## 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

UC Berkeley Haas