Exercise 1. Let $x$ be a strictly positive real number such that $\frac{\sqrt{x^{2}+4 x+4}}{\sqrt{x^{2}+6 x+9}}=\frac{37}{39}$. Determine the value of $10 x$.
Solution to Exercise 1 We have $\frac{\sqrt{(x+2)^{2}}}{\sqrt{(x+3)^{2}}}=\frac{37}{39}$, so $\frac{x+2}{x+3}=\frac{37}{39}$, which gives $39(x+2)=$ $37(x+3)$. This simplifies to $(39-37) x=37 \times 3-39 \times 2=111-78=33$, or $x=33 / 2$. Finally, $10 x=33 \times 5=165$.