Question

solve for x and y: x+4y =27xy &x+2y=21xy

Answer 1

x+4y=27xy                                                           (1)
x+2y=21xy                                                           (2)

 divide both the equation by xy

(x+4y =27xy)/xy

x/xy+4y/xy =27xy/xy

1/y+4/x =27

1(1/y)+4(1/x)=27

substitute
p = 1/x
q =1/y

     p+4q =27                          (3)

-     p+2q =21                        (4)
-----------------------
         2q    = 6
 
 q =6/2
    =3
 now ,substitute q's value in equation     (4)
  
 p+2q =21
  
 p+2 * 3  = 21

p + 6 =21
 
p =21-6
   
   =15             

therefore
 we know that
q= 1/y

1/y =1/3

p =1/x
 
1/x = 1/15

Answer 2

Answer:

Value of $x=\frac{1}{3}$ and $y=\frac{1}{15}$

Step-by-step explanation:

Given the expressions, $x+4y =27xy$ and $x+2y =21xy$

Substracting 2nd equation from the first,

                  $x+4y=27xy\\\underline { x+2y=21xy}$

 we get            $2y=6xy\implies2=6x$

                                        $\implies x=\frac{2}{6} =\frac{1}{3}$  

To get the value of $y$, let us consider the second equation

                            $x+2y=21xy$        

Dividing the equation with $xy$,

                          $\frac{x}{xy} +\frac{2y}{xy} =\frac{21xy}{xy} \implies \frac{1}{y} +\frac{2}{x} =21$

Substituting $x$ , we get

                        $\implies \frac{1}{y} +\frac{2}{\frac{1}{3}} =21\implies \frac{1}{y} +6=21$

                                                    $\implies \frac{1}{y} =15\implies y=\frac{1}{15}$

Hence, the values of $(x,y)=(\frac{1}{3},\frac{1}{15} )$