4.3.2. Finding Multiple Solutions

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 = (2 × 3) + (3 × 2) = 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...

In this chunk, we learn about solutions in linear equations with two variables. A linear equation can have multiple solutions rather than just one, as seen in equations involving only one variable. In our specific example, the equation 2x + 3y = 12 can be satisfied by different pairs of (x, y). Here, for x = 3, y must be 2 to satisfy the equation, while for x = 0, y must be 4. Written as ordered pairs, these solutions appear as (3, 2) and (0, 4). The term 'ordered pair' signifies the sequence they must be considered in: the first number refers to x, and the second refers to y.

Imagine scoring in basketball, where x represents points scored by one player and y by another. The total points scored need to equal 12. If player A scores 3 points, then player B must score 2 points to reach the total of 12 (as in our example). However, if player A scores 0 points, player B would need to score all 12 points! Just like this basketball game, many combinations of scores (or solutions) can reach the total goal.

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 gives you y = 8/3, providing another solution (2, 8/3) for 2x + 3y = 12. Similarly, choosing x = -5 leads us to y = 22/3, yielding another solution (-5, 22/3). 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.

In this chunk, we continue exploring how to find different solutions for the linear equation 2x + 3y = 12. Starting with already known solutions, we can substitute different x values to yield corresponding y values. Choosing x = 2 gives us y = 8/3, creating a new ordered pair (2, 8/3). Similarly, if we select x = -5, we can find another y value of 22/3, leading us to the ordered pair (-5, 22/3). This illustrates that we can generate an infinite number of pairs (solutions) just by varying x.

Consider a garden where you can plant various types of flowers in certain pairs (like x flowers of type A and y flowers of type B) while aiming for a specific total number of flowers. If you plant 2 of type A, you can calculate exactly how many of type B you still need for the total. However, you could also decide to plant fewer of type A and adjust the number of type B accordingly. As a result, many combinations will still achieve your target total, much like finding countless solutions in our equation.

Example 3: Find four different solutions of the equation x + 2y = 6. By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2, x + 2y = 2 + 4 = 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).

In this final chunk, we review how we can derive multiple solutions for the equation x + 2y = 6. Starting by plugging in x = 2, we quickly determine corresponding values of y, achieving a solution of (2, 2). If we instead set x to 0, we find y must equal 3, forming another solution (0, 3). Each of these steps reveals a new pair until we have reached four distinct solutions. This method illustrates a pattern and shows the systematic approach to finding multiple solutions.

Think of organizing a meeting where there are a total of six participants represented by x (individuals) and pairs of tasks represented by y (each requiring two people). Depending on how you assign individuals to tasks, you can achieve a viable combination of participants to fulfill each role, resulting in multiple configurations. Just like our equation, each arrangement leads to a unique outcome however we describe it!

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