Question

Solve the following equations for $x$ and $y$:
${\log }_{100}|x+y|=\frac{1}{2},{\log }_{10}y-{\log }_{10}\left|x\right|={\log }_{100}4$

• A. $(x,y)=(\frac{10}{3},\frac{20}{3}),(-10,20)$
• B. $(x,y)=(\frac{-10}{3},\frac{20}{3}),(-10,20)$
• C. $(x,y)=(\frac{-20}{3},\frac{-10}{3}),(-10,20)$
• D. $(x,y)=(\frac{10}{3},\frac{20}{3}),(-20,-10)$

Answer 1

The correct option is A $(x,y)=(\frac{10}{3},\frac{20}{3}),(-10,20)$

${\log }_{100}|x+y|=\frac{1}{2}={\log }_{100}10$
$\Rightarrow |x+y|=10$ and $|x+y|>0$
${\log }_{10}y-{\log }_{10}|x|={\log }_{100}4={\log }_{10}2$
Here, $y>0$ and $|x|>0$
$\Rightarrow {\log }_{10}\frac{y}{|x|}={\log }_{10}2$
$\Rightarrow y=2|x|$
Case 1: $x>0$
$\because y>0$
$x+y=10$ and $y=2x$, gives $x=\frac{10}{3},y=\frac{20}{3}$

Case 2: $x>-y$ and $x<0$
$x+y=10$ and $y=-2x$, gives $x=-10,y=20$
$\therefore (x,y)=(\frac{10}{3},\frac{20}{3})$ and $(-10,20)$
Hence, option A.