Question

solve 57/x+y +6/x-y =5 and 38/x+y + 21/x-y =9

Answer 1

 [57/x+y]+[6/x-y]=5 ......(1)

[57*(1/x+y)]+[6*(1/x-y)]=5

 

[38/x+y]+[21/x-y]=9 ......(2)

[38*(1/x+y)]+[21*(1/x-y)]=9



Let 1/x+y = A & 1/x-y = B

therefore,

57A+6B=5          .......(3)

38A=21B=9        .......(4)

Now,

(3)*7 & (4)*2

=

399A+42B=35

  76A+42B=18

(-)     (-)        (-)

___________

323A      = 17

A=17/323

A=1/19    .....(5)

 

put (5)in (4)=

38*(1/19)+21B=9

2+21B=9

21B=9-2

21B=7

B=7/21

B=1/3  ....(6)

1/x+y = A & 1/x-y = B=

x+y=1/A=19

x+y=19  ....(7)

x-y=1/B=3

x-y=3   ....(8)

Add (7)+(8)=

x+y=19

x -y=  3

______

2x = 22

x=11 &

y=8

Answer 2

Answer:

x=11,y=8

Step-by-step explanation:

$\frac{57}{x+y} +\frac{6}{x-y} =5----(1)\\\\\frac{38}{x+y} +\frac{21}{x-y} =9----(2)$

Let us consider u=$\frac{1}{x+y}$ and v=$\frac{1}{x-y}$

$57u+6v=5---(1)\\\\38u+21v=9---(2)$

The equation is of the form ax+by+c=0

 $a_{1}=57, a_{2}=38,b_{1}=6 ,b_{2}=21 ,c_{1}=-5 ,c_{2} =-9$

$\frac{u}{b_{1} c_{2} -b_{2} c_{1} } =\frac{-v}{a_{1} c_{2} -a_{2} c_{1} } =\frac{1}{a_{1} b_{2} a_{2} b_{1} }$

$\frac{u}{6*-9-21*-5} =\frac{-v}{57*-9-38*-5} =\frac{1}{57*21-38*6}$

          $u=\frac{51}{969}=\frac{1}{19} } \\\\v=\frac{323}{969} =\frac{1}{3}$

Now we have

           $\frac{1}{x+y} =\frac{1}{19} \\\\\\x+y-19=0\\\\----(3)\frac{1}{x-y} =\frac{1}{3} \\x-y-3=0-----(4)$

Here $a_{1}=1, a_{2}=1,b_{1}=1 ,b_{2}=-1 ,c_{1}=--19 ,c_{2} =-3$

$\frac{x}{b_{1} c_{2} -b_{2} c_{1} } =\frac{-y}{a_{1} c_{2} -a_{2} c_{1} } =\frac{1}{a_{1} b_{2} a_{2} b_{1} }$

$\frac{u}{1*-3-2(-1)*-19} =\frac{-v}{1*-3-1*-19} =\frac{1}{1*-1-1*1}$

$\frac{x}{-22} =\frac{-y}{16} =\frac{1}{-2} \\\\x=11,y=8$

The project code is #SPJ2