Question

7/2x+1 +13/y+2=27; 13/2x+1 + 7/y+2=33 sol it ​

Answer 1

To solve:

                $\frac{7}{2x} +1+\frac{13}{y}+2=27$

                $\frac{13}{2x} +1+\frac{7}{y} +2=33$

solution:

               $\frac{7}{2x} +1+\frac{13}{y}+2=27$

                $\frac{13}{2x} +1+\frac{7}{y} +2=33$

assume,

let $'\frac{1}{2x} '$ be $a$

let $'\frac{1}{y}'$be $b$

${7}{a} +1+{13}{b}+2=27$

${7}{a} +{13}{b}=30$_____(i)

${13}a +1+{7}{b} +2=33$

${13}a +{7}{b} =36$_____(ii)

from subtracting (i) and (ii)

${7}{a} +{13}{b}=30$

${13}a +{7}{b} =36$

___________

$-6a+6b=-6$

$-6(a-b)=-6$

$a-b=1$

$a=1+b$______(iii)

substitute (iii) in (i)

$7(1-b)+13b=30$

$7-7b+13b=30\\7+6b=30\\6b=23\\b=\frac{23}{6}$_(iv)

substitute (iv) in (ii)

${13}a +{7}({\frac{23}{6}) } =36$

$13a+\frac{161}{6} =36\\\\\frac{78a+161}{6} =36$

${78a+161} =216$

$78a=216-161\\78a=55\\a=\frac{55}{78}$

as per our assumption $'\frac{1}{2x} '=a$  and $'\frac{1}{y}'=b$

$\frac{1}{2x}=\frac{55}{78}$                          

$110x=78\\x=\frac{78}{110} \\\\x=\frac{39}{55}$

$\frac{1}{y} =\frac{23}{6} \\\\23y=6\\y=\frac{6}{23}$

∴Answer:   $x=\frac{39}{55} ; y=\frac{23}{6}$