## 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

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

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

- 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

- 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

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## Special Equations

2x + 3y = 24.Method 1:Hit and Trial (similar to the one we did above)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?

Method 2:Manipulate the Equation\(\Rightarrow 2x=120-3y\)

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

y has to be a multiple of 2## Form of 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:\(\Rightarrow b=\frac{31-5g}{4}\)

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

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

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

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