Linear Equations In Two Variables, Class 10, Math Detailed Chapter Notes PDF Download

Introduction - Pair of Linear Equations in Two Variables, CBSE, Class 10, Mathematics

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

INTRODUCTION

In class IX, we have read about linear equations in two variables. A linear equation is a rational and integral
equation of the first degree.
For example, the equations : 3x + 2y = 7, 2x–  are linear equations in two variables, since in
each case

(i) Neither x nor y is under a radical sign i.e., x and y are rational.
(ii) Neither x nor y is in the denominator.
(iii) The exponent of x and y in each term is one.
In general, ax + by +c = 0; a, b, c R ; a ≌ 0 and b≌0 is a linear equation in two variables. A linear equation in two variables has an infinite number of solutions. The graph of a linear equaton in two variables is always a straight line. In this chapter, we shall study about systems of linear equations in two variables, solution of a system of linear equations in two variables and graphical and algebraic methods of solving a system of linear equations in two variables. In the end of the chapter, we shall be discussing some applications of linear equations in two variables in solving simple problems from different areas.

HISTORICAL FACTS

Diophentus, the last genius of Alexandria and the best algebraic mathematician of the Greeko Roman Era, has made a unique contribution in the development of Algebra and history of mathematics. He was born in the 3rd century and lived for 84 years. Regarding his age it has been told in a VINODIKA of Greek collections. "He spent one sixth of his life in childhood, his beared grew after one twelfth more, after another one-seventh he married, five years later his son was born, the son lived to half the father's age, and Diophentus died four years after his son."
i.e. 
He was known as the father of Algebra. Arithimetica is his famous book.

i) Equation : A statement of equality of two algebraic expressions which involve one or more unknown quantities
is known as an equation.

(ii) Linear Equation : An equation in which the maximum power of variable is one is called a linear equation.
(iii) Linear Equation in One Variable : An equation of the form ax + b = 0. where x is a variable, a,b are real numbers and a ≌ 0 is called a linear equation in one variable.

Any value of the variable that satisfies the given equation is called a solution or root of the equation.
Ex.x = 3 is a solution of 3x +2 = 11.

(iv) Linear Equation in Two Variables : An equation of the form ax + by + c = 0, where a,b,c are real numbers, a≌ 0, b ≌ 0 and x, y are variables is called linear equation in two variables.
Any pair of values of x & y which satisfies the equation ax + by + c = 0 is called a root or solution of it.
Ex.(x = 1, y = 1) is a solution of 4x – y – 3 = 0.

Remark : A linear equation in two variables have infinite number of solutions.

(v) Graph of a Linear Equation in Two Variables : Assume y – x = 2 be a linear equation in two variables. The following table exhibits the abscissa and ordinates of points on the line represented by the equation, y – x = 2

Plotting the points (1, 3), (2, 4) and (3, 5) on the graph paper and drawing the line joining them we obtain the graph of the line represented by the given equation as shown in fig.

SIMULTANEOUS 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 : a₁x + b₁y + c₁ = 0 and a_​2 x + b_​2 y + c₂ = 0 
Where a₁, a₂, b₁, b₂, c₁ are c₂ real number ;    and x, y are variables.

Ex. Each of the following pairs of linear equations form a system of two simultaneous linear equations in two variables.

(i) x – 2y = 3, 2x + 5y = 5
(ii) 3x + 5y + 7 = 0, 5x + 2y + 9 = 0

SOLUTION OF THE SYSTEM OF EQUATIONS

Consider the system of simultaneous linear equations : a₁x + b₁y + c₁ = 0 and a_​2 x + b_​2 y + c₂ = 0 
A pair of values of the variables x and y satisfying each one of the equations in a given system of two simultaneous linear equations in x and y is called a solution of the system.
Ex. x = 2, y = 3 is a solution of the system of simultaneous linear equations.
2x + y = 7, 3x + 2y = 12
The given equations are 2x + y = 7 ........(i)
3x + 2y = 12 .......(ii)
Put x = 2, y = 3 in LHS of equation (i), we get
LHS = 2 × 2 + 3 = 7 = RHS
Put x = 2, y = 3 in LHS of equation (ii), we get
LHS = 3 × 2 + 2 × 3 = 12 = RHS
The value x = 2, y = 3 satisfy both equations (i) and (ii).
Hence x = 2, y = 3 is a solution of the given system.

Remark : An equation involving two variables cannot give value of both the variables. For values of both the variables we require two equations. Similarly for three variables we require three equations and so on, i.e. to find n variables we need n equations.

HOMOGENEOUS SYSTEM OF EQUATIONS

A system of simultaneous equations is said to be homogenous, if all of the constant terms are zero.

General Form : a₁x + b₁y + c₁ = 0 and a_​2 x + b_​2 y + c₂ = 0 

Homogeneous equation of the form ax + by = 0 is a line passing through the origin.

Therefore, the system is always consistent.

(i) the system of equation has only one solution.
(ii)  the system of equation has infinitely many  solution.

Ex.1 On comparing the ratios ,find out whether the following points of linear equations are consistent or inconsistent.

(i) 3x + 2y = 5, 2x – 3y = 7
(ii) 2x – 3y = 8, 4x – 6y = 9

Sol. (i) We have, 3x + 2y = 5 ⇒ 3x + 2y – 5 = 0 and 2x – 3y = 7 ⇒ 2x – 3y – 7 = 0

∴ 

Therefore, the given pair of linear equations is consistent.

(ii) We have, 2x – 3y = 8 ⇒ 2x – 3y – 8 = 0 and 4x – 6y = 9 ⇒4x – 6y – 9 = 0

∴ 

Therefore, the given pair of linear equations is inconsistent.

Ex.2 For what value of k, the system of equations x + 2y = 5, 3x + ky + 15 = 0 has

(i) a unique solution (ii) No solution?

Sol. We have,x + 2y = 5 ⇒ x + 2y – 5 = 0 and 3x + ky + 15 = 0.

(i)  The required condition for unique solution is  : 

∴

Hence, for all real values of k except 6, the given system of equations will have a unique solution.

(ii) Hie required ccndition for no solution is

∴ 

Hence the given system of equations will have no solution when k = 6.

Ex.3 Find the value of k for which the system of equations 4x + 5y = 0, kx + 10y = 0 has infinitely many solutions.

Sol. The given system is of the form a₁x + b₁y = 0, a₂x + b_​2y = 0
a₁= 4, a_​2 = k, b₁ = 5, b_​2 = 10
a_1/a_{2 =}b_1/b_​2  , the system has infinitely many solutions.
4/k = 5 /10 ⇒ k = 8

Ex.4 Find the value of a and b for which the given system of equations has an infinite number of solutions :

2x + 3y = 7 ; (a + b + 1) x + (a + 2b + 2)y = 4 (a + b) + 1

Sol. We have 2x + 3y = 7 ⇒ 2x + 3y – 7 = 0
and (a + b + 1) x + (a + 2b + 2)y = 4 (a + b) + 1

⇒ (a + b + 1) x + (a + 2b + 2)y – {4 (a + b) + 1} = 0
The required condition for an infinite number of solutions is 

⇒  2a + 4b + 4 = 3a + 3b + 3 and 12a + 12b + 3 = 7a + 14b + 14
⇒  a - b - 1 = 0 and 5a - 2b = 11
⇒ a - b = 1    ..(i)
and 5a - 2b = 11    ...(ii)
Multiplying (i) by 2 we get 2a - 2b = 2    ...(ni)

Subtracting (did) frcm (ii) we get 3a = 9 ⇒ a = 9/3 = 3

Put a = 3 in (i), we get 3 – b = 1 ⇒ b = 2

Hence, the given system of equations will have infinite number of solutions when a = 3 and b = 2.

GRAPHICAL METHOD OF SOLVING A SYSTEM OF SIMULTANEOUS LINEAR EQUATIONS

To solve a system of two linear equations graphically,
(i) Draw graph of the first equation.
(ii) On the same pair of axes, draw graph of the second equation.
(iii) (a) If the two lines intersect at a point, read the coordinates of the point of intersection to obtain the solution and verify your answer.
(b) If the two lines are parallel, there is no point of intersection, write the system as inconsistent. Hence, no solution
(c) If the two lines have the same graph, then write the system as consistent with infinite number of solutions.

Ex.5 Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.

(i) x + 2y – 3 = 0, 4x + 3y = 2
(ii) 3x + y = 1, 2y = 2 – 6x
(iii) 2x – y = 2, 2y – 4x = 2

the graph, we see that the two lines intersect at a point (–1,2).

So the solution of the pair of linear equations is x = – 1, y = 2

i.e., the given pair of equations is consistent.

The two equations have the same graph. Thus system is consistent with infinite number of solutions, i.e. the system is dependent.

The graph of the system consists of two parallel lines. Thus, the system is inconsistent. It has no solution.

COMPETITION WINDOW

DISTANCE BETWEEN TWO PARALLEL LINES

Consider pair of parallel lines

a₁X + b1y + c₁ - 0    ....(i)
a₂x + b₂y + c₂ - 0    ....(ii)

∴ The lines are; parallel.

∴  a₁ = a₂ k & b₁ = b₂ k

Putting the.se values in (i), we get : a₂kx + b₂ky + c₁ = 0 or   

or a₂x + b₂y + c₂ = 0    ... (iii) 

Clearly in equation (ii) and (iii), coefficients of x and y are same but the constant term is different in both the equations. The perpendicular distance (d) between the two lines can be calculated by using the following formula :

e.g, The distance between the parallel lines 3x – 4y + 9 = 0 and 6x – 8y – 15 = 0 can be calculated as follows:

3x – 4y + 9 = 0 ...(i), 6x – 8y – 15 = 0 or 3x – 4y – 15/2 = 0 ...(ii)

Required perpendicular distance,

ALGEBRAIC METHOD OF SOLVING A PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Some times, graphical method does not give an accurate answer. While reading the co-ordinate of a point on a
graph paper we are likely to make an error. So we require some precise method to obtain accurate result. The
algebraic methods are given below :
(i) Method of elimination by substitution. (ii) Method of elimination by equating the coefficients.
(iii) Method of cross multiplication.

ALGEBRAIC 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-I : Write the given equations
a₁x + b₁y + c₁ = 0 ...(i)
and a₂x + b₂y + c₂ = 0 ...(ii)
Step-II : 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-III : Substitute this value of y obtained in step-II, in the other equation to get a linear equation in x.
Step-IV : Solve the linear equation obtained in step-III and get the value of x.
Step-V : Substitute this value of x in the relation obtained in step-II and find the value of y.
Ex.6 Solve for x and y : 4x + 3y = 24, 3y – 2x = 6.
Sol. 4x + 3y = 24 ...(i)
3y – 2x = 6 ...(ii)
From equation (i), we get

Substituting in equation (ii), we get

⇒ 24 – 4x – 2x = 6
⇒ – 6x = – 24 + 6
⇒ 6x = 18
⇒ x = 3
Substituting x = 3 in (iii), we get

Hence, x = 3, y = 4.

ALGEBRAIC SOLUTION BY ELIMINATION METHOD

To solve a pair of linear equations in two variables x and y by elimination method, we follow the following steps:
Step-I : Write the given equations
a₁x + b₁y + c₁ = 0 ...(i)
and a₂x + b₂y + c₂ = 0 ...(ii)
Step-II : Multiply the given equations by suitable numbers so that the coefficient of one of the variables are numerically equal.
Step-III : If the numerically equal coefficients are opposite in sign, then add the new equations otherwise subtract.
Step-IV : Solve the linear equations in one variable obtained in step-III and get the value of one variable.
Step-V : Substitute this value of the variable obtained in step-IV in any of the two equations and find the value of
the other variable.

Ex.7 Solve the following pair of linear equations by elimination method : 3x + 4y = 10 and 2x – 2y = 2.

Sol. We have, 3x + 4y = 10 ...(i)
and 2x – 2y = 2 ...(ii)
Multiplying (ii) by 2, we get 4x – 4y = 4 ...(iii)
Adding (i) and (iii), we get 7x = 14 ⇒ x = 2
Putting x = 2 in equation (ii), we get 2 × 2 – 2y = 2⇒ 2y = 4 – 2  y = 1
Hence, the solution is x = 2 and y = 1

Ex.8 Solve : ax + by = c, bx + ay = 1 + c

Sol. ax + by = c ...(i)
bx + ay = 1 + c ...(ii)
Adding (i) and (ii), we get
(a + b) x + (a + b) y = 2c + 1

Subtracting (ii) and (i), we get.

Adding (iii) and (iv), we get

Subtracting (iv) from (iii) we get

ALGEBRAIC SOLUTION BY CROSS-MULTIPLICATION METHOD

Consider the system of linear equations

a₁x + b₁y + c₁ = 0 ...(i)
a₂x + b₂y + c₂ = 0 ...(ii)

To solve it by cross multiplication method, we follow the following steps :

Step-I : Write the coefficients as follows :

The arrows between the two numbers indicate that they are to be multiplied. The products with upward arrows are to be subtracted from the products with downward arrows.

To apply above formula, all the terms must be in left to the equal sign in the system of equations –

Now, by above mentioned rule, equation (i) reduces to

Case-I : If a₁b₂ – a₂b₁ ⇒0  x and y have some finite values, with unique solution for the system of equations.
Case-II : If a₁b₂ – a₂b₁ = 0 ⇒a₁/a₂= b₁/b₂

Here two cases arise :

Put. these values in equation a₁x + b₁y + c₁ = 0     .....(i)
⇒ a₂λx + b₂λy + c₂λ = 0
⇒  λ(a₂x + b₂y + c₂) = 0 but λ ≠ 0
⇒ a₂x + b₂y + c₂ = 0       ...(ii)

So (i) and (ii) ace cependant, so there are infinite number of solutions ,

So system of equations is inconsistent.

Ex.9 Solve by cross-multiplicaiton method : x + 2y + 1 = 0 and 2x – 3y – 12 = 0

Sol. We have,x + 2y + 1 = 0 and 2x – 3y – 12 = 0

By cross-multiplication method, we have

Hence the solution is x = 3 and y = –2.

 Ex.10

Solve for x and y 

.......(i)
.....(ii)

7u - 2v = 5 ...(iii)
8u + 7v = 15 ...(iv)

Multiplying (iii) by 7 and. (iv) by 2 and adding ve get

49u - 14v = 35 and 16u + 14v = 30

65 u = 65 ⇒ u = 1 

Substituting u = 1 in (iii) we get : 7 - 2v = 5 ⇒ v = 1 ⇒ 1/x or x = 1

Hence, x = 1, y = 1.