Question

Solve for x and y by substitution method:

$\frac{1}{2x}+\frac{1}{3y}=2(x\ne 0,y\ne 0)\frac{1}{3x}+\frac{1}{y}=\frac{11}{3}$

• A.
  $\frac{1}{2}and\frac{1}{3}$
• B.
  $\frac{1}{4}and\frac{1}{5}$
• C.
  
  $\frac{1}{2}and\frac{1}{4}$
• D.
  $\frac{1}{6}and\frac{1}{3}$

Answer 1

The correct option is A

$\frac{1}{2}and\frac{1}{3}$
Let $\frac{1}{x}=a,\frac{1}{y}=b\frac{1}{2x}+\frac{1}{3y}=2\Rightarrow \frac{a}{2}+\frac{b}{3}=\mathrm{2.......}\left(1\right)\frac{1}{3x}+\frac{1}{y}=\frac{11}{3}\Rightarrow \frac{a}{3}+b=\frac{11}{3}....\left(2\right)fromequation\left(2\right)\frac{a}{3}+b=\frac{11}{3}\Rightarrow b=\frac{11}{3}-\frac{a}{3}.....\left(3\right)Substitutingb=\frac{11}{3}-\frac{a}{3}in\left(1\right)\frac{a}{3}+\frac{b}{3}=2\Rightarrow \frac{a}{2}+\frac{1}{3}(\frac{11}{3}-\frac{a}{3})=2\Rightarrow \frac{a}{2}+\frac{11}{9}-\frac{a}{9}=2\Rightarrow \frac{a}{2}-\frac{a}{9}=2-\frac{11}{9}\Rightarrow \frac{9a-2a}{18}=\frac{18-11}{9}\Rightarrow \frac{7a}{18}=\frac{7}{9}\therefore a=2Substitutinga=2inequation\left(3\right),b=\frac{11}{3}-\frac{a}{3}=\frac{11}{3}-\frac{2}{3}=\frac{9}{3}\therefore b=3a=2\to \frac{1}{x}=2\Rightarrow x=\frac{1}{2}b=3\Rightarrow \frac{1}{y}=3\Rightarrow y=\frac{1}{3}$