Type Here to Get Search Results !

# Pair of Linear Equations in Two Variables I Maths I CBSE Class 10 I Chapter 3

CONTENT LIST

## General form of linear equation in two variables

General form of linear equation in two variables is ax + by + c, where a, b and c are real numbers  a and  b are not equal to zero.

The numbers a and b are called the coefficient of x and y and c is called constant term.

Examples: 2x + 3y = 5. represent linear equation in two variables.

A linear equation has highest power or exponent value of the variable as 1, not greater than 1 in any of the cases.

A pair of values, each for x and y to make both the sides of an equation equal is called solution of linear equation of two variables.

The graph of a linear equation in two variables is always straight line.

Any liner equation containing two variables has infinitely many solutions.

The solution of such equations is a point on the line representing the equation.

## General form of pair of linear equations

Two linear equations have a same set of variables are called a pair of linear equations.

General form of pair of linear equations are

a1x+ b1y + c1=0

a2x+ b2y+c2=0

Graphical representation of  a pair straight lines are:

1. Parallel

2. Interesting

3. Conceding

1. Parallel

When pair of lines are parallels. Then they have not unique solution.

Suppose we a pair of linear equations

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

They have not unique solution. If

a1/a2 = b1/b2 ≠ c1/c2

2. Interesting

When pair of lines are interesting then they have unique solution.

Suppose we a pair of linear equations

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

They have unique solution if

a1/a2 ≠ b1/b2 ≠ c1/c2

3. Conceding

When pair of lines are conceding. Then they have infinite many solutions.

Suppose we a pair of linear equations

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

They have infinite many solutions. If

a1/a2 = b1/b2  = c1/c2

Note: A pair of linear equations is called consistent, if the lines are intersecting or coincident.

## Solution of linear pair of equations

To find the solution to linear pairs of equations algebraically there are three methods namely, Substitution method, Elimination method, and Cross Multiplication method.

Substitution method

A pair of Linear Equations with two variables x and y, then w following steps are required to solve them with the substitution method-

Step 1: Select any one equation and find the value of one variable in terms of other variable i.e. y in terms of x.

Step 2:  Substitute the calculated value of y in terms of x in another equation.

Step 3: Now solve this Linear Equation in terms of x then it becomes a linear equation in one variable only i.e. x.

Step 4:  Calculate value of x in the given equations and then find the value of y.

Elimination method

By this method, we solve the equations by eliminating any one of the variables from both the equations.

Step 1: Multiply both the equations by such a number so that the coefficient of any one variable becomes equal in both equations.

Step 2: Add or subtract the equations so that the one term containing same variable will get eliminated as the coefficients of same variable are equal in both equations.

Step 3: Solve the equation in with leftover variable to find its value.

Step 4: Substitute the calculated value of variable in any given equations to find the value of the other variable.

Cross multiplication method

Given two equations in the form of

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, where

We write it in general form as

## Problem based on Trigonometry Class 10

Question 1.
For what value of k, the pair of equations 4x – 3y = 9, 2x + ky = 11 has no solution?

Solution :

Given, pair of linear equations
=> 4x – 3y = 9 ---(i)
=> 2x + ky = 11 --(ii)

For no solution of equation (i) and (ii),
=> 4/2 = -3/k (a1/a2 = b1/b2 ≠ c1/c2)
=> 2 = -3/k
=> k = -3/2
Question 2.
Calculate the area bounded by the line 2x + 5y = 10 and both the co-ordinate axes.

Solution :

Given, linear equations => 2x + 5y = 10

∴ Area enclosed by the equation 2x + 5y = 10 = 1/2 x 2 x 5 = 5 sq. unit
Question 3.
Find whether the following pair of linear equations is consistent or inconsistent: 3x + 2y = 8, 6x – 4y = 9

Solution :

Given, linear equations => 3x + 2y = 8, 6x – 4y = 9

=> a1/a2 = 3/6 = 1/2
=> b1/b2 = 2/-4 = -1/2
=> c1/c2 = 8/9

∵ a1/a2 ≠ b1/b2 ≠ c1/c2
So, the given linear pairs are intersecting.
∴ They are consistent.
Question 4.
Check graphically whether the pair of equations 3x – 2y + 2 = 0 and (3/2)x – y + 3 = 0, is consistent.

Solution :

Given, linear equations => 3x – 2y + 2 = 0, (3/2)x – y + 3 = 0

=> a1/a2 = 3 x 2/3 = 2
=> b1/b2 = -2/-1 = 2
=> c1/c2 = 2/3

∵ a1/a2 = b1/b2 ≠ c1/c2
So, the given linear pairs are parallel.
∴ They are inconsistent.
Question 5.
Solve the following pair of linear equations for x and y:
141x + 93y = 189
93x + 141y = 45

Solution :

Given, linear equations => 141x + 93y = 189 ---(i)
=> 93x + 141y = 45 ---(ii)

Multipling eqaution (i) and (ii) with 93 and 141 respectively, we get
=> 93(141x) + 93(93y) = 93(189) ---(iii)
=> 141(93x) + 141(141y) = 141(45) ---(iv)

Substracting equation (iii) from (iv), we get
=> [141(93x) + 141(141y)] - [93(141x) + 93(93y)] = 141(45) - 93(189)
=> 141(93x) + 141(141y) - 93(141x) - 93(93y) = 141(45) - 93(189)
=> 141(141y) - 93(93y) = 141(45) - 93(189)
=> 47(47y) - 31(31y) = 47(15) - 31(63)
=> 2209y - 961y = 705 - 1953
=> 1248y = -1248
=> y = -1

Putting y = -1 in equation (ii), we get
=> 93x + 141(-1) = 45
=> 93x - 141 = 45
=> 93x = 45 + 141 = 186
=> x = 186/93 = 2

So, the solution of the given equations are x = 2 and y = -1
Question 6.
Solve the following pair of linear equations for x and y:
(b/a)x + (a/b)y = a² + b²
x + y = 2ab

Solution :

Given, linear equations => (b/a)x + (a/b)y = a² + b² ---(i)
=> x + y = 2ab ---(ii)

Multiply equation (ii) by b/a, we get
=> (b/a)x + (b/a)y = 2b² ---(iii)

Substracting equation (iii) from (i), we get
=> [(b/a)x + (a/b)y] - [(b/a)x + (b/a)y] = [a² + b²] - [2b²]
=> (b/a)x + (a/b)y - (b/a)x - (b/a)y = a² + b² - 2b²
=> (a/b)y - (b/a)y = a² - b²
=> (a²y - b²y)/ab = a² - b²
=> y(a² - b²)/ab = a² - b²
=> y/ab = 1
=> y = ab

putting y = ab in equation (ii), we get
=> x + ab = 2ab
=> x = 2ab - ab = ab

So, the solution of the given equations are x = ab and y = ab.
Question 7.
Solve by elimination:
3x = y + 5
5x – y = 11

Solution :

Given, linear equations => 3x = y + 5 => 3x - y = 5 ---(i)
=> 5x – y = 11 ---(ii)

Substracting equation (ii) from (i), we get
=> (3x - y) - (5x – y) = 5 - 11
=> 3x - y - 5x + y = -6
=> -2x = -6
=> x = 3

Putting x = 3 in equation (i), we get
=> 3(5) - y = 5
=> 15 - y = 5
=> -y = -10
=> y = 10

So, the solution of the given equations are x = 3 and y = 10