Demystifying Special Equations

Special Equations

• Let us begin this article with a very simple equation 2x + 3y = 24. 
  • 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:

• 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 to 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

• 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

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 \(2x+3y=120\)

\(\Rightarrow 2x=120-3y\)

\(\Rightarrow x=60-\frac{3y}{2}\)

• 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 

• 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, …, 3. 
    • 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.

• 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 \(4b=31-5g\)

\(\Rightarrow b=\frac{31-5g}{4}\)

\(\Rightarrow b=\frac{28+3-4g-g}{4}\)

\(\Rightarrow b=\frac{28-4g}{4}+\frac{3-g}{4}\)

\(\Rightarrow b=7-g+\frac{3-g}{4}\)

• 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)

• 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