1. **State the problem:** We need to find the value of $A$ given three balancing scales involving bananas, pears, lemons, and peas. 2. **Assign variables:** Let - $B$ = weight of one banana - $P$ = weight of one pear - $L$ = weight of one lemon - $Pe$ = weight of one pea 3. **Translate the scales into equations:** - Top-left scale: One pair of bananas balanced with one pair of pears means $$2B = 2P \implies B = P$$ - Top-right scale: One pair of bananas and two lemons balanced with one pair of peas and seven bananas means $$2B + 2L = 2Pe + 7B$$ Rearranged: $$2L = 2Pe + 7B - 2B = 2Pe + 5B$$ - Bottom scale: One banana balanced with one lemon labeled as $x A$ means $$B = L = xA$$ 4. **Use $B = P$ and $B = L$ to substitute:** From step 3, $L = B$, so substitute $L$ with $B$ in the second equation: $$2B = 2Pe + 5B \implies 2Pe = 2B - 5B = -3B \implies Pe = -\frac{3}{2}B$$ 5. **Interpretation:** The pea weight is negative relative to banana weight, which is unusual but mathematically consistent. 6. **Find $A$:** From the bottom scale, $B = L = xA$, so $$A = \frac{B}{x}$$ Since $x$ is the label on the lemon in the bottom scale and the banana weight equals lemon weight, $A$ equals the banana weight divided by $x$. **Final answer:** $$A = \frac{B}{x}$$ Without numerical values for $B$ or $x$, this is the expression for $A$ based on the given scales.