## 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 :**

=> 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.

## Question 3.

Find whether the following pair of linear equations is consistent or inconsistent: 3x + 2y = 8, 6x – 4y = 9

**Solution :**

=> 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 :**

=> 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 :**

=> 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 :**

=> 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 :**

=> 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