Welcome dear students! Today we are going to learn about Linear Equations in Two Variables from Class 9 Maths. The principal use of the Analytic Art is to bring Mathematical Problems to Equations and to exhibit those Equations in the most simple terms that can be. This profound thought was shared by Edmund Halley.
In earlier classes, you have studied linear equations in one variable. Can you write down a linear equation in one variable? You may say that x + 1 = 0, x + 2 = 0 and 2y + 3 = 0 are examples of linear equations in one variable. You also know that such equations have a unique, that is, one and only one solution. You may also remember how to represent the solution on a number line. In this chapter, the knowledge of linear equations in one variable shall be recalled and extended to that of two variables. You will be considering questions like: Does a linear equation in two variables have a solution? If yes, is it unique? What does the solution look like on the Cartesian plane? You shall also use the concepts you studied in Chapter 3 to answer these questions.
[CHECKPOINT]
Let us first recall what you have studied so far. Consider the following equation: 2x + 5 = 0. Its solution, i.e., the root of the equation, is -5/2. This can be represented on the number line as shown in Figure 4.1. In this figure, we see a horizontal number line with a marked origin at zero. A point is marked to the left of zero, exactly halfway between -2 and -3, representing the value -2.5. While solving an equation, you must always keep the following points in mind. The solution of a linear equation is not affected when the same number is added to or subtracted from both the sides of the equation. Also, the solution is not affected when you multiply or divide both the sides of the equation by the same non-zero number.
Let us now consider the following situation. In a One-day International Cricket match between India and Sri Lanka played in Nagpur, two Indian batsmen together scored 176 runs. Express this information in the form of an equation. Here, you can see that the score of neither of them is known, i.e., there are two unknown quantities. Let us use x and y to denote them. So, the number of runs scored by one of the batsmen is x, and the number of runs scored by the other is y. We know that x + y = 176, which is the required equation. This is an example of a linear equation in two variables. It is customary to denote the variables in such equations by x and y, but other letters may also be used.
[CHECKPOINT]
Some examples of linear equations in two variables are: 1.2s + 3t = 5, p + 4q = 7, πu + 5v = 9 and 2x – 7y = 3. Note that you can put these equations in the form 1.2s + 3t – 5 = 0, p + 4q – 7 = 0, πu + 5v – 9 = 0 and 2x – 7y – 3 = 0, respectively. So, any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables. This means that you can think of many such equations.
Let us now work through Example 1. Write each of the following equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case. Part (i): 2x + 3y = 4.37. This can be written as 2x + 3y – 4.37 = 0. Here a = 2, b = 3 and c = –4.37. Part (ii): x – 4 = 3y. This can be written as x – 3y – 4 = 0. Here a = 1, b = –3 and c = –4. Part (iii): 4 = 5x – 3y. This can be written as 5x – 3y – 4 = 0. Here a = 5, b = –3 and c = –4. Do you agree that it can also be written as –5x + 3y + 4 = 0? In this case a = –5, b = 3 and c = 4. Part (iv): 2x = y. This can be written as 2x – y + 0 = 0. Here a = 2, b = –1 and c = 0. Equations of the type ax + b = 0 are also examples of linear equations in two variables because they can be expressed as ax + 0.y + b = 0. For example, 4 – 3x = 0 can be written as –3x + 0.y + 4 = 0.
[CHECKPOINT]
Now, Example 2. Write each of the following as an equation in two variables. Part (i): x = –5. This can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0. Part (ii): y = 2. This can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0. Part (iii): 2x = 3. This can be written as 2x + 0.y – 3 = 0. Part (iv): 5y = 2. This can be written as 0.x + 5y – 2 = 0.
Let us now move to Exercise 4.1. Question 1: The 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. Solution: The cost of a notebook is x and the cost of a pen is y. Since the notebook costs twice as much as the pen, we write x = 2y. To put this in standard form, we subtract 2y from both sides, giving x – 2y = 0. This is the required linear equation in two variables.
Question 2: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case. (i) 2x + 3y = 9.35. Rewriting gives 2x + 3y – 9.35 = 0. Here a = 2, b = 3, c = –9.35. (ii) x – 5y – 10 = 0. It is already in the required form. Here a = 1, b = –5, c = –10. (iii) –2x + 3y = 6. Rewriting gives –2x + 3y – 6 = 0. Here a = –2, b = 3, c = –6. (iv) x = 3y. Subtracting 3y from both sides gives x – 3y = 0. Adding zero for c, we get x – 3y + 0 = 0. Here a = 1, b = –3, c = 0. (v) 2x = –5y. Adding 5y to both sides gives 2x + 5y = 0. In standard form: 2x + 5y + 0 = 0. Here a = 2, b = 5, c = 0. (vi) 3x + 2 = 0. We write this with a y term: 3x + 0.y + 2 = 0. Here a = 3, b = 0, c = 2. (vii) y – 2 = 0. We write this with an x term: 0.x + y – 2 = 0. Here a = 0, b = 1, c = –2. (viii) 5 = 2x. Subtracting 2x from both sides gives –2x + 5 = 0. Adding the y term: –2x + 0.y + 5 = 0. Here a = –2, b = 0, c = 5.
[CHECKPOINT]
Section 4.3: Solution of a Linear Equation. You have seen that every linear equation in one variable has a unique solution. What can you say about the solution of a linear equation involving two variables? As there are two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation. Let us consider the equation 2x + 3y = 12. Here, x = 3 and y = 2 is a solution because when you substitute x = 3 and y = 2 in the equation above, you find that 2x + 3y equals 2 × 3 + 3 × 2, which equals 6 + 6, giving 12. This solution is written as an ordered pair (3, 2), first writing the value for x and then the value for y. Similarly, (0, 4) is also a solution for the equation above. On the other hand, (1, 4) is not a solution of 2x + 3y = 12, because on putting x = 1 and y = 4 we get 2x + 3y equals 2 × 1 + 3 × 4, which is 2 + 12, giving 14, which is not 12. Note that (0, 4) is a solution but not (4, 0).
You have seen at least two solutions for 2x + 3y = 12, i.e., (3, 2) and (0, 4). Can you find any other solution? Do you agree that (6, 0) is another solution? Verify the same. In fact, we can get many solutions in the following way. Pick a value of your choice for x, say x = 2, in 2x + 3y = 12. Then the equation reduces to 4 + 3y = 12, which is a linear equation in one variable. On solving this, you get 3y = 8, so y = 8/3. So the ordered pair (2, 8/3) is another solution of 2x + 3y = 12. Similarly, choosing x = –5, you find that the equation becomes –10 + 3y = 12. This gives 3y = 22, so y = 22/3. So, the ordered pair (–5, 22/3) is another solution of 2x + 3y = 12. So there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.
[CHECKPOINT]
Let us look at Example 3. Find four different solutions of the equation x + 2y = 6. Solution: By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2, x + 2y equals 2 + 4, which equals 6. Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6. Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the given equation. So four of the infinitely many solutions of the given equation are: (2, 2), (0, 3), (6, 0) and (4, 1). Remark: Note that an easy way of getting a solution is to take x = 0 and get the corresponding value of y. Similarly, we can put y = 0 and obtain the corresponding value of x.
Now, Example 4. Find two solutions for each of the following equations. Part (i): 4x + 3y = 12. Taking x = 0, we get 3y = 12, i.e., y = 4. So, (0, 4) is a solution of the given equation. Similarly, by taking y = 0, we get 4x = 12, so x = 3. Thus, (3, 0) is also a solution. Part (ii): 2x + 5y = 0. Taking x = 0, we get 5y = 0, i.e., y = 0. So (0, 0) is a solution of the given equation. Now, if you take y = 0, you again get (0, 0) as a solution, which is the same as the earlier one. To get another solution, take x = 1, say. Then you can check that the corresponding value of y is found by solving 2(1) + 5y = 0, which gives 5y = –2, so y = –2/5. So the ordered pair (1, –2/5) is another solution of 2x + 5y = 0. Part (iii): 3y + 4 = 0. Writing the equation 3y + 4 = 0 as 0.x + 3y + 4 = 0, you will find that y = –4/3 for any value of x. Thus, two solutions can be given as (0, –4/3) and (1, –4/3).
[CHECKPOINT]
Let us proceed to Exercise 4.2. Question 1: Which 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. Solution: The correct option is (iii) infinitely many solutions. This is because y = 3x + 5 is a linear equation in two variables. For every real number value of x that we choose, we get a corresponding real number value for y. Since there are infinitely many real numbers, there are infinitely many pairs (x, y) that satisfy the equation.
Question 2: Write four solutions for each of the following equations. (i) 2x + y = 7. Let x = 0, then y = 7. Solution: (0, 7). Let x = 1, then 2 + y = 7, so y = 5. Solution: (1, 5). Let x = 2, then 4 + y = 7, so y = 3. Solution: (2, 3). Let x = 3, then 6 + y = 7, so y = 1. Solution: (3, 1). (ii) πx + y = 9. Let x = 0, then y = 9. Solution: (0, 9). Let x = 1, then π + y = 9, so y = 9 – π. Solution: (1, 9 – π). Let x = –1, then –π + y = 9, so y = 9 + π. Solution: (–1, 9 + π). Let x = 2, then 2π + y = 9, so y = 9 – 2π. Solution: (2, 9 – 2π). (iii) x = 4y. Let y = 0, then x = 0. Solution: (0, 0). Let y = 1, then x = 4. Solution: (4, 1). Let y = 2, then x = 8. Solution: (8, 2). Let y = –1, then x = –4. Solution: (–4, –1).
[CHECKPOINT]
Question 3: Check which of the following are solutions of the equation x – 2y = 4 and which are not. (i) (0, 2). Substitute x = 0, y = 2. Left side is 0 – 2(2) = –4. This is not equal to 4. So (0, 2) is not a solution. (ii) (2, 0). Substitute x = 2, y = 0. Left side is 2 – 2(0) = 2. This is not equal to 4. So (2, 0) is not a solution. (iii) (4, 0). Substitute x = 4, y = 0. Left side is 4 – 2(0) = 4. This equals the right side. So (4, 0) is a solution. (iv) (√2, 4√2). Substitute x = √2, y = 4√2. Left side is √2 – 2(4√2) = √2 – 8√2 = –7√2. This is not equal to 4. So this pair is not a solution. (v) (1, 1). Substitute x = 1, y = 1. Left side is 1 – 2(1) = –1. This is not equal to 4. So (1, 1) is not a solution.
Question 4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. Solution: Substitute x = 2 and y = 1 into the equation. We get 2(2) + 3(1) = k. Calculating this gives 4 + 3 = k, so k = 7.
[CHECKPOINT]
Section 4.4: Summary. In this chapter, you have studied the following points. First, an equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables. Second, a linear equation in two variables has infinitely many solutions. Third, every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]