Pair Of Linear Equations In Two Variables

Hello students, welcome to today's mathematics lesson. I'm so excited to be teaching you this important chapter from your NCERT textbook. Today we're going to learn about Pair of Linear Equations in Two Variables, which is Chapter 3 of your Class 10 Mathematics syllabus. This is a fundamental chapter that will help you solve many real-life problems, and I promise by the end of this lesson, you will have a complete understanding of every concept.

So let's begin with a real-life situation that you can relate to. Imagine there's a fair in your village. There's a Giant Wheel ride that costs ₹3 per ride, and there's a game called Hoopla that costs ₹4 per game. Now suppose your friend Akhila went to this fair. She played both the Giant Wheel and Hoopla. The number of times she played Hoopla was exactly half the number of rides she had on the Giant Wheel. And she spent a total of ₹20 on both the rides and the games. Can we find out how many rides she took and how many times she played Hoopla?

This is exactly the kind of problem we can solve using a pair of linear equations in two variables. Let me show you how to set this up.

Let the number of rides on the Giant Wheel be denoted by x, and the number of times she played Hoopla be denoted by y. Now, from the problem statement, we know two things. First, the number of Hoopla games is half the number of rides, so we can write this as y equals one-half of x, or y = (1/2)x. That's our first equation. Second, the total amount spent is ₹20, and each ride costs ₹3 while each game costs ₹4. So the total cost would be 3x + 4y, and this equals 20. That's our second equation: 3x + 4y = 20.

So we have two equations: y = (1/2)x and 3x + 4y = 20. Together, these form a pair of linear equations in two variables. Now the question is, how do we find the values of x and y that satisfy both these equations? That's exactly what this chapter teaches us. There are several methods to solve such pairs of equations, and we will learn all of them step by step.

Now let's move on to the first method: the Graphical Method.

When we have a pair of linear equations in two variables, each equation represents a straight line when we draw it on a graph. The solution to the pair of equations is simply the point where these two lines intersect, or meet. That's the basic idea behind the graphical method.

But before we go further, students, there are some important terms you need to understand. A pair of linear equations that has no solution is called an inconsistent pair of linear equations. On the other hand, a pair of linear equations that has at least one solution is called a consistent pair of linear equations. Now here's an interesting case: sometimes the two equations are actually equivalent to each other, meaning one equation can be obtained by multiplying the other equation by some constant. In such cases, the two lines coincide, which means they lie on top of each other. This gives us infinitely many common solutions because every point on the line is a solution to both equations. Such a pair is called a dependent pair of linear equations. And here's an important point to remember: a dependent pair of linear equations is always consistent because it does have solutions, in fact infinitely many solutions.

Now let's think about what can happen when we draw two lines on a graph. There are three possibilities. First, the lines may intersect each other at a single point. In this case, the pair of equations has a unique solution, meaning there is exactly one pair of values for x and y that satisfies both equations. This is a consistent pair. Second, the lines may be parallel to each other, meaning they never meet. In this case, the equations have no common solution at all. This is an inconsistent pair. Third, the lines may be coincident, meaning they lie exactly on top of each other. In this case, every point on the line is a common solution, so there are infinitely many solutions. This is a dependent or consistent pair.

Let me summarize what we just learned: intersecting lines mean exactly one solution, parallel lines mean no solution, and coincident lines mean infinitely many solutions. These three cases cover all possibilities when we have a pair of linear equations.

Now let's look at some examples to understand this better. Consider these three pairs of equations:

First pair: x – 2y = 0 and 3x + 4y – 20 = 0. When we draw these on a graph, the lines will intersect at one point.

Second pair: 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0. Notice that the second equation is exactly twice the first equation. So these two equations represent the same line. They are coincident.

Third pair: x + 2y – 4 = 0 and 2x + 4y – 12 = 0. These two lines are parallel to each other and will never meet.

Now, to understand this more systematically, let's compare the coefficients of these equations. Any linear equation in two variables can be written in the form ax + by + c = 0, where a, b, and c are constants. For a pair of equations, let's denote the first equation as a₁x + b₁y + c₁ = 0 and the second as a₂x + b₂y + c₂ = 0.

Now let's compare the ratios a₁/a₂, b₁/b₂, and c₁/c₂ for each pair.

For the first pair, x – 2y = 0 and 3x + 4y – 20 = 0, we have a₁ = 1, b₁ = -2, c₁ = 0 for the first equation, and a₂ = 3, b₂ = 4, c₂ = -20 for the second equation. So a₁/a₂ = 1/3, b₁/b₂ = -2/4 = -1/2, and c₁/c₂ = 0/(-20) = 0. Notice that a₁/a₂ is not equal to b₁/b₂. This tells us the lines intersect at a single point, and there is exactly one solution.

For the second pair, 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0, we have a₁ = 2, b₁ = 3, c₁ = -9 for the first equation, and a₂ = 4, b₂ = 6, c₂ = -18 for the second equation. So a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, and c₁/c₂ = (-9)/(-18) = 1/2. All three ratios are equal. This means the lines are coincident, and there are infinitely many solutions.

For the third pair, x + 2y – 4 = 0 and 2x + 4y – 12 = 0, we have a₁ = 1, b₁ = 2, c₁ = -4 for the first equation, and a₂ = 2, b₂ = 4, c₂ = -12 for the second equation. So a₁/a₂ = 1/2, b₁/b₂ = 2/4 = 1/2, and c₁/c₂ = (-4)/(-12) = 1/3. Here, a₁/a₂ equals b₁/b₂, but this is not equal to c₁/c₂. This means the lines are parallel and there is no solution.

So students, here's the rule we can state: For the pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, if the lines are intersecting, then a₁/a₂ is not equal to b₁/b₂. If the lines are coincident, then a₁/a₂ equals b₁/b₂ equals c₁/c₂. And if the lines are parallel, then a₁/a₂ equals b₁/b₂ but does not equal c₁/c₂. This is a very useful test to determine the nature of the solutions without actually drawing the graph. And the converse is also true: if you see these ratio relationships, you can predict whether the lines will intersect, be parallel, or be coincident.

Now let's work through some examples to see how this graphical method actually works in practice.

Example 1 asks us to check graphically whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent, and if so, solve them graphically.

To draw the graph of x + 3y = 6, we need to find two points that satisfy this equation. When x = 0, we get 3y = 6, so y = 2. When y = 0, we get x = 6. So the points (0, 2) and (6, 0) lie on this line.

For the second equation 2x – 3y = 12, when x = 0, we get -3y = 12, so y = -4. When y = 0, we get 2x = 12, so x = 6. So the points (0, -4) and (6, 0) lie on this line.

Now when we plot these points and draw both lines, we see that both lines pass through the point (6, 0). This is the point of intersection. So the solution is x = 6 and y = 0. This means the pair of equations is consistent, and there is exactly one solution.

Let me verify this by substituting x = 6 and y = 0 into both equations. For the first equation, x + 3y = 6 + 3(0) = 6, which equals 6, so it's satisfied. For the second equation, 2x – 3y = 2(6) - 3(0) = 12, which equals 12, so it's also satisfied. Perfect!

Now let's look at Example 2. We need to graphically find whether the pair of equations 5x - 8y + 1 = 0 and 3x - (24/5)y + (3/5) = 0 has no solution, unique solution, or infinitely many solutions.

At first glance, these equations look different. But let's simplify the second equation. If we multiply the second equation by 5/3, we get 5x - 8y + 1 = 0, which is exactly the same as the first equation! So both equations represent the same line. Therefore, they are coincident, and there are infinitely many solutions.

This is a great example that shows how sometimes two equations that look different might actually be equivalent. That's why it's important to check the ratios of coefficients.

Now let's look at a word problem that we can solve using the graphical method.

Example 3: Champa went to a sale to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased." Help her friends to find how many pants and skirts Champa bought.

Let the number of pants be x and the number of skirts be y. From the first statement, the number of skirts is two less than twice the number of pants, so y = 2x - 2. From the second statement, the number of skirts is four less than four times the number of pants, so y = 4x - 4.

Now let's find points for the first equation y = 2x - 2. When x = 2, y = 2(2) - 2 = 2. When x = 0, y = 2(0) - 2 = -2. So the points (2, 2) and (0, -2) lie on this line.

For the second equation y = 4x - 4, when x = 0, y = -4. When x = 1, y = 4(1) - 4 = 0. So the points (0, -4) and (1, 0) lie on this line.

When we draw these lines, they intersect at the point (1, 0). So the solution is x = 1 and y = 0. This means Champa bought 1 pant and 0 skirts. In other words, she bought only one pant and no skirts at all!

Let me verify this. If she bought 1 pant, then according to the first statement, the number of skirts would be 2(1) - 2 = 0, which matches. According to the second statement, the number of skirts would be 4(1) - 4 = 0, which also matches. So our solution is correct.

Now students, I want to pause here and summarize what we've learned so far about the graphical method. We learned that a pair of linear equations in two variables can be represented by two lines on a graph. The solution is the point where these lines intersect. We also learned the three possible cases: if the lines intersect, there is exactly one solution; if the lines are coincident, there are infinitely many solutions; and if the lines are parallel, there is no solution. We learned how to determine these cases by comparing the ratios of coefficients. And we worked through some examples to see how this works in practice.

Now let's move on to the algebraic methods. Sometimes the graphical method is not convenient, especially when the solution involves non-integer coordinates like square roots or fractions. In such cases, it's better to use algebraic methods. There are two main algebraic methods: the substitution method and the elimination method. Let's learn both.

The Substitution Method: As the name suggests, in this method we substitute the value of one variable from one equation into the other equation. Let me explain this with an example.

Example 4: Solve the following pair of equations by substitution method: 7x - 15y = 2 and x + 2y = 3.

Step 1: We pick one of the equations and express one variable in terms of the other. Looking at the two equations, the second one, x + 2y = 3, is simpler, so let's express x in terms of y. From x + 2y = 3, we get x = 3 - 2y.

Step 2: Now we substitute this value of x in the first equation. So we replace x with (3 - 2y) in 7x - 15y = 2. This gives us 7(3 - 2y) - 15y = 2. Simplifying, we get 21 - 14y - 15y = 2, which is 21 - 29y = 2. So -29y = 2 - 21 = -19. Therefore, y = 19/29.

Step 3: Now we substitute this value of y back into the expression we found for x. So x = 3 - 2(19/29) = 3 - 38/29 = (87/29) - (38/29) = 49/29.

So the solution is x = 49/29 and y = 19/29. We can verify this by substituting these values into both original equations. For the first equation, 7x - 15y = 7(49/29) - 15(19/29) = (343/29) - (285/29) = 58/29 = 2, which is correct. For the second equation, x + 2y = 49/29 + 2(19/29) = 49/29 + 38/29 = 87/29 = 3, which is also correct.

Now let me summarize the steps of the substitution method. First, find the value of one variable in terms of the other from either equation, whichever is more convenient. Second, substitute this value in the other equation to get an equation in one variable. Solve this equation to find the value of that variable. Third, substitute this value back into the expression from step 1 to find the value of the other variable.

Now let's look at another example that involves a real-life situation.

Example 5: This is an interesting problem. Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." We need to represent this situation algebraically and solve it by substitution.

Let s be Aftab's age and t be his daughter's age, both in years. Now let's translate the statements into equations.

Seven years ago, Aftab's age was s - 7, and his daughter's age was t - 7. At that time, Aftab was seven times as old as his daughter was then. So we have s - 7 = 7(t - 7). Simplifying, s - 7 = 7t - 49, which gives s - 7t + 42 = 0. That's our first equation.

Three years from now, Aftab's age will be s + 3, and his daughter's age will be t + 3. At that time, Aftab will be three times as old as his daughter will be. So we have s + 3 = 3(t + 3). Simplifying, s + 3 = 3t + 9, which gives s - 3t = 6. That's our second equation.

Now let's solve these by substitution. From the second equation, s - 3t = 6, we can write s = 3t + 6. Substitute this in the first equation: (3t + 6) - 7t + 42 = 0. Simplifying, -4t + 48 = 0, so 4t = 48, which gives t = 12. Then s = 3(12) + 6 = 36 + 6 = 42. So Aftab is 42 years old and his daughter is 12 years old.

Let me verify this. Seven years ago, Aftab was 42 - 7 = 35 years old, and his daughter was 12 - 7 = 5 years old. Indeed, 35 is 7 times 5. Three years from now, Aftab will be 42 + 3 = 45 years old, and his daughter will be 12 + 3 = 15 years old. Indeed, 45 is 3 times 15. Perfect!

Now let's look at some special cases in the substitution method.

Example 6: In a shop, the cost of 2 pencils and 3 erasers is ₹9, and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser.

Let the cost of one pencil be ₹x and the cost of one eraser be ₹y. Then the equations are: 2x + 3y = 9 and 4x + 6y = 18.

Now let's solve by substitution. From the first equation, 2x + 3y = 9, we can write x = (9 - 3y)/2. Substitute this in the second equation: 4((9 - 3y)/2) + 6y = 18. Simplifying, 4(9 - 3y)/2 + 6y = 18 gives us 2(9 - 3y) + 6y = 18, which is 18 - 6y + 6y = 18, which gives 18 = 18.

Now this is interesting. We got a statement that is always true, 18 = 18, regardless of the value of y. This means there is no unique solution. This situation arose because the second equation is actually just twice the first equation. In other words, both equations represent the same line. So there are infinitely many solutions. We cannot find a unique cost for a pencil and an eraser because there are many possible combinations that would satisfy both equations.

This is an important observation, students. When you substitute and get a true statement like 18 = 18 with no variables, it means the equations are dependent and have infinitely many solutions. When you get a false statement like 0 = 5, it means the equations are inconsistent and have no solution.

Let's verify this with Example 7.

Example 7: Two rails are represented by the equations x + 2y - 4 = 0 and 2x + 4y - 12 = 0. Will the rails cross each other?

Let's solve by substitution. From the first equation, x + 2y - 4 = 0, we get x = 4 - 2y. Substitute this in the second equation: 2(4 - 2y) + 4y - 12 = 0. Simplifying, 8 - 4y + 4y - 12 = 0 gives us 8 - 12 = 0, which is -4 = 0.

This is a false statement. This means the equations have no common solution. So the two lines are parallel and will never cross each other. In the context of the problem, the two rails will never meet.

Now students, let me recap what we've learned about the substitution method. We express one variable in terms of the other from one equation, substitute this into the second equation, solve for one variable, and then find the other variable. We also learned that if we get a true statement with no variables after substitution, the equations have infinitely many solutions. If we get a false statement with no variables, the equations have no solution.

Now let's move on to the second algebraic method: the Elimination Method.

The elimination method is sometimes more convenient than substitution, especially when the coefficients of one variable are already simple or can be made equal easily. The idea is to eliminate one variable by adding or subtracting the equations after making the coefficients equal.

Let me explain with an example.

Example 8: The ratio of incomes of two persons is 9:7, and the ratio of their expenditures is 4:3. If each of them manages to save ₹2000 per month, find their monthly incomes.

Let the incomes be 9x and 7x (since the ratio is 9:7). Let the expenditures be 4y and 3y (since the ratio is 4:3). Now, income minus expenditure equals savings. So for the first person, 9x - 4y = 2000. For the second person, 7x - 3y = 2000.

So we have the equations: 9x - 4y = 2000 and 7x - 3y = 2000.

Now let's use the elimination method. We want to eliminate one variable. Looking at the coefficients of y, we have -4 and -3. If we multiply the first equation by 3 and the second equation by 4, we get 27x - 12y = 6000 and 28x - 12y = 8000. Now the coefficients of y are both -12, which are equal.

Now if we subtract the first equation from the second, the y terms will cancel out. So (28x - 27x) + (-12y - (-12y)) = 8000 - 6000, which gives x = 2000.

Now substitute x = 2000 into the first equation: 9(2000) - 4y = 2000, which gives 18000 - 4y = 2000, so 4y = 18000 - 2000 = 16000, so y = 4000.

So the solution is x = 2000 and y = 4000. This means the incomes are 9x = 9(2000) = ₹18,000 and 7x = 7(2000) = ₹14,000. And the expenditures are 4y = 4(4000) = ₹16,000 and 3y = 3(4000) = ₹12,000. Let's verify: the first person saves 18000 - 16000 = ₹2000, and the second person saves 14000 - 12000 = ₹2000. Both save ₹2000, as given in the problem. Also, the income ratio is 18000:14000 = 9:7, and the expenditure ratio is 16000:12000 = 4:3. Perfect!

Now let me summarize the steps of the elimination method. First, multiply both equations by suitable non-zero constants to make the coefficients of one variable numerically equal. Second, add or subtract one equation from the other to eliminate that variable. If you get an equation in one variable, go to the next step. If you get a true statement with no variable, there are infinitely many solutions. If you get a false statement with no variable, there is no solution. Third, solve the equation in one variable to get its value. Fourth, substitute this value in either of the original equations to get the value of the other variable.

Now let's look at another example to see how the elimination method handles inconsistent equations.

Example 9: Use the elimination method to find all possible solutions of the following pair of linear equations: 2x + 3y = 8 and 4x + 6y = 7.

Let's multiply the first equation by 2 to make the coefficients of x equal. We get 4x + 6y = 16 and the second equation is 4x + 6y = 7. Now subtract the second equation from the first: (4x - 4x) + (6y - 6y) = 16 - 7, which gives 0 = 9.

This is a false statement. This means there is no solution to this pair of equations. The lines are parallel to each other.

Now let's look at a more complex example that involves a two-digit number.

Example 10: The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Let the tens digit be x and the units digit be y. Then the number is 10x + y. When we reverse the digits, we get 10y + x. According to the problem, (10x + y) + (10y + x) = 66. Simplifying, 11x + 11y = 66, which gives x + y = 6. That's our first equation.

We are also told that the digits differ by 2. This could mean either x - y = 2 or y - x = 2. Let's consider both cases.

Case 1: x - y = 2. We have x + y = 6 and x - y = 2. Adding these two equations, we get 2x = 8, so x = 4. Then y = 6 - x = 6 - 4 = 2. So the number is 10(4) + 2 = 42.

Case 2: y - x = 2, which means x - y = -2. We have x + y = 6 and x - y = -2. Adding these, we get 2x = 4, so x = 2. Then y = 6 - x = 6 - 2 = 4. So the number is 10(2) + 4 = 24.

So there are two such numbers: 42 and 24. Let's verify: 42 + 24 = 66, and the digits of 42 differ by 4 - 2 = 2. Also, 24 + 42 = 66, and the digits of 24 differ by 4 - 2 = 2. So both numbers satisfy the conditions.

Now students, I want to take a moment to summarize everything we've learned in this chapter.

We started with a real-life situation and learned how to represent it as a pair of linear equations in two variables. Then we learned two main methods for solving such pairs: the graphical method and the algebraic methods.

In the graphical method, we learned that the solution is the point where the two lines intersect. We learned about three cases: if the lines intersect at a point, there is exactly one solution and the equations are consistent. If the lines are coincident, there are infinitely many solutions and the equations are dependent but consistent. If the lines are parallel, there is no solution and the equations are inconsistent.

We also learned a quick test using the ratios of coefficients. For the pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, if a₁/a₂ is not equal to b₁/b₂, the lines intersect and there is a unique solution. If a₁/a₂ equals b₁/b₂ equals c₁/c₂, the lines are coincident and there are infinitely many solutions. If a₁/a₂ equals b₁/b₂ but does not equal c₁/c₂, the lines are parallel and there is no solution.

Then we learned the algebraic methods. The substitution method involves expressing one variable in terms of the other from one equation and substituting into the second equation. The elimination method involves making the coefficients of one variable equal and then adding or subtracting to eliminate that variable.

We saw that both methods can handle all three cases: unique solution, infinitely many solutions, and no solution. When we get a true statement with no variables, it means infinitely many solutions. When we get a false statement with no variables, it means no solution.

We also worked through several examples, including real-life problems involving ages, costs, incomes, and two-digit numbers. Each example helped reinforce the concepts and methods we learned.

This chapter is very important because the skills you learn here will be used in many other chapters in mathematics, including those involving quadratic equations and coordinate geometry. The methods of forming and solving linear equations are fundamental to problem-solving in mathematics.

So students, make sure you practice these methods thoroughly. Understand when to use substitution versus elimination. Remember the ratio test for determining the nature of solutions. And most importantly, always verify your solutions by substituting them back into the original equations.

That's all for today's lesson. Thank you for your attention, and keep practicing!