Question

Solve for x and y where $x\ne 0,y\ne 0:$
$\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=\frac{-3}{2}\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}$

• A.
  $\frac{1}{3}and\frac{4}{5}$
• B.
  $\frac{1}{8}and4$
• C.
  $\frac{1}{3}and\frac{4}{5}$
• D.
  $\frac{1}{2}and\frac{5}{4}$

Answer 1

The correct option is D

$\frac{1}{2}and\frac{5}{4}$
$\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=\frac{-3}{2}....\left(1\right)\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}....\left(2\right)Let\frac{1}{x+2y}=a,\frac{1}{3x-2y}=bFrom\left(1\right),\frac{a}{2}+\frac{5b}{3}=\frac{-3}{2}\Rightarrow \frac{3a+10b}{6}=\frac{-3}{2}\Rightarrow 3a+10b=-\frac{3}{2}\times 6\Rightarrow 3a+10b=-\mathrm{9...}\left(3\right)from\left(3\right),\frac{5a}{4}=\frac{3b}{5}=\frac{61}{60}\Rightarrow \frac{25a-12b}{20}=\frac{61}{60}\Rightarrow 75a-36b=61...\left(4\right)3a+10b=-\mathrm{9...}\left(3\right)75a-36b=61...\left(4\right)Multiplyby25in\left(3\right),weget:25(3a+10b)=25(-9)\Rightarrow 75a+250b=-\mathrm{225...}\left(5\right)Subtracting\left(4\right)from\left(5\right),weget:75a+250b=-22575a-36b=61-+---------286b=-286\Rightarrow b=\frac{-286}{286}\therefore b=-1Substitutingb=-1inequation\left(3\right)3a+10b=-9\Rightarrow 3a+10(-1)=-9\Rightarrow 3a-10=-9\Rightarrow 3a=-9+10\Rightarrow 3a=1\therefore a=\frac{1}{3}a=\frac{1}{3}andb=-1\Rightarrow \frac{1}{x+2y}=\frac{1}{3}\Rightarrow \frac{1}{3x-2y}=-1\therefore x+2y=\mathrm{3..}\left(6\right)\therefore 3x-2y=-\mathrm{1...}\left(7\right)Adding\left(6\right)and\left(7\right),weget:x+2y=33x-2y=-1-------4x=2\Rightarrow 4x=2\Rightarrow x=\frac{2}{4}\Rightarrow x=\frac{1}{2}Substitutingn=\frac{1}{2}in\left(6\right),weget:x+2y=3\Rightarrow \frac{1}{2}+2y=3\Rightarrow 2y=3-\frac{1}{2}\Rightarrow 2y=\frac{5}{2}\Rightarrow y=\frac{5}{2}\times \frac{1}{2}\therefore y=\frac{5}{4}$