# CONSISTENCY AND INCONSISTENCY OF LINEAR EQUATIONS IN TWO VARIABLES

## About "Consistency and Inconsistency of Linear Equations in Two Variables"

Consistency and Inconsistency of Linear Equations in Two Variables :

Here we are going to see some example problems of testing consistency and inconsistency of linear equations in two variables.

## Consistency and Inconsistency of Linear Equations in Two Variables - Practice questions

Question 1 :

The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number

Solution :

Let "xy" be the two digit number.

xy + yx  =  110

10x + y + 10y + 1x  =  110

11x + 11y  =  110

Divide it by 11,

x + y  =  10 --------(1)

xy - 10  =  5(x + y) + 4

10x + y - 10  =  5x + 5y + 4

10x - 5x + y - 5y  =  4 + 10

5x - 4y  =  14 --------(2)

Let us use elimination method to solve these equations.

(1) x 4 + 2 ==>  4x + 4y  =  40

5x - 4y  =  14

----------------

9x  =  54

x  =  54/9  =  6

By applying the value of x in (1), we get

6 + y  =  10

y  =  10 - 6  =  4

Hence the required two digit number is 64.

Question 2 :

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.

Solution :

Let "x/y" be the fraction

x + y  =  12 ------(1)

x/(y + 3)  =  1/2

2x  =  y + 3

2x - y  =  3 ------(2)

(1) + (2)

x + y  =  12

2x - y  =  3

---------------

3x  =  15

x  =  5

By applying the value of x in (1), we get

5 + y  =  12

y  =  12 - 5

y  =  7

Hence the required fraction is 5/7.

Question 3 :

ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)° , ∠B = (3y –5)°, ∠C =(4x) ° and ∠D = (7x + 5)° . Find the four angles.

Solution :

The sum of interior angles of a cyclic quadrilateral   =  360

4y + 20 + 3y - 5 + 4x + 7x + 5  =  360

11x + 10y  =  360 - 20

11x + 10y  =  340  ----(1)

<A + <C  =  180

4y + 20 + 4x  =  180

4x + 4y  =  160

x + y  =  40 -------(1)

∠B = (3y –5)° ∠D = (7x + 5)°

<B + <D  =  180

3y - 5 + 7x + 5  =  180

7x + 3y  =  180 ----------(2)

(1) x 3 - (2)

3x + 3y  =  120

7x + 3y  =  180

(-)   (-)      (-)

-----------------

-4x  =  -60

x  =  15

By applying the value of x in (1), we get

15 + y  =  40

y  =  40 - 15

y  =  25

∠A = (4y + 20)°  =  120

∠B = (3y –5)°  =  70

∠C =(4x)°  =  60

∠D = (7x + 5)°  =  110

Hence the required angles are 120, 70, 60 and 110. After having gone through the stuff given above, we hope that the students would have understood, "Solving Equations Using Cross Multiplication Method"

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