Question

Solve the following equation for $(x,y)$:

${\log }_{100}|x+y|=\frac{1}{2},{\log }_{10}y-{\log }_{10}|x|={\log }_{100}4$

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

Answer 1

Answer: (A)

$(-10,20),(\frac{10}{3},\frac{20}{3})$

${\log }_{100}|x+y|=\frac{1}{2},{\log }_{10}y-{\log }_{10}|x|={\log }_{100}4={\log }_{10}2$
$\Rightarrow |x+y|=10$ and $\frac{y}{|x|}=2$
$\Rightarrow x+y=\pm 10$ and $y=2|x|$

From these two, $x+2|x|=\pm 10$
Now when $x>0,x+2x=\pm 10\Rightarrow x=10/3\Rightarrow y=\frac{20}{3}$
and when $x<0,x-2x=\pm 10\Rightarrow x=-10\Rightarrow y=20$
Hence, the solutions are $(-10,20)$ and $(\frac{10}{3},\frac{20}{3})$