Question

Solve : 4/x+y -3/x-y =5 9/x+y -5/x-y =4where x + y ≠ 0 and x – y ≠ 0​

Answer 1

Given: The equations 4/x+y  -  3/x-y =  5 and 9/x+y  -  5/x-y  =  4

To find: Solve the equations.

Solution:

• Now we have given the equations as: 4/x+y - 3/x-y = 5 and 9/x+y  -  5/x-y  =  4

• Lets solve by substitution method.

• Consider 1 / x+y = a and 1 / x-y = b, then:

               4a - 3b = 5 and 9a - 5b = 4

• Now solving these equations, we get:

               4a - 3b = 5  ........ x 5   ........(i)

               9a - 5b = 4 ......... x 3   ........(ii)

• So the equations becomes:

                          20a - 15b = 25

                    ( - ) 27a - 15b =  12

                          -7a            = 13

               -7a = 13

               a = -13/7 .........(iii)

• Putting (iii) in (i), we get:

               4(-13/7) - 3b = 5

               -52/7 - 3b = 5

               -52/7 - 5 = 3b

               -52 - 35 / 7 = 3b

               -87/7 = 3b

               b = -87/21 = -29/7 .........(iV)

• Again re substituting, we get:

               a = -13/7 = 1/x+y

               x+y = -7/13

               13x + 13y = -7     ........(v)

               b = -29/7 = 1/x-y

               x - y = =-7/29

               29x - 29y = -7    .......(vi)

               So (v)x29/13 + (vi), we get:

               29x + 29y = -7(29/13)

               29x - 29y = -7        

               58x           =  -7(29/13 + 1)

                58x = -7(42/13)

               x =  (-7x42/13x58) = -294/754 = - 0.389

• Putting x in (v), we get:

               13(-0.389) + 13y = -7

               -5.057 + 13y = -7

               13y = -1.943

               y = -1.943/13

Answer:

            So the value of x is - 0.389 and y is -1.943/13.

Answer 2

$Given \: \frac{4}{x+y} - \frac{3}{x-y} = 5 \: --(1) \\and \: \frac{9}{x+y} - \frac{5}{x-y} = 4 \: --(2)$

$Let \: a = \frac{1}{x+y} \: and \: b = \frac{1}{x-y} \\now , Rewrite \:the \: equations$

$\implies 4a - 3b = 5 \: --(3)$

$\implies 9a - 5b = 4 \: ---(4)$

/* Multiplying equation (3) with 5 and equation (4) with 3 , we get */

$\implies 20a - 15 b = 25 \: --(5) \:and \\</p><p>27a - 15b = 12 \: ---(6)$

/* Subtract (5) from (6) we get */

$\implies 7a = -13$

$\implies a = \frac{ -13}{7}\: --(7)$

$Put \: a = \frac{ -13}{7}\:in \: equation \: (3) , we \\get$

$\implies 4 \times \frac{ -13}{7} - 3b = 5$

$\implies \frac{- 52}{7} - 3b = 5$

$\implies -3b = \frac{52}{7} + 5$

$\implies -3b = \frac{ 52+35}{7}$

$\implies -3b = \frac{ 87}{7}$

$\implies b = \frac{ -29}{7} \: --(8)$

$Now, a = \frac{ -13}{7} = \frac{1}{x+y}$

$\implies x + y = \frac{-7}{13} \: --(9)$

$Now, b = \frac{ -29}{7} = \frac{1}{x-y}$

$\implies x - y = \frac{-7}{29} \: --(9)$

/* Add Equations (8) and (9) , we get */

$2x = \frac{-7}{13} - \frac{7}{29}$

$\implies 2x = \frac{(-7) \times 42 }{377}$

$\implies x = \frac{-147}{377} \: --(10)$

/* Subtract equation (10) from (9) , we get */

$2y = \frac{-7}{13} + \frac{7}{29}$

$\implies 2y = \frac{7 \times (-16) }{377}$

$\implies y = \frac{-112}{377} \: --(11)$

Therefore.,

$\red{ Value \:of \: x } \green { = \frac{-147}{377}}$

$\red{ Value \:of \: y } \green { = \frac{-112}{377}}$

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