CONSISTENCY AND INCONSISTENCY OF LINEAR EQUATIONS IN TWO VARIABLES

Example 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.

Example 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.

Example 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.

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