Simple Equations - Concepts & Tricks

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All placement test and competitive exams consist directly or indirectly test your basic skills in the area of algebra through simple equations. Here we explain various concepts of simple equations along with the ways to solve them using examples.

This topic covers following aspects.
  • Definition of simple equation.
  • The awareness about the nature of solutions.
  • Understand various forms of simple equations.
  • Algebraic translation: the skill to frame equations from the given data.
  • Data adequacy: check whether the given data is sufficient to find solutions.
  • Solving system of equations: method of substitution, elimination and determinate approach.
  • Determinacy and consistency of a system of equations.

 

These are the conventional requirements in this topic, even though the student needs to be able to understand the logical dimension of most of the questions from this topic and responsible to give the quick responses within a strategically restricted time allotment. The thorough understanding of the fundamental concept will give you extra mileage to use your logical thinking ability to tackle most of the tricky questions from this area.

Definition of Simple Equation

As simple as Simple Equation is a first degree equation without any restrictions in the number of unknowns which are presented in the equation.
Examples:
x + 4 = 10
2x + 3y = 15
5p + 7q + 11 r =13

Geometrically it is an equation which can be expressed as a straight line. Therefore simple equations are also called Linear Equations. Standard form of a Linear Equation is ax + by + c = 0, where a, b and c are constants, x and y are variables and at least a or b not equal to zero.

Nature of solutions/ determinacy of a system of equations.

A system of simple equations is referred as a group of similar equations.

System 1: Simple equation in one variable.

Consider a one variable simple equation.

2x + 3 = 11.

What is the total number of solutions possible for this system?

Obviously it's not a challenging task to find out the suitable value for 'x', x = 4.

Whether there any more suitable values for 'x' exist?

There is one and only one value of 'x' is satisfying the condition framed by the system. So we can generalize this result.

"A simple equation in single variable should have exactly one solution".

System 2: Simple equation in two variables.

Find the number of solutions for the equation 3m + 2n = 15?

Is there any unique solution?

If m = 1 then n = 6. Can we conclude, this is the only one solution?

Consider some more other possibilities.

m = 0, n = 7.5
m = 3, n = 3
m = 5, n = 0
m = -1, n = 9 and many more in the form of a pair of real numbers.

Therefore "a simple equation in two (or more than two) variables has infinitely many solutions".

Understand this concept thoroughly, because most of the competitive exams frequently using this as a testing point to evaluate your theoretical skills in the area of simple equations. We can expect this testing point more often in the Data Sufficiency type questions and rarely in the problem solving types.

Example 1. Problem Solving.

Individual prices of an apple and an orange are Rs.4 and Rs.3 respectively. How many possible combinations of apples and oranges can include in a total purchase worth Rs.35?

A)0  B)1  C)3  D)can't be determined.

Solution :

By a simple analysis on the structure of the question, an algebraically sound student may reach a wrong interpretation of this question. Structurally it's a question which consisting only one equation with two unknowns. Therefore theoretically this in determinate system of equations has infinitely many solutions. That is the exact trick which is applicable here.

Just think about the equation which we can furnish as per the given data.

Let 'a' and 'b' are respectively the number of apples and oranges that he bought. Then the equation will be,

4a + 3b = 35

Normally this equation has infinitely many solutions when the variables are irrespective in nature. But in this context 'a' and 'b' are natural numbers, because they are representing the number of fruits. Therefore find out the natural number solutions.

Solution1. a = 2, b = 9
Solution2. a = 5, b = 5
Solution3. a = 8, b = 1

These are the possible natural number values for 'a' and 'b'.

Hence the answer is option C.

Review of the above concept in exam point of view.

Above discussed simple equations in two variables 4a + 3b = 35 has infinitely many solutions. Even though as per the additional restrictions implied in the nature of unknowns, we may get a finite (countable) number of solutions.

Some more interesting factors:

  • From the above equation, 35 is not a multiple of 3 or 4 but a multiple of 3+4 = 7. When an equation pertain such a relation then a = b is one of the solutions. Here this as a = b = 5 is one among the three possible solutions. It will give you a start to catch the remaining set of solutions.
  • In the above three set of solutions the values of 'a' and 'b' are in an Arithmetic Progression. Values of 'a' start as 2 and sequencing as an arithmetic progression with a common difference of 3(this is the coefficient the other unknown 'b'), while that of 'b' start from 9 and decreasing with a common difference of - 4 (4 is the coefficient the other unknown 'a').

Illustration of application of the above concept:

Find the natural number solutions for 5p + 3q = 88

Here 88 is not a multiple of 5 or 3 but 5+3, so p = q is one of the solutions.

If p = q, then the value of p = q = 11, we can start to catch the further values from 11, as given below.

So the given equation has 6 natural number solutions.

System 3: Determinate system of two variable equations.

What is a determinate system of equation?

If the number of equations in a system is equal to the number of variables, then the system is called a determinate system.

Example 1:

Solve,

5p + 4q = 37
3p + 2q = 21

Here we have to recall the methods of solving equations. Solving equations means to find the suitable values for the variables in the given equations. There are the following different methods to solve a system.

  • Substitution.
  • Elimination.
  • Determinate method - Cramer's rule.
  • Cross multiplication.

Out of the above mentioned methods, the commonly following and easy to understand is the Method of Elimination. In this method we are eliminating the like variable from the system. Please find below the application of this method in the given system.

For eliminating any particular variable, first we have to make the coefficient of the variables are same in both the equations. Here it's easy to equalize the coefficient of 'q' in both the equations.

For the simplification purpose we can number both equations.

5p + 4q = 37 ...... ( Eq 1 )
3p + 2q = 21.........(Eq 2)

Multiply equation 2 by two.

(Eq2) * 2 → 6p + 4q = 42..........(Eq 3)

Subtract equation 1 from equation 3.

(Eq3) - (Eq1) → p = 5

Substitute the value of p = 5 in any of the equations to get the value of q.

Consider (Eq2) → 3(5) + 2q = 21 2q = 21 - 15 q = 6/2 = 3. Hence , p = 5 and q = 3.

Example 2: Determinate system of three variable equations.

Solve,

3x + 2y - z = 4 ......(Eq 1)
7x + 3y + 2z = 19 ...... (Eq 2)
2x - y + 3z = 9 ......(Eq 3)

Solution:

Step 1: Through elimination process, frame two simple equations in two variables.

(Eq 1) * 2 → 6x + 4y - 2z = 8 ......(Eq 4)
(Eq2) + (Eq 4) → 13x + 7y = 27 ......(Eq5)
(Eq 1) * 3 → 9x + 6y - 3z = 12......(Eq 6)
(Eq3) + (Eq 6) → 11x + 5y = 21 ......(Eq7)

Then solve equations 5 and 7.

X = 1 and y = 3.

Substitute these values in any of the equations.

Let us consider (Eq1) → 3(1) + 2 (2) - z = 4
7 - z = 4
Hence z = 3.
Ans: x = 1, y = 2 and z = 3.

Determinacy of the system is a minimum requirement to get a unique solution for the system, even though it's a minimum condition only. Means, in a system if the number of equations is equal to the number of variables, then it's not sure that the system has a unique solution. There are again three conditions about the solutions.

Consistency of a system of equations.

Consistency means the availability of unique or infinite solutions for a system. As per consistency concepts there are three classifications in the nature of solutions of a system.

  • Consistent and independent system of equations means the system has a unique solution.
  • Consistent and dependent system of equations means the system has infinite solutions.
  • Inconsistent system of equations means the system doesn't have any solutions.

Let us briefly consider the conditions of the above mentioned natures of a system.

Type 1: Consistent and independent system of equations.

We can call it as a healthy system of equations because this system has a unique solution.

In a general form, determinate system of two variable equations with respect to x and y;

a1x + b1y = c1 and
a2x + b2y = c2
where a1, a2, b1 and b2 are the coefficients of x and y correspondingly and c1 and c2 are the constants.

If a1/a2 ≠ b1/b2,

Then the system is Consistent and independent, hence the system has a unique solution. This condition is fully irrespective of the nature of the constant terms c1 and c2.

Example:

5m + 7n = 21
4m + 9n = 34
In this system, a1/a2 = 5/4 and b1/b2 = 7/9
5/4 ≠ 7/9, hence the system has a unique solution.

Type 2: Consistent but dependent system of equations.

In a general form of a determinate system of two variable equations with respect to x and y;

a1x + b1y = c1 and
a2x + b2y = c2
where a1, a2, b1 and b2 are the coefficients of x and y correspondingly and c1 and c2 are the constants.

If a1/a2 = b1/b2 = c1/c2, then the system is Consistent but dependent, hence the system has infinite solutions.

Example:

3p - 5q = 7

9p - 15q = 21

In this system, a1/a2 = 3/9, b1/b2 = -5/-15 and c1/c2 = 7/21

3/9 = -5/-15 = 7/21, hence the system has infinite solutions.

Type 3: Inconsistent system of equations.

In a general form of a determinate system of two variable equations with respect to x and y;

a1x + b1y = c1 and
a2x + b2y = c2
where a1, a2, b1 and b2 are the coefficients of x and y correspondingly and c1 and c2 are the constants.

If a1/a2 = b1/b2 ≠ c1/c2, then the system is inconsistent, hence the system doesn't have a solution.

Example:

2a + 3b = 5

4a + 6b = 7

In this system a1/a2 = 2/4, b1/b2 = 3/6 and c1/c2 = 5/7

2/4 = 3/6 ≠ 5/7, hence the system doesn't have any solution.

Additional conditions applicable in a system of simple equations.

Condition I: Find the result as a combination of variables.

Sometimes the question may ask to find the answer from an determinate or indeterminate system of equations, in the form of a combination of presented variables. For those types of questions, it does not require to find the value for each variable, instead give the answer in the form of a combination of variables.

Example1:

The total expenditure for 5 tables and 7 chairs is Rs. 1.2 Lakhs and that for 7 tables and 5 chairs is Rs. 1.5 Lakhs. Find the total cost for one table and one chair.

  • Rs. 55,000.
  • Rs. 23,000.
  • Rs. 20,000.
  • Rs. 22,500.

Sol:

5T + 7C = 1,20,000 and

7T + 5C = 1,50,000

Adding both the equations, then;

12 T + 12 C = 2,70,000

12 (T+C) = 2,70,000

So, T+C = 2,70,000 / 12

= 22,500.

Hence the cost for one table and chair togethe is Rs. 22,500.

Example 2:

The sum of four times the salary of Arvind, three times the salary of Bala and two times the salary of Charan is Rs. 1.8 Lakhs. If 5 times the total salary of Bala and Charan together and 7 times the salary of Arvind sum up to Rs. 2.6 Lakhs then the salary of Arvind and Bala together is more than that of Charan by…

  • Rs. 1 Lakhs.
  • Rs. 1.25 Lakhs.
  • Rs. 0.75 Lakhs
  • Rs. 0.8 Lakhs

Sol:

From the given data, we can furnish the following equations:

Let a, b and c represents the salary of Arvind, Bala and Charan espectively.

Then;

4a + 3b + 2c = Rs. 1.8 Lakhs...... (Eq 1)

7a + 5b + 5c = Rs. 2.6 Lakhs...... (Eq 2)

Multiply the first equation by 2, then the new equation is;

8a + 6b + 4c = Rs. 3.6 Lakhs......(Eq 3)

Subtract Equation 2 from Equation 3,

a + b - c = Rs.1 Lakhs.

Means, the total salary of Arvind and Bala together is Rs. 1 Lakh more than that of Charan.

Hence the answer is A.

Condition 2: Find the value of one variable from an indeterminate system of equations.

Here in this chapter we already discussed about the limitations of an indeterminate system of equations. Theoretically it's clear that, it's not possible to find a unique solution for the system which is indeterminate. Even though under some certain conditions, it's possible to find a unique value for any one among the given variables. I.e., if we have a system of two equations in three variables, depend on the nature of equations, it may possible to determine the value of one variable uniquely but the other two variables have infinitely many possible values. It will happen when the ratio of other two variables in one equation is same as that in the other equation.

Illustrated example:

Consider the system of equation:

4x + 5y + 7z = 21 ...........(Eq:1)

12x + 10y + 21z = 58..........(Eq:2)

Can we find out a unique value for any one among the variables?

Basically it's an indeterminate system, because the number of variables (three variables) more than the number of equations (two equations), so it's not possible to find a unique solution for the entire system.

As per the given system the ratio of the coefficients of variables x, y and z respectively from both the equations are 4/12, 5/10 and 7/21. Here it's clear that the ratios of the coefficients of variables x and z are equal, so it's possible to find a unique value for the third variable 'y'.

Multiply the first equation by 3, and then we will get the following equation;

12x + 15y + 21z = 63.

Subtract Eq:2 from this newly formed equation, then 5y = 5.

So y = 1.

However it's not possible to find a unique value for the other two variables, x and z.

Example 1: Data sufficiency.

Ram spent a total of Rs. 512 for 20 books, 12 pens and 5 erasers. If the price of each eraser is Rs.2 less than that of a pen, find the unit price of an eraser?

Statement I: Total cost for 10 books, 2 erasers and 6 pens is Rs.251.

Statement II: Total cost for 40 books, 10 erasers and 20 pens is Rs.1000.

  • Statement I alone sufficient but statement II is not.
  • Statement II alone sufficient but statement I is not.
  • Statements are not sufficient alone but the clubbing of both statements will give a unique conclusion.
  • Each statement alone sufficient.
  • Statements are not sufficient alone, even after clubbing also can't make a unique conclusion.

Solution:

Remember....!!!!!

It is not a problem solving question, so our approach towards this question should in the manner of testing the sufficiency of the given data. Don't waste your precious to find any kind of numeral value as an answer for this question.

Let 'b, p and e' are the respective prices for a book, a pen and an eraser.

From question statement: 20b + 12p + 5e = 512.......(Eq:1)

From statement I : 10b + 6p + 2e = 251......(Eq:2)

From statement II : 40b + 20p + 10e = 1000........(Eq:3)

While considering Equation 1 and Equation 2 together, it is clear that the ratios of the coefficients of the variables 'b' and 'p' are equal and which are not equal to that of the variable 'e'. Hence it is possible to find a unique value for 'e'. So statement I alone is sufficient.

While considering Equation 1 and Equation 3 together, it is clear that the ratios of the coefficients of the variables 'b' and 'e' are equal and which are not equal to that of the variable 'p'. Hence it is possible to find a unique value for 'p' and question itself consists a direct relation between the prices of a pen and an eraser and it will help you to find a unique value as the price for an eraser. So statement II alone is also sufficient.

Hence the answer is D.

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