If $\log_{10} x + \log_{10} y = 3$ and $\log_{10} x - \log_{10} y = 1$, then $x$ and $y$ are respectively

### Concept & Strategy This is a system of linear equations disguised as a logarithms problem. Treat $\log_{10}(x)$ and $\log_{10}(y)$ as individual variables (like $A$ and $B$) and solve using the standard elimination method. ### Step-by-Step Solution Let $A = \log_{10}(x)$ and $B = \log_{10}(y)$. The given equations become: Equation 1: $A + B = 3$ Equation 2: $A - B = 1$ Use the elimination method. Add Equation 1 and Equation 2 together: $(A + B) + (A - B) = 3 + 1$ $2A = 4$ $A = 2$ Now substitute $A = 2$ back into Equation 1 to find $B$: $2 + B = 3$ $B = 1$ Now, substitute the logarithmic definitions back for $A$ and $B$: $\log_{10}(x) = 2$ $\log_{10}(y) = 1$ Convert to exponential form to solve for $x$ and $y$: $x = 10^2 = 100$ $y = 10^1 = 10$ So, $x$ and $y$ are respectively $100$ and $10$. ### Exam Strategy & Shortcut You don't even need to solve the linear equations. Use the product and quotient rules directly on the values. $A + B = 3$ means $\log_{10}(xy) = 3 \Rightarrow xy = 1000$. $A - B = 1$ means $\log_{10}(x/y) = 1 \Rightarrow x/y = 10$. Look at the options. (a) $10 \times 100 = 1000$ (Pass), $10/100 = 0.1$ (Fail) (b) $100 \times 10 = 1000$ (Pass), $100/10 = 10$ (Pass) Option (b) is the correct answer. ### Common Pitfall The main pitfall is swapping the order of $x$ and $y$ in the final answer. The question asks for "$x$ and $y$ respectively". Since $x = 100$ and $y = 10$, selecting Option (a) "$10$ and $100$" is a careless mistake that costs full marks despite doing the math correctly. ### Final Answer **Therefore, the correct answer is 100 and 10.**