Special Equations

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So far we have discussed about simultaneous linear equations in many variables. In this special module we are going to consider the extended properties of simple equations, which is generally called 'Special Equations'.

Objectives of this article is:
  • An in-depth understanding about the system of equations.
  • Determinate System.
  • Indeterminate system.
  • Different versions of indeterminate systems of simple equations.
  • Nature of solutions of various special equations.
  • Solving processes of special equations.

 

If we want to group a set of simple equations, then there are mainly two classifications as per the nature of the system.
  • Determinate System of simple equations.
  • Indeterminate system of simple equations.

Determinate system

If we can find out a unique set of solutions (unique values for variables which are presented in the system) then the system is called determinate system. The minimum required condition for such a system is " the number of equations in the system is equal to the number of variables presented in the system".

i.e. If we are considering a system of two variable simple equations, then there should be minimum two equations for getting a unique value for the variables.

Note: The number of simple equations is equal to the number of variables presented in the system is the minimum condition only to get a unique solution, means even if the system satisfies the mentioned condition then also there are chances for not getting a unique solution. i.e. There were many solutions or no solution and it is depends upon the consistency of the system, which we discussed earlier in this chapter.

Indeterminate system

If the number of equations in a system is less than the number of variables presented then it is an indeterminate system. Hence it is not possible to get a unique solution.

Example:

Solve, x + y = 5

Here it is only one equation in two unknowns.

When x = 0, then y = 5.
When x = 1, then x = 4
When x = - 1 , then y = 6, and so on.

Similarly we can find out infinitely many solutions for this equation.

In this particular section, we are going to consider the properties and nature of solutions of indeterminate system of equations. These kinds of equations generally referred as "Special Equations".

This is one among the tricky concept under the Quantitative Aptitude syllabus of Competitive entrance exams. Here we are mainly considering the special equations in first degree(simple equations).

Type I : All of the coefficients equal to '1'

This is basic form of special equations. This same concept dealt in the topic 'Permutations and Combinations', we can recall those results too. Just go through the following illustrated examples.

Example 1

Find the number of natural number solutions of the equation; x + y = 10.

Solution:

It is possible to catch all of the solutions through continuous substitutions.

When x = 1, y = 9

Here x = 1, is the least possible natural number value for x and y = 9, is the largest possible natural number value of y. In the similar way, we can find all of the possible values for x and y, which are given below.

xy
1 9
2 8
3 7
4 6
5 5
6 4
7 3
8 2
9 1

i.e. There are 9 such pairs of natural number solutions are possible.

Task for you!!!  Find the number of natural number solutions for, x + y = 100

TRICK - The number of natural number solutions of x + y = k, where k is a natural number greater than 1, is 'k - 1'

Example 2

Find the number of whole number solutions of the equation, x + y = 7.

Here we are going to find out the whole number solutions, so the least possible value which we can substitute for x is zero. As the same method we applied for the former question, we can find the set of all possible whole number solutions can find out through substitutions and the solutions are;

xy
0 7
1 6
2 5
3 4
4 3
5 2
6 1
7 0

TRICK : The number of whole number solutions of x + y = k, where k is a whole number, is 'k + 1'

Example 3

Find the number of natural number solutions of x + y + z = 5.

Solution:

When x = 1, y + z = 4  4 - 1 = 3 natural number solutions are possible.
When x = 2, y + z = 3  3 - 1 = 2 natural number solutions are possible.
When x = 3, y + z = 2  2 - 1 = 1 natural number solution is possible.

Hence the total number of natural number solutions = 3 + 2 + 1 = 6

TRICK : The number of natural number solutions of x + y + z = k, where k is a natural number greater than 2 is,
1 + 2 + 3 + 4 + .... + (k - 2 )
i.e. The sum of first (k - 2 ) natural numbers.
i.e. (k - 2)(k - 1) / 2
This result we can elevated to the application of Combinations for generalization.
(k - 2)(k - 1) / 2 = (k - 1) C2 = (k - 1) C(3-1)

Result : The number of positive integral (natural number) solutions for;
x1 + x2 + x3 + .... + xn = k, where k ≥ n > 0 is
(k - 1) C(n-1)

Example 4

Find the number of non-negative integral (whole number) solutions of x + y + z = 5.

Solution:

When x = 0, y + z = 5, 5 + 1 = 6 whole number solutions are possible.
When x = 1, y + z = 4, 4 + 1 = 5 whole number solutions are possible.
When x = 2, y + z = 3, 3 + 1 = 4 whole number solutions are possible.
When x = 3, y + z = 2, 2 + 1 = 3 whole number solutions are possible.
When x = 4, y + z = 1, 1 + 1 = 2 whole number solutions are possible.
When x = 5, y + z = 0, 0 + 1 = 1 whole number solution is possible.
Hence the total number of whole number solutions = 6 + 5 + 4 + 3 + 2 + 1 = 21
TRICK
The number of whole number solutions of x + y + z = k, where k is a whole number is,
1 + 2 + 3 + 4 + .... + (k + 1 )
i.e. The sum of first (k + 1 ) natural numbers.
i.e. 1 + 2 + 3 + 4 + .... + (k + 1 ) = (k + 1 )(k + 2) / 2
This result we can elevated to the application of Combinations for generalization.
(k + 1 )(k + 2) / 2 = (k + 2) C2 = (k+3-1) C(3-1)

Result:

The number of non-negative integral (whole number) solutions for;

x1 + x2 + x3 + ...... + xn = k, where k is any whole number is,
(k+n-1) C(n-1)

Type II: Coefficients are not always be equal to '1'

This is the exact form of Special Equations. The solving process of these kinds of equations is bit logical. Sometimes in data sufficiency type of questions also we can see the application of this concept.

Here we are going to familiar with the solving process of these kinds of equations with examples.

Example:

Find all possible natural number solutions of the equation, 2x + 3y = 50.

Solution:

2x + 3y = 50

This system contains only one equation in two unknowns. Possibly there are many natural number solutions for this system and through the substitution method we can catch those sets of solutions.

First of all convert the equation in a suitable form for substitution.

i.e. 2x = 50 - 3y

Here, LHS = 2x, where x is any natural number.

i.e. For any natural number value of x, LHS is a multiple of 2.

RHS = 50 - 3y.

As the LHS is a multiple of 2, RHS also should be a multiple of 2. Hence find for which values of y, RHS become a multiple of 2.

By observation the least possible value of y, which will make 50 - 3y is a multiple of 2 is, y = 2.

When y = 2, RHS = 50 - 3y = 50 - 3(2) = 44
Therefore LHS, 2x = 44
x = 22.

In the similar way through the further substitutions, we can find all the possible natural number solutions of the given equations, and those solutions are given below.

XY
22 2
19 4
16 6
13 8
10 10
7 12
4 14
1 16

There are 8 pairs of natural number solutions are possible.

Remark:

In the above set of solutions, we can see some special and interesting properties.

  • The subsequent values of x form an Arithmetic Progression with a common difference '3' and this common difference is the coefficient of y.
  • Similarly the subsequent values of y form an Arithmetic Progression with a common difference '2' and this common difference is the coefficient of x.
  • These two AP's are in opposite trends, i.e. when the values of x are in ascending order then the corresponding values of y are in descending order.

These are the common conditions of the nature of the solutions of the natural and whole number solutions of the equations in type of ax + by = k.

Hence, the availability of any one among the possible solutions will help us to find out the remaining solutions easily.

We can classify the types of equations in the form of ax + by = k, in the following models.

Model 1: At least one among the coefficients is a factor of the constant.

Example

Consider the equation 5x + 7y = 45

Here the coefficient of x is 5, which is a factor of the constant 45. Hence it is easy to start the substitution from x = 9, then y = 0.

Now we got one among the available whole number solutions, then we can find out the further solutions by the following way. Here we are using the already mentioned findings about the Arithmetic Progression nature of the solutions.

 

The next such whole number solutions are , x = 2 and y = 5.

If we are preceding the above operation further, the values of x become negative. So we can stop the operation at this step. Hence there are two such possible whole number solutions for the equation and which are (9, 0) and (2, 5).

And there is only one natural number solution, (2, 5).

Illustrated example 1:

Find the total number of whole number solutions of 34x + 38y = 816.

Solution:

Step 1: Simplify the equation in its lowest form.

i.e. Divide the equation by 2.
Then, 17x + 19y = 408

Step 2: By observation, we will get 408 is a multiple of 17.

i.e. 17 * 24 = 408
Therefore, when x = 24, y = 0

Step 3: Catch the further whole number solutions by applying the concept of AP.

 

While preceding the further steps of the above operation, the value of 'x' becomes negative.

Hence stop the operation.

i.e. There are two whole number solutions for the given equation and which are,

(24, 0) and (5, 17)

Illustrated example 2:

Find the total number of whole number solutions of 7x + 5y = 140.

Solution:

Here both the coefficients 7 and 5 are factors of 140.

i.e. 7 * 20 = 5 * 28 = 140
When x = 20, y = 0

Find the further solutions through the above discussed method.

 

Hence the total number of whole number solutions = 5

Model II : Sum of the coefficients is a factor of the constant.

Example:

Consider the equation, 3x + 8y = 132

Here the sum of the coefficients of x and y is 3 + 8 = 11, which is a factor of the constant 132.

i.e. 11 * 12 = 132

Hence x = y = 12 is one among the possible whole number solutions. If we will get such a solution, then we can find out the remaining solutions in the following way.

 

Hence there are 6 such pairs of whole number solutions are possible.

Result:

In an equation of the form ax + by = k, if a + b is a factor of k, then x = y is one among the possible whole number solutions.

There is a third model special equation too, in which the conditions of model 1 or model 2 are not satisfying. In such an equation, we have to apply the general method for finding the whole number solutions, which we have already explained in the first general example of this section.

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