linear equations in two variables based on solutions.

1. Consistent System: A pair of linear equations is said to be a consistent system if it has at least one solution. It means, that after solving this type of equation, you get at least only one solution.
   • x-y=10 and x+y=20
   • 2x+3y=10 and x+3y=3
2. In-consistent System: A pair of linear equations is said to be an in-consistent system if it has no solution. It means, that after solving this type of equation, you get no solution.
• x+2y-4=0 and 2x+4y-12=0
• 2x+4y=10 and 3x+6y=12

The behaviour of lines of Pair linear equations in two variables in Graphical representation.

1. When two lines intersect at one point then equations are known as the consistent system and the behaviour of the equations in the graph is an intersecting line. 
   
   
   
2. When the two lines are parallel, they do not intersect however far they extend and equations are known as an inconsistent system. It means the behaviour of the equations in the graph is a parallel line.

   

   
   
3. When the two lines are coincident if one line overlaps the other line. That means it has many solutions. this type of Pair of linear equations in two variables is known as a consistent system. The behaviour of equations in the graph is coincident lines.

   

   
   

To Check the behaviour of lines in a graph of a pair of linear equations in two variables.

Let's linear equations of two variables  a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.

Graphical method to solve Linear equations in two variables.

1. Obtain the given system of a Pair of linear equations in x and y i.e. a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.
2. Draw the graphs of the equations. let the lines l₁ and l₂ represent the graphs of the equations.
3. If the lines l₁ and l₂ intersect at a point and (a, b) are the coordinates of an intersecting point, then the given system has a unique solution given by x=a, y=b. Or
4. If the lines l₁ and l₂ are coincident, then the system is consistent and it has infinitely many solutions. In this case, every solution of an equation is a solution of the system. Or
5. If the l₁ and l₂ are parallel, then the given system of the equations is inconsistent i.e. It has no solution.

Questions 2x-y-4=0 and x+y+1=0

we have,

⇒2x-y=4 ..........(1)

⇒x+y=1  ...........(2)

Take equation (1)

When x=1,

⇒2x-y=4

⇒y= -4+2x

⇒y=-4+2(1)

⇒y=-2

When x=2,

⇒2x-y=4

⇒y= -4+2x

⇒y= -4+2(2)

⇒y=-4 +4

⇒y= 0

Now, Take equation (2)

When x=1

⇒x+y= -1

⇒y= -1- x

⇒y= -1-(1)

⇒y= -2

When x= 2

⇒y= -1-x

⇒y= -1-2

⇒y= -3

Plots these points on the same graph and observe intersecting points on both equations' lines. The Intersecting point in the graph is the solution of these equations.

Hence, the solution of equations with the help of a graph is (1, -2).

In the graphical method, the solution of an equation is done by the above process. Solving a pair of linear equations in two variables is a simple process. Solve the equations by the above process, plot the graph and observe the intersecting point which is the solution of the equations.

Algebraic methods of solving a pair of linear equations in two variables.

Various methods to solve a pair of linear equations in two variables are given below.

1. Substitution method
2. Elimination method
3. Cross-multiplication method.

Substitution method to solve a pair of linear equations in two variables.

In this method, we express one of the variables in terms of the other variable from either of the two equations and then this expression puts the other equation to obtain one variable equation. Some steps help you to solve the equations by Substitution method.

Step 1: Obtain the two equations. let the equations be x+2y+1=0 and 2x-3y-12=0.

Solution

x+2y+1=0  ..........(1)

2x-3y-12=0  ...........(2)

Step 2: Choose either of the two equations, here I choose equation (1), and find the value of one variable x, in terms of the other i.e. y or vice versa.

From (1), we get

⇒ x = -1 - 2y 

Step 3: Substitute the value of x obtained in step (2), in the other equation means the equation (2) to get an equation in y. 

Step 4: Solve the equation obtained in Step (3) to get the value of y.

Substituting x = -1- 2y in (2), we get

⇒2(-1-2y) - 3y - 12=0 

⇒-2-4y-3y-12=0

⇒-2-7y-12=0

⇒-7y-14=0

⇒-7y=14

⇒y= -2

Step 5: Substitute the value of y obtained in Step (4), in the expression for x in terms of y obtained in Step (2) to get the value of x.

Put y= -2,

⇒x = -1 - 2(-2)

⇒x= -1 +4

⇒x= 3

Step 6: The values of x and y obtained in steps (4) and (5) respectively constitute the solution of the given system of two linear equations.

Elimination method by equating the coefficients to solve a pair of linear equations in two variables.

In this method, we eliminate one of the two variables to obtain an equation in one variable which can be easily solved. Put the obtained value of this variable in any one of the given equations, the value of the other variable can be obtained.

Step 1: Obtain the two equations. let the equations be 3x+2y=11 and 2x+3y=4.

Solutions: 

3x+2y=11  ................(1)

2x+3y=4  ..................(2)

Step 2: Multiply the equations to make the coefficients of the variable to be eliminated equal. It means multiplying by numbers such that both coefficients of the same variable(anyone variable term) will be the same in both equations.

Let's eliminate x from the given equations. The coefficients of x in the given equations are 2 and 3 respectively. The LCM of 2 and 3 is 6. so, we make the make coefficients of x equal to 6 in these equations.

Multiply (1) by 3 and (2) by 2, we get

⇒3(3x+2y)=3(11)

⇒9x+6y=33    .......... (3)

⇒2(2x+3y)=2(4)

⇒4x+6y= 8  ............(4)

Step 3: Add or subtract the equations obtained in Step 2 according to the terms having the same coefficients of opposite or of the same sign. It means you have to cancel out the same coefficient and variable term to convert equations to a single variable term.

Step 4: Solve the equation in one variable obtained in Step 3

Subtracting the eq..(4) from eq..(3)

Step 5: Substitute the value found in Step 4 in any one of the equations and find the value of the other variable. Put the known value of the variable in a given system of equations to find the value of another variable.

Put x= 5 in eq..(1)

⇒3(5)+2y=11

⇒15+2y=11

⇒2y=11-15

⇒2y= -4

⇒y= -2

Cross-Multiplication Method to solve a pair of linear equation

Theorem Let a₁x+b₁y+c₁=0

                 and a₂x+b₂y+c₂=0.

be a pair of linear of equations in two variable x and y such that a₁/a₂ ≠ b₁/b₂ i.e. a₁b₂-a₂b₁ ≠ 0. Then the pair has a unique solution given by

Now, Proof Cross-Multiplication Method The given pair of equations is:

a₁x+b₁y+c₁=0

a₂x+b₂y+c₂=0

Multiplying equation (1) by b₂, (2) by b₁ and subtracting, we get

⇒b₂(a₁x+b₁y+c₁) - b₁(a₂x+b₂y+c₂) = 0

⇒x(a₁b₂ - a₂b₁) = (b₁c₂ - b₂c₁)      ..........(1)

Multiplying equation (1)  by a₂, (2) by a₁ and subtracting, we get

⇒a₂(a₁x+b₁y+c₁) - a₁(a₂x+b₂y+c₂) = 0

⇒y(a₁b₂ - a₂b₂) = (c₁a₂ - c₂a₁)      ..........(2)

From equation (1) & (2)

Step 1: Obtain the two equations. Let x+y=7 and 5x+12y=7

Step 2: Shift all terms on LHS in the two equations to introduce zeroes on RHS. i.e, write the two equations in the following form:

a₁x+b₁y+c₁=0

a₂x+b₂y+c₂=0

Now, equations:

x + y -7 = 0

5x + 12y - 7 = 0

Step 3: In the above pair of equations, to obtain the solutions, write x, -y and 1 separated by equality signs as shown below:

The arrow between two numbers indicates that the numbers are to be multiplied.

Step 4: To obtain the denominators of x, -y and 1, multiply the numbers with a downward arrow and subtract the product of the numbers with an upward arrow from their product. 

Question 1 Determine the value of k so that the following linear equations have no solution:

(3k + 1)x + 3y - 2 = 0

(k² + 1)x + (k - 2)y - 5 = 0

[CBSE 2001]

Solution the given pair of linear equations will have no solution,

Question 2 Find the value of a so that the following pair of linear equations has infinitely many solutions:

2x + 3y = 7

(a - 1)x + (a + 2)y = 3a

[CBSE 2001]

Solution: the given pair of linear equations will have infinitely many solutions,

Question 3 The sum of digits of a two-digit number 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

[CBSE 2010]

Solution:

Let the unit digit be x and the tens digit be y,

the sum of digits of a two-digit number 15.

⇒y + x = 15  .........(1)

Similarly,

After reversing the number's order of digits, it exceeds 9.

⇒10y + x = 10x + y + 9

⇒10y + x - 10x - y = 9

⇒9y - 9x = 9

⇒y - x = 1  .........(2)

Using the substitution method,

From eq.(2)

⇒y = 1 + x

 Substitute the value of y in eq.(1)

⇒(1 + x) + x = 15

⇒1 +2 x = 15

⇒2x = 15 - 1

⇒2x = 14

⇒x = 7

 

Put the value of x in eq.(1)

⇒y + 7 = 15

⇒y = 15 - 7

⇒y = 8

If x = 7, y = 8 

The original number is 10x7 + 8 = 78

Question 4 A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

[CBSE 2005]

Solution

Let the unit digit be x and the tens digit be y,

Product of digits = 20

⇒ xy = 20  ...........(1)

Similarly,

If 9 is added, the number is obtained by interchanging their places.

⇒10y + x + 9 = 10x + y

⇒10y + x - 10x - y = -9

⇒9y - 9x = -9

⇒y - x = -1  

⇒ x - y = 1    ...........(2)

Using the substitution method,

From eq.(2)

⇒x = 1 + y

Put the value of x in eq.(1)

⇒y(1 + y) = 20

⇒y + y² = 20

⇒y² + y - 20 = 0

⇒y² + y - 20 = 0

⇒y² +5y - 4y - 20 = 0

⇒y (y + 5) - 4(y + 5) = 0

⇒ (y + 5 ) (y - 4).

⇒y = -5, 4

Here we take y = 4

Put y = 4 in eq.(1)

⇒x(4) = 20

⇒x = 5

If x = 5 and y = 4, the original number is 45.

Question 5 The denominator of a fraction is 4 more than twice the numerator, when both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine the fraction.

[CBSE 2005]

Solution:

Let the numerator and denominator of the fraction be x and y respectively,

Then,

The fraction = x/y

It is given that

Denominator = 2 (Numerator) + 4

⇒y = 2x + 4

⇒y - 2x = 4    .............(1)

According to the second given conditions, we have.

⇒y - 6 = 12(x - 6)

⇒y - 6 = 12x - 72

⇒y - 6 - 12x + 72 = 0

⇒y - 12x + 66 = 0

⇒y - 12x = -66   ..................(2)

Using the substitution method,

From eq.(2)

⇒y = -66 + 12x

 Put this in eq.(1)

⇒(-66 + 12x) - 2x = 4

⇒-66 + 12x -2x = 4

⇒10x = 4 + 66

⇒10x = 70

⇒ x = 7

Put x = 7 in equ.(1)

⇒y - 2(7) = 4

⇒y - 14 = 4

⇒y = 4 + 14

⇒y = 18

If x = 7 and y = 18, then the fraction will be 7/18.

Question 6 The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2:3. Determine the fraction.

[CBSE 2001C, 2010]

Solution:

Let the numerator and denominator of the fraction be x and y respectively,

Then,

The fraction = x/y

It is given that

Denominator = 2 (Numerator) + 4

⇒x + y = 2x + 4

⇒y - x = 4    .............(1)

According to the second given conditions, we have.

Using the substitution method,

From eq.(1)

⇒y = 4 + x

Put this value in eq.(2)

⇒3x - 2(4 + x) = -3

⇒3x - 8 - 2x = -3

⇒3x - 2x = -3 + 8

⇒x = 5

Put x = 5 in eq.(2)

⇒3(5) - 2y = -3

⇒15 - 2y = -3

⇒-2y = -3 -15

⇒-2y = -18

⇒y = 9

If x = 5 and y = 9, the fraction will be 5/9.

Question 7 The Father's age is three times the sum of the ages of his two children. After 5 years his age will be twice the sum of the ages of the two children. Find the age of the father.

[CBSE 2003]

Solution

Let the present age of the father be x years and the present age of the two children be y and z years.

It is given that 

⇒x = 3(y + z)

⇒y + z = x/3   ...........(1)

After the 5 years, the father's age will be (x + 5) years and the children's age will be (y + 5) and (z + 5) years.

⇒ x + 5 = 2{(y + 5) + (z + 5)}

⇒x + 5 = 2y + 10 + 2z + 10

⇒x = 2y + 2z + 15   

⇒x = 2(y + z) +15...........(2)

Using the substitution method,

Put the value y + z  in eq.(2)

 The present age of the Father is 45.

Question 8 The father was five times as old as his son two years ago. Two years later, his age will be 8 more than three times the age of his son. Find the present ages of the father and son.

[CBSE 2004]

Solution

Let the present age of the father be x and the age of his son be y,

It is given that,

⇒x - 2 = 5(y - 2)

⇒x - 2 = 5y - 10

⇒x - 5y = -10 + 2

⇒x - 5y = -8    ................(1)

 

Two years later,

⇒x + 2 = 3(y +2) + 8

⇒x + 2 = 3y + 6 + 8

⇒x - 3y = 14 - 2

⇒ x - 3y = 12    .............(2)

Using the substitution method,

From eq.(2)

⇒x = 12 + 3y

Put this value in eq.(1)

⇒(12 + 3y) - 5y = -8

⇒12 + 3y -5y = -8

⇒12 - 2y = -8

⇒-2y = -8 -12

⇒-2y = 20

⇒y = 10

Put y=10 in eq.(2)

⇒x - 3(10) = 12

⇒x - 30 = 12

⇒x = 30 + 12

⇒x = 42

The father's present age is 42 years and the son is 10 years.

Question 9 Places A and B are 100km apart on the highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. what are the speeds of two cars?

[NCERT, CBSE 2009]

Solution

Let X and Y be two cars starting from points A and B respectively.

Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case1: When two cars move in the same direction:

Suppose two cars meet at point Q, then,

Distance travelled by car X = AQ

Distance travelled by car Y = BQ

It is given that two cars meet in 5 hours. 

Distance travelled by car X in 5 hours = 5x km , AQ = 5x

Distance travelled by car Y in 5 hours = 5y km BQ = 5y

Clearly AQ - BQ = AB 5x - 5y = 100  

⇒x - y = 20    .........(1)

Case 2: When two cars move in the opposite direction

Suppose two cars meet at a point P, then

Direction travelled by car X = AP

Direction travelled by car Y = BP

In this case, two cars meet in 1 hour

Therefore,

Distance travelled by car X in 1 hour = 1x km

Distance travelled by car Y in 1 hour = 1y km

AP + BP = AB

⇒1x + 1y = 100

⇒x + y =100    ..........(2)

Using the substitution method,

From eq.(2)

⇒x = 100 - y

 Put this value in eq.(1) we get

⇒(100 - y) - y = 20

⇒100 - y - y = 20

⇒ 100 - 2y = 20

⇒-2y = 20 - 100

⇒-2y = - 80

⇒y = 40

Put y = 40 in eq.(1), we get

⇒x - 40 = 100

⇒x = 100 - 40

⇒x = 60

Hence, The speed of car X is 60km/hr, and The speed of car Y is 40km/hr.

Question 10 Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours and 30 minutes. but if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. find the speed of the train and that of the taxi.

[CBSE 2006]

Solution

Let the speed of the train be x km/hr and the speed of the taxi be y km/hr.

Case 1: When Abdul travels 300 km by train and 200 km by taxi,

The time taken by Abdul to travel 300 km by train = 300/x hrs

The time taken by Abdul to travel 200 km by taxi = 200/y hrs

Total time taken by Abdul to cover 500 km = 300/x +200/y

It is given that the total time taken is 5 hours and 30 minutes

From Case 1, we can write

Case 2: When Abdul travel travels 260 km by train and 240 km by taxi,

Time taken by Abdul to travel 260 km by train = 260/x hrs

Time taken by Abdul to travel 240 km by taxi = 240/y hrs

In this case, the total time of the journey is 5 hours and 36 minutes

The Speed of a train is 100 km/hr and the speed of a taxi is 80 km/hr.

Question 11 The value of k for which the pair of linear equations 5x + 2y - 7 = 0 and 2x + ky + 1 =0 don't have a solution, is:

[CBSE 2024]

Solution

 The given pair of linear equations will have no solution,

Question 12 Solve the following pair of linear equations for x and y algebraically:

x + 2y = 9 and y - 2x = 2

[CBSE 2024]

Solution

Given,

⇒x + 2y = 9     ................(1)

⇒y - 2x = 2     ..................(2)

Using the Substitution method:

⇒x = 9 - 2y 

Put this value in eq.(2)

⇒y - 2(9 - 2y) = 2

⇒y - 18 + 4y = 2

⇒5y = 2 + 18

⇒5y = 20

⇒y = 4

Put y = 4 in eq. (2)

⇒4 - 2x = 2

⇒-2x = 2 - 4

⇒-2x = -2

⇒x = 1

Question 13 Check whether the point (-4, 3) lies on both the lines represented by the linear equations x + y + 1 = 0 and x - y = 1.

[CBSE 2024]

Solution

Given,

⇒x + y = -1  .............(1)

⇒x - y = 1    ..............(2)

Now,

Put x = -4 in equ. (1), we get

⇒-4 + y = -1

⇒y = -1 +4

⇒y = 3

Similarly,

Put x = -4 in eq.(2) we get

⇒-4 - y = 1

⇒-y = 1 + 4

⇒-y = 5

⇒y = -5

From the above conclusion, we can say that (-4, 3) is the solution of x + y +1 but (-4, 3) is not the solution of x - y = 1.

It means (-4, 3) lies on the x + y + 1.

Question 14 The value of k for which the pair of linear equations x + y - 4 = 0 and 2x + ky - 8 = 0 has infinitely many solutions, is

[CBSE 2024]

Solution

The given pair of linear equations will have infinitely many solutions,

Question 15 The sum of two natural numbers is 70 and their difference is 10. Find the natural numbers.

[CBSE 2024]

Solution

Let two natural numbers are x and y,

Case 1:

⇒x + y =70    .........(1)

⇒x - y = 10    ..........(2)

Using the Substitution method,

From the equ.(2)

⇒x = 10 + y

Put this value in equ. (1)

⇒(10 + y) + y = 70

⇒2y + 10 = 70

⇒2y = 70 -10

⇒2y = 60

⇒y = 30

Put y = 30 in equ.(1)

⇒x - 30 = 10

⇒x = 10 + 30

⇒x = 40

The natural numbers are 30 and 40.

Question 16 Solve for x and y:

x - 3y = 7

3x - 3y = 5

[CBSE 2024]

Solution

Given,

⇒x - 3y = 7 .................(1)

⇒3x - 3y = 5    .............(2)

Using the substitution method,

From equ. (1)

⇒x = 7 +3y

Put this value in equ. (2), we get

⇒3(7 + 3y) - 3y = 5

⇒21 + 9y - 3y = 5

⇒21 + 6y = 5

⇒6y = 5 - 21

⇒6y = -16

⇒y = -16/6

⇒y = -8/3

Put y = -8/3 in eq.(2), we get

Question 17 The pair of linear equations 2kx + 5y =7, 6x + 5y = 11 have a unique solution, if

[CBSE 2024]

Solution

The given pair of linear equations will have a unique solution,

Question 18 The value of k for which the pair of equations kx = y + 2 and 6x = 2y + 3 has infinitely many solutions,

[CBSE 2024]

Solution

The given pair of linear equations will have infinitely many solutions,

⇒kx - y - 2 = 0

⇒6x - 2y - 3 = 0

Question 19 Solve the pair of equations x = 3 and y = -4 graphically.

[CBSE 2024]

Solution

Given,

⇒x = 3 and y = 0 ................. (1)

⇒ y = -4 and x = 0 ...............(2)

From the above conditions:

The solution of equations is (3, -4) using the graphical method.

Question 20 Using the graphical method, find whether the following system of linear equations is consistent or not:  x =0 and y = -7.

[CBSE 2024]

Solution

Given,

⇒ x = 0, y = -7

⇒y = -7, x = 0

The solution of equations is (0, -7) using the graphical method and has at least one solution.

 So, these equations are consistent.

Question 21 If the pair of equations 3x - y + 8 = 0 and 6x - ry + 16 = 0 represent coincident lines, then the value of r is :

[CBSE 2024]

Solution

The given pair of linear equations is a coincidental line. that means, it will have infinitely many solutions.

Question 22 The pair of linear equations as ax + 2y = 9 and 3x + by = 18 represent parallel lines, where a, and b are integers, if:

[CBSE 2024]

Solution

The given pair of linear equations will have no solution,

Conclusion  

I have provided tough questions about linear equations in two variable and digestible solutions. I hope you learn easily and enjoy it. Most of the questions are based on the algebraic methods that are listed above in detail. If you have any doubts regarding solutions. you can contact us immediately. Our expert reach you very soon.