Question

(x+y/4) – (x-y/3) = 11 ; (x+y/2) + (x-y/6) = 12

Answer 1

Given equations are ( x + y ) / 4 - ( x - y ) / 3 = 11 and ( x + y ) / 2 + ( x - y ) / 6 = 12.

Let the algebraic value of x + y and x - y be a and b respectively. Thus, now, equations are :

• a / 4 - b / 3 = 11 ...( i )
• a / 2 + b / 6 = 12 ...( ii )

Simplifying ( i ) & ( ii ) : -

= > a / 4 - b / 3 = 11

= > ( 3a - 4b ) / 12 = 11

= > 3a - 4b = 11 x 12

= > 3a - 4b = 132

= > 3a = 132 + 4b ...( 1 )

= > a / 2 + b / 6 = 12

= > 1 / 2 { a + b / 3 } = 12

= > ( 3a + b ) / 3 = 12 x 2

= > 3a + b = 12 x 2 x 3

= > 3a + b = 72

= > 132 + 4b + b = 72 { substituting the value of 3a from ( 1 ) }

= > 132 - 72 = - 4b - b

= > 60 = - 5b

= > - 12 = b

Substitute the value of b in ( 1 ) :

= > 3a = 132 + 4b

= > 3a = 132 + 4( - 12 )

= > 3a = 132 - 48

= > 3a = 84

= > a = 28

Then, a = x + y = 28

b = x - y = - 12

Adding both ( a & b ) :

= > a + b = 28 - 12

= > x + y + x - y = 16

= > 2x = 16

= > x = 8

Thus,

= > x + y = 28

= > 8 + y = 28

= > y = 20

Hence the required value of x is 8 and y is 20.

Answer 2

solving eq (1)

3x+3y-4x+4y=11*12

7y-x=132

eq(2)

3x+3y+x-y=12*6

2x-y=36

mul (1) by 2 and add (1)&(2)

14y-2x=264

-y+2x=36

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13y=300

y=23.07

put y in (2)

-y+2x=36

2x=36+y

=36+23.07

2x=59.07

x=29.53