NEETClass 11thClass 12thClass 12th PlusJEEClass 11thClass 12thClass 12th PlusClass 6-10Class 6thClass 7thClass 8thClass 9thClass 10thOnline CoursesDistance LearningInternational OlympiadNEETClass 11thClass 12thClass 12th PlusJEE (Main+Advanced)Class 11thClass 12thClass 12th PlusJEE MainClass 11thClass 12thClass 12th PlusClass 6-10Class 6thClass 7thClass 8thClass 9thClass 10thKCET/MHT-CETKCETMHT-CETNEET2025202420232022JEE20262025202420232022Class 6-10JEE MainPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DatePercentile PredictorAnswer KeyCounsellingEligibilityExam PatternJEE MathsJEE ChemistryJEE PhysicsJEE AdvancedPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DateAnswer KeyEligibilityExam PatternRank PredictorNEETPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DateCollege PredictorAnswer KeyRank PredictorCounsellingEligibilityExam PatternBiologyNCERT SolutionsClass 6Class 7Class 8Class 9Class 10Class 11Class 12TextbooksCBSEClass 12Class 11Class 10Class 9Class 8Class 7Class 6SubjectsSyllabusNotesSample PapersQuestion PapersICSEClass 10Class 9Class 8Class 7Class 6State BoardBiharKarnatakaMadhya PradeshMaharashtraTamilnaduWest BengalUttar PradeshOlympiadMathsScienceEnglishSocial ScienceNSOIMONMTCASATInstant Online ScholarshipAIOT(NEET)TALLENTEXALLEN for SchoolsAbout ALLENBlogsNewsCareersRequest a call backBook a demo
Pair of Linear Equations in Two Variables
1.0Simultaneous linear equations in two variables
A pair of linear equations in two variables is said to form a system of simultaneous linear equations. General form : and , where and are real non zero numbers; and and , are variables. Here is an example of a system of simultaneous linear equations.
- is a solution a linear equation , because values of satisfies the equation. i.e. LHS = 6-2 = RHS
Solution of simultaneous linear equations in two variables
A pair of values of the variables and satisfying each one of the equations in a given system of two simultaneous linear equations in and is called a solution of the system. E.g. is a solution of the system of simultaneous linear equations. ..(1) ..(2) Put in L.H.S of equation (1), we get L.H.S R.H.S Put in L.H.S of equation (2), we get L.H.S R.H.S The value of satisfy both equations (1) and (2).
2.0Investigating graphs of a system of equations
Recall from class IX that the geometrical representation of a linear equation in two variables is a straight line. Each solution ( ) of a linear equation in two variables, , corresponds to a point on the line representing the equations, and vice versa. As such, there are infinitely many solutions of a linear equation in 2 variables. Since a linear equation in two variables represents a straight line. Therefore, a pair of linear equations in two variables will be represented by two straight lines, both to be considered together. You know that for given two lines in a plane, only one of the following three possibilities can happen: (i) The two lines intersect at one point. (ii) The two lines are parallel. (iii) The two lines are coincident lines.
Thus, the graphical representation of a pair of simultaneous linear equations in two variables will be in one of the forms. (see figure below)
Building Concepts 1 In one week a music store sold 7 violins for a total of ₹ . Two different types of violins were sold. One type cost ₹ 200 and the other type cost ₹ 300 . Represent this situation algebraically and graphically. Explanation:
points are
points are The graphical representation is shown in the figure which shows intersecting lines.
Building Concepts 2 Two railway tracks are represented by the equations and . Represent this situation graphically. Explanation
Points are Points are
The graphical representation is shown in the figure which shows parallel lines.
Building Concepts 3 Lorea is moving along the path and Johana is moving along the path . Represent this situation graphically. Explanation:
- At least 3 solutions are required to draw a graph of a linear equation in two variables.
3.0Solving system of equations by graphing
A system of two linear equations can be solved graphically, by graphing both equations in the same co-ordinate plane. Every point on the line of an equation is a solution of that equation.
The point at which the two lines cross lies on both lines and so it is the solution of both equations, it is also known as unique solution.
If the two lines coincide, they have infinitely many common points, and hence the system has infinitely many solutions.
Finally, if the lines represented by the equations are parallel, they do not have a common point and so the system has no solution.
Building Concepts 4 Solve the following system of equations by graphing. (i) (ii) (iii) Solution:
(i)
Points are
Points are From the graph, (see fig.) we see that the two lines intersect at a point ( ). So unique solution of the pair of linear equations is . (ii)
Points are
Points are The graph is shown in figure. The two equations have the same graph. Thus, system has infinite number of solutions.
(iii)
Points are
Points are The graph is shown alongside (see fig.). The graph of the system consists of two parallel lines. So, it has no solution.
- Plot three points for drawing a graph of linear equation in two variables. If these three points do not lie on same line that means at least one of the points (solution) is not correct.
4.0Types of System of Equation
Consistent system: A system of equations with at least one solution is called a consistent system (Unique and infinite many solutions).
Inconsistent system: A system of equations with no solution is called an inconsistent system (No solutions).
Dependent system: A system of equations with an infinite number of solutions is said to be dependent (Infinite many solutions).
Independent system: A system of equations with only one solution is said to be independent (Unique solutions).
Algebraic conditions for consistency/inconsistency of the system
- Linear equation in two variables is written in standard form as
Numerical Ability 1 In each of the following pairs of equations, determine whether the system has a unique solution, no solution or infinitely many solutions : (i) (ii) (iii) Solution: The given pair of equations can be rewritten as The given pair of equations has a unique solution. (ii) The given pair of equations is Here and and and Clearly . The given pair of equations has no solution. (iii) The given pair of equations can be rewritten as Here and and and . The given pair of equations has infinitely many solutions.
Numerical Ability 2 For what value of , the system of equations has (i) A unique solution (ii) No solution? Solution: We have, and . (i) The required condition for unique solution is: Hence, for all real values of except 6 , the given system of equations will have a unique solution. (ii) The required condition for no solution is: and and and Hence the given system of equations will have no solution when .
Numerical Ability 3 Find the value of for which the system of equations has infinitely many solutions. Solution: The given system is of the form If the system has infinitely many solutions,
- For finding , take coefficients with signs before them in the equations.
- If the constant terms are zero in a simultaneous system of equations, the system will always be consistent. Consider the following system : As the constant terms are zero, equation (1) and (2) will pass through the origin. As such, they will always intersect at one point i.e. the origin. Hence, such a system will always be consistent.
Building Concepts 5 Consider the following system of equations: and (a) If the system is consistent, how many solutions are possible, find them. (b) If the coefficient of in second equation is replaced by 10 , will there be any change in the number of solutions? Explain your answer. Explanation (a) ... (1) ; ... (2) Hence, . and Given system is consistent and has unique solution.
Points are
From the graph, we see that the two lines intersect at a point . So, the solution of the pair of linear equations are . (b) On replacing the coefficient of y in second equation by 10 , we have :
Here and Now the system has infinite number of solutions.
Points are
Points are
From the graph (see fig.), we see that the two lines are coincident. So, the system has infinitely many solutions.
5.0Algebraic solution by substitution method
To solve a pair of linear equations in two variables x and y by substitution method, we follow the following steps:
Step-1: Write the given equations Step- 2: Choose one of the two equations and express y in terms of x (or x in terms of y ), i.e., express one variable in terms of the other. Step-3: Substitute this value of obtained in step-2, in the other equation to get a linear equation in x . Step-4: Solve the linear equation obtained in step-3 and get the value of . Step-5 : Substitute this value of x in the relation obtained in step- 2 and find the value of y .
Numerical Ability 4 Solve for x and . Solution: From equation (1), we get Substituting in equation (2), we get Substituting in (3), we get Hence, .
A homogeneous system is always consistent.
Building Concepts 6 In one day the Museum admitted 321 adults and children and collected ₹ 1590 . The price of admission is ₹ 6 for an adult and ₹ 4 for a child. How many adults and how many children were admitted to the museum that day? Explanation:
- While applying substitution method, do not put the value of (in term of ) in the same equation from which the value of is obtained in terms of .
6.0Algebraic solution by elimination method
To solve a pair of linear equations in two variables and by elimination method, we follow the following steps: Step-1: Write the given equations. Step-2: Multiply the given equations by suitable numbers so that the coefficient of one of the variables are numerically equal. Step-3: If the numerically equal coefficients are opposite in sign, then add the new equations otherwise subtract. Step-4: Solve the linear equations in one variable obtained in step-3 and get the value of one variable. Step-5 : Substitute this value of the variable obtained in step-4 in any of the two equations and find the value of the other variable.
Numerical Ability 6 Solve the following pair of linear equations by elimination method and . Solution We have, and Multiplying equation (2) by 2 , we get Adding (1) and (3), we get Putting in equation (2), we get Hence, the solution is and .
- In elimination method while multiplying any one (or both) the equations by the constant, multiplication is done with each term of the equations.
7.0Algebraic solution by cross-multiplication method
Consider the system of linear equations. To solve it by cross multiplication method, we follow the following steps : Step-1 : Write the coefficients as follows :
Case-1 : If and have some finite values, with unique solution for the system of equations.
Case-2 : If Here two cases arise : (a) If . Then
Put these values in equation but So, (i) and (ii) are dependent, so there are infinite number of solutions.
(b) If . Then Put these values in equation but [from equation (ii)] Put this value in equation , we get As and we are getting So, system of equations is inconsistent and it has infinite many solution.
- In cross multiplication method try to apply basic method rather than applying formula directly because it might possible that you are remembering wrong formula.
Numerical Ability 6 Solve by cross-multiplication method: and Solution: We have, and By cross-multiplication method, we have Hence, the solution is and .
Building Concepts 7 Bus fare from Bangalore bus stand, if we buy 2 tickets to Malleswarm and 3 tickets to Yeshwanthpur, the total cost is ₹ 46; but if we buy tickets to Malleswarm and 5 tickets to Yeshwanthpur the total cost is ₹ 74. Find the fares from Bangalore to Malleswarm and to Yeshwanthpur. Explanation:
- Before applying cross multiplication method write both linear equations in two variables in standard form.
Interchanged Coefficients Method
Equations of the form and , where . To solve the equations of the form: and where , we follow the following steps : Step-1: Add (1) and (2) and obtain , i.e., Step-2 : Subtract (2) from (1) and obtain , Step-3: Solve (3) and (4) to get x and y .
Building Concepts 8 Solve for x and . Solution We have, and Adding (1) and (2), we get : Subtracting (2) from (1), we get : Now, adding (3) and (4), we get : Putting in (3), we get : Hence, the solution is and
8.0Equations reducible to linear equations in two variables
Equations which contain the variables, only in the denominators, are called reciprocal equations. These equations can be of the following types and can be solved by the under mentioned method:
Type-I: and (where are real number). Put and and find the value of and by any method described earlier. Then and Type-II : and a'u + b'v = c'uv (where ', ' are real number). Divide both equations by uv and equations can be converted in the form explained in (I). Type-III : (where are real number) Put and Then equations are and Find the values of and and put in and Again, solve for x and y , by any method explained earlier.
Numerical Ability 7 Solve for and and Solution We have, and Let and . Then, the given equations can be written as and Multiplying (1) by 3 and (2) by 2 , we get and Adding (3) and (4), we get Therefore, and Hence the solution is and .
Numerical Ability 8 Solve for x and and Solution: Putting and , we get Multiplying (3) by 7 and (4) by 2 and adding, we get and or Substituting in (3) we get : or Hence, .
Numerical Ability 9 Solve : and . Solution: We have and Let and Then, the given equations can be rewritten as and By cross-multiplication method, we have Adding (3) and (4), we get Put in (3), we get Hence, the solution is and .
- Word problems are mainly dependent on formation of equations i.e. decoding of language given in word problem. Word Problems based on articles and their costs
Building Concepts 9 37 pens and 53 pencils together cost ₹ 320 , while 53 pens and 37 pencils together cost . Find the cost of a pen and that of a pencil. Explanation: Let the cost of a pen be ₹ and that of a pencil be ₹ . Then, and Adding equations (1) and (2), we get Subtracting equation (1) from (2), we get Adding equations (3) and (4), we get Substituting in equation (3), we get Hence, cost of one pen = ₹ 6.50 and cost of one pencil .
Numerical Ability 10 and each have certain number of oranges. A says to , "if you give me 10 of your oranges, I will have twice the number of oranges left with you". B replies, "if you give me 10 of your oranges, I will have the same number of oranges as left with you". Find the number of oranges with A and B separately. Solution Suppose A has x oranges and B has y oranges. According to the given conditions, we have and, Subtracting equation (2) from equation (1), we get Putting in equation (1), we get Hence, A has 70 oranges and B has 50 oranges.
Word Problems based on numbers
- Recall that the two-digit number having a and b as units and ten's digits respectively is equal to and the number obtained by reversing the order of digits is .
Building Concepts 10 In a two-digit number, the unit's digit is twice the ten's digit. If 27 is added to the number, the digits interchange their places. Find the number. Explanation: Let the digit in the unit's place be x and digit in the ten's place be y . Then, [Given] and, Number Number obtained by reversing the digits It is given that the digits interchange their places if 27 is added to the number. i.e., Number Number obtained by interchanging the digits Putting in equation (2), we get Putting in equation (2), we get Hence, the number is
Numerical Ability 11 The sum of a two digit number and the number obtained by reversing the order of its digits is 121 , and the two digits differ by 3 . Find the number. Solution: Let the digit in the units place be x and the digit at the ten's place be y . Then, number The number obtained by reversing the order of the digits is . According to the given conditions, we have and Difference of digits is 3 ] Thus, we have the following sets of simultaneous equations. and or On solving equation (1) and (2), we get On solving equations (3) and (4), we get When , we have Number When , we have Number Hence, the required number is either 47 or 74 . Word Problems based on fractions
Building Concepts 11 A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3 , we get , but if the numerator is increased by 8 and the denominator is doubled, we get . Find the fraction. Explanation: Let the fraction be . Then, according to the given conditions, we have Multiplying equation (1) by 2 and (2) by 3 and subtracting (2) from (1), Putting in equation (2) Hence, the fraction is .
Numerical Ability 12 The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine the fraction. Solution: Let the numerator and denominator of the fraction be x and y respectively. Then, Fraction It is given that, Denominator (Numerator) +4 According to the given condition, we have Thus, we have the following system of equations. Subtracting equation (1) from equation (2), we get Putting in equation (1), we get Hence, required fraction .
Word Problems based on ages
Building Concepts 12 If twice the son's age in years is added to the father's age, the sum is 70. But if twice the father's age is added to the son's age, the sum is 95 . Find the ages of father and son. Explanation: Suppose father's age (in years) be and that of son's age be . Then, and This system of equations may be written as Multiplying equation (1) by 2 and subtracting from (2) Putting in equation (1) Hence, father's age is 40 years and the son's age is 15 years.
Numerical Ability 13 Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son. Find their present ages. Solution: Let present age of father and his son (in years) be and respectively. Ten years ago, Father's age years Son's age years Ten years later, Father's age years. Son's age Subtracting (2) from (1), we get Putting in (1), we get Thus, present age of father is 34 years and the present age of son is 12 years.
Word Problems based on time, distance and speed
In solving problems based on time, distance and speed, we use the following formulae :
- Distance = Speed Time
- Time and Speed
9.0Upstream & Downstream concept
If Speed of a boat in still water (or stream) Speed of the current Then, Speed in upstream Speed in downstream Following examples will illustrate the use of these formulae.
Building Concepts 13 A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream. Explanation: Let the speed of the boat in still water be and the speed of the stream be . Then, Speed in upstream Speed in downstream Case-I : Now, Time taken to cover 32 km upstream Time taken to cover 36 km downstream But, total time of journey is 7 hours. Case-II : Time taken to cover 40 km upstream Time taken to cover 48 km downstream In this case, total time of journey is given to be 9 hours. Putting and in equations (1) and (2), we get Multiplying equation (3) by 5 and (4) by 4 and subtracting Putting in equation (3) and Now, and, Solving equations (5) and (6), we get and Hence, Speed of the boat in still water Speed of the stream
Numerical Ability 14 A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car. Solution
Word Problems based on geometry
Building Concepts 14 The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and the breadth is increased by 3 units. If we increase the length by units and breadth by 2 units, the area is increased by 67 square units. Find the length and breadth of the rectangle. Explanation: Let the length and breadth of the rectangle be x units and y units respectively. Then, Area = xy sq. units If length is reduced by 5 units and the breadth is increased by 3 units, then area is reduced by 9 square units When length is increased by 3 units and breadth by 2 units, the area is increased by 67 sq. units Thus, we get the following system of linear equations: Multiplying equation (1) by 2 and (2) by 3 and subtracting
Putting in equation (1) Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.