If $3^{2x-y} = 3^{x+y} = \sqrt{27}$, the value of $y$ is

### Concept & Strategy When a continuous equality chain involves exponential terms, convert every term to the same base. Here, $\sqrt{27}$ must be expressed as a power of $3$. Once the bases are identical, set up simultaneous linear equations by equating the exponents. ### Step-by-Step Solution * **Step 1:** Express the constant term $\sqrt{27}$ as a base of $3$. $ 27 = 3^3 $ $ \sqrt{27} = (3^3)^{\frac{1}{2}} = 3^{\frac{3}{2}} $ * **Step 2:** Rewrite the full chain of equations with the common base. $$ 3^{2x-y} = 3^{x+y} = 3^{\frac{3}{2}} $$ * **Step 3:** Break the chain into two separate linear equations by dropping the bases. Equation (1): $ 2x - y = \frac{3}{2} $ Equation (2): $ x + y = \frac{3}{2} $ * **Step 4:** Solve the system of linear equations for $y$. First, find $x$ in terms of $y$ by setting the left sides equal (since both equal $\frac{3}{2}$): $ 2x - y = x + y $ $ x = 2y $ * **Step 5:** Substitute $x = 2y$ into Equation (2). $ 2y + y = \frac{3}{2} $ $ 3y = \frac{3}{2} $ $ y = \frac{3}{2 \cdot 3} $ $ y = \frac{1}{2} $ ### Exam Strategy & Shortcut **Simultaneous Equation Elimination:** Once you extract $2x - y = 1.5$ and $x + y = 1.5$, standard elimination is lightning fast. We want $y$, so let's eliminate $x$. Multiply the second equation by $2$: $2x + 2y = 3$. Subtract the first equation from this new equation: $ (2x + 2y) - (2x - y) = 3 - 1.5 $ $ 3y = 1.5 $ $ y = 0.5 $ (or $\frac{1}{2}$). This skips substitution steps and avoids fraction miscalculations. ### Common Pitfall Students often correctly identify that $27$ is $3^3$, but mishandle the square root, erroneously writing $3^{\frac{1}{3}}$ instead of $3^{\frac{3}{2}}$. Remember that a square root always corresponds to a power of $\frac{1}{2}$. ### Final Answer Therefore, the correct answer is **$\frac{1}{2}$**.