Special Equation Let us begin this article with a very simple equation 2x + 3y = 2 Let’s say a question asks how many pairs of values of x and y will satisfy the equation How would you proceed? You might think to yourself that there are an infinite number of possible solutions to this problem For every value of x we put, we will get a different value of y For example: SE1.png And honestly, you won’t be wrong to think so Now, let us make a small change to the above equation Let’s say the question now asks how many pairs of positive integral values of x and y will satisfy the equation So now we have 2x + 3y = 24 such that x and y are positive integers. Do you still think there will be an infinite number of solutions? Let’s see this in detail. Let’s explore a couple of methods to find the solutions for this equation. Method 1: Hit and Trial (similar to the one we did above) As the name suggests, we will put different values of x (as per the constraint) and get corresponding values of y But since now we know some additional information about the variable (i.e., they are positive integers), a lot of the possibilities will get filtered out SE2.png There is no need to go further as the further values of y will all be negative So, we get three pairs of values of (x, y) and those are (3, 6), (6, 4), and (9, 2) This is exactly what makes an equation special A special equation is a type of equation in which we have a linear equation of two or more variables along with some constraints which make the equation solvable and we no longer have an infinite number of solutions Having gained an understanding of what a special equation is, we will now proceed to discuss the second method for finding the solution to such an equation Always Hit and Trial We previously used the hit-and-trial method to solve the equation 2x + 3y = 24 However, if the equation is more complex, such as 2x + 3y = 120, then this method may no longer be feasible. For such an equation, we may need to manipulate the equation in a certain way. Let's explore this further Method 2: Manipulate the Equation We have 2x+3y=120 ⇒2x=120−3y ⇒x=60-\frac{3y}{2} Because of the constraint of x and y is a positive integer, we can infer, from the above-obtained manipulation, that y has to be a multiple of 2 SE3.png Now, do we need to find out all these values? No. An interesting thing to notice is that: the value of x is getting decreased by the coefficient of y (i.e., 3) whereas the value of y is getting increased by the coefficient of x (i.e., 2) This pattern is not a coincidence and is observed in all cases of special equation ax + by = c where one value gets increased by the coefficient of the other and the other gets decreased So now if we go ahead with this pattern, we can say that y will have all the even values starting from 2 and the values of y will start from 57 and drop by 3 To find the maximum value of ‘y’ we can put x = 3 and we get y = 38. So the values of y will be 2, 4, 6, ..., 38, and respective values for x will be 57, 54, 51, ... It’s easy to see that there are 19 multiples of 2 in 2 to 38 so it must have 19 solutions. Form of Question Questions testing the concept of special equations will not be as straightforward as the ones we have shown above Instead of algebra, questions come in the form of word problems. Let us take one simple example Let’s solve such a question Question: There are a total of 31 chocolates. Each boy in the class gets 4 chocolates and each girl gets 5 chocolates. How many boys are there in the class? Solution: let us assume the number of boys and girls be b and g respectively Notice that b and g being positive integers is not given but it is obvious with context to the question Thus according to the question, we can form the equation 4b+5g=31 such that b and g are positive integers and we need the value of b We have 4b=31-5g ⇒b=\frac{31-5g}{4} ⇒b=\frac{28+3-4g-g}{4} ⇒b=\frac{28-4g}{4}+\frac{3-g}{4} ⇒b=7-g+\frac{3-g}{4} g=3 will make the term \frac{3-g}{4} an integer Therefore plugging g=3 , we get b=7-3+\frac{3-3}{4}=4 So, one of the values of b and g when 4b + 5g = 31 is b = 4 and g = 3 For further values, we can take advantage of the pattern observed earlier in the article b = 4 + 5 (adding the coefficient of g) = 9 g = 3 – 4 (subtracting the coefficient of b) = -1 (not valid) SE4.png We do not need to go any further because the value of g will keep getting negative like -9, -13, -17, etc Thus, the solution to the above question is that the number of boys in the class is 4 .

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Demystifying Special Equations

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Special Equation

Let us begin this article with a very simple equation $2x + 3y = 2$
- Let’s say a question asks how many pairs of values of x and y will satisfy the equation
- How would you proceed?
You might think to yourself that there are an infinite number of possible solutions to this problem
- For every value of x we put, we will get a different value of y
- For example:
  Attachment:
  
  SE1.png [ 8.12 KiB | Viewed 2794 times ]
- And honestly, you won’t be wrong to think so

Now, let us make a small change to the above equation
- Let’s say the question now asks how many pairs of positive integral values of x and y will satisfy the equation
- So now we have 2x + 3y = 24 such that x and y are positive integers.
- Do you still think there will be an infinite number of solutions?
- Let’s see this in detail.
Let’s explore a couple of methods to find the solutions for this equation.

Method 1: Hit and Trial (similar to the one we did above)

As the name suggests, we will put different values of x (as per the constraint) and get corresponding values of y
But since now we know some additional information about the variable (i.e., they are positive integers), a lot of the possibilities will get filtered out
Attachment:

SE2.png [ 10.12 KiB | Viewed 2780 times ]
There is no need to go further as the further values of y will all be negative
So, we get three pairs of values of (x, y) and those are (3, 6), (6, 4), and (9, 2)
This is exactly what makes an equation special

A special equation is a type of equation in which we have a linear equation of two or more variables along with some constraints which make the equation solvable and we no longer have an infinite number of solutions

Having gained an understanding of what a special equation is, we will now proceed to discuss the second method for finding the solution to such an equation

Always Hit and Trial

We previously used the hit-and-trial method to solve the equation 2x + 3y = 24
However, if the equation is more complex, such as 2x + 3y = 120, then this method may no longer be feasible.
For such an equation, we may need to manipulate the equation in a certain way. Let's explore this further

Method 2: Manipulate the Equation

We have $2x+3y=120$
$⇒2x=120−3y$
$⇒x=60-\frac{3y}{2}$
Because of the constraint of x and y is a positive integer, we can infer, from the above-obtained manipulation, that y has to be a multiple of 2
Attachment:

SE3.png [ 5.86 KiB | Viewed 2757 times ]
- Now, do we need to find out all these values? No.
An interesting thing to notice is that:
- the value of x is getting decreased by the coefficient of y (i.e., 3)
- whereas the value of y is getting increased by the coefficient of x (i.e., 2)
This pattern is not a coincidence and is observed in all cases of special equation ax + by = c where one value gets increased by the coefficient of the other and the other gets decreased
- So now if we go ahead with this pattern, we can say that y will have all the even values starting from 2 and the values of y will start from 57 and drop by 3
- To find the maximum value of ‘y’ we can put x = 3 and we get y = 38.
- So the values of y will be 2, 4, 6, ..., 38, and respective values for x will be 57, 54, 51, ...
  - It’s easy to see that there are 19 multiples of 2 in 2 to 38 so it must have 19 solutions.

Form of Question

Questions testing the concept of special equations will not be as straightforward as the ones we have shown above
- Instead of algebra, questions come in the form of word problems. Let us take one simple example
Let’s solve such a question

Question: There are a total of 31 chocolates. Each boy in the class gets 4 chocolates and each girl gets 5 chocolates. How many boys are there in the class?
Solution:
- let us assume the number of boys and girls be b and g respectively
  - Notice that b and g being positive integers is not given but it is obvious with context to the question
- Thus according to the question, we can form the equation $4b+5g=31$ such that b and g are positive integers and we need the value of b
- We have $4b=31-5g$
  $⇒b=\frac{31-5g}{4}$
  $⇒b=\frac{28+3-4g-g}{4}$
  $⇒b=\frac{28-4g}{4}+\frac{3-g}{4}$
  $⇒b=7-g+\frac{3-g}{4}$
- $g=3$ will make the term $\frac{3-g}{4}$ an integer
- Therefore plugging $g=3$, we get $b=7-3+\frac{3-3}{4}=4$
- So, one of the values of b and g when 4b + 5g = 31 is
  - b = 4 and g = 3
- For further values, we can take advantage of the pattern observed earlier in the article
  - b = 4 + 5 (adding the coefficient of g) = 9
  - g = 3 – 4 (subtracting the coefficient of b) = -1 (not valid)
    Attachment:
    
    SE4.png [ 6.75 KiB | Viewed 2687 times ]
- We do not need to go any further because the value of g will keep getting negative like -9, -13, -17, etc
- Thus, the solution to the above question is that the number of boys in the class is 4.

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