solve each of the following pairs of equation by reducing them to a pair of linear equation
5/x-y + 3/y-2 = 1
6/x-1 – 3

1 thought on “solve each of the following pairs of equation by reducing them to a pair of linear equation <br />5/x-y + 3/y-2 = 1<br />6/x-1 – 3”

1. Step-by-step explanation:

   Given Question:–

   5/x-y + 3/y-2 = 1

   6/x-1 – 3/y-2 = 1

   Correction:–.

   5/x-1 + 3/y-2 = 1

   6/x-1 – 3/y-2 = 1

   To find:–

   Solve each of the following pairs of equation by reducing them to a pair of linear equation ?

   Solution:–

   Given that:

   5/x-1 + 3/y-2 = 1 ———(1)

   6/x-1 – 3/y-2 = 1 ——–(2)

   Put 1/(x -1) = a and 1/(y-2) = b then

   5a + 3b = 1 ————-(3)

   6a -3b = 1—————(4)

   On adding (3)&(4) then

   5a + 3b = 1

   6a -3b = 1

   (+)

   ___________

   11 a + 0 = 2

   ____________

   => 11 a = 2

   => a = 2/11 ——–(5)

   On Substituting the value of a in (3) then

   5(2/11)+3b = 1

   =>(10/11)+3b = 1

   => 3b = 1-(10/11)

   => 3b = (11-10)/11

   => 3b = 1/11

   => b = 1/(3×11)

   => b = 1/33

   Now ,

   a = 1/(x-1)

   => 2/11 = 1/(x-1)

   => 11/2 = x-1

   => x = (11/2)+1

   => x = (11+2)/2

   => x = 13/2

   and

   b = 1/(y-2)

   => 1/33 = 1/(y-2)

   => y-2 = 33

   => y = 33+2

   => y = 35

   Answer:–

   The value of x = 13/2

   The value of y = 35

   The solution for the given lines = ( 13/2 , 35 )

   Check:–

   x = 13/2 and y = 35

   5/x-1 + 3/y-2

   => 5/(13/2-1) +3/(35-2)

   => 5/(11/2)+ 3/33

   => (10/11)+(1/11)

   => (10+1)/11

   =>11/11

   => 1

   LHS = RHS is true for x = 13/2 and y = 35

   and

   6/x-1 – 3/y-2

   => 6/(13/2-1) – 3/(35-2)

   => 6/(11/2) – 3/33

   => 12/11 – 1/11

   => (12-1)/11

   => 11/11

   = 1

   LHS = RHS is true for x = 13/2 and y = 3

   Verified the given relations.

   Used Method:–

   Reducing them to a pair of linear equation

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