1. **Problem statement:** We need to find values of $x$ and $y$ such that the two right triangles in item 6 are congruent by the HL (Hypotenuse-Leg) theorem. 2. **HL theorem:** Two right triangles are congruent if their hypotenuses are equal and one corresponding leg is equal. 3. **Identify sides:** - Triangle 1 legs: $3y + x$ and $y - x$ - Triangle 2 legs: $x + 5$ and $y + 5$ 4. **Find hypotenuses:** - Hypotenuse of Triangle 1: $$h_1 = \sqrt{(3y + x)^2 + (y - x)^2}$$ - Hypotenuse of Triangle 2: $$h_2 = \sqrt{(x + 5)^2 + (y + 5)^2}$$ 5. **Set hypotenuses equal:** $$\sqrt{(3y + x)^2 + (y - x)^2} = \sqrt{(x + 5)^2 + (y + 5)^2}$$ Square both sides: $$ (3y + x)^2 + (y - x)^2 = (x + 5)^2 + (y + 5)^2 $$ 6. **Expand each term:** $$ (3y + x)^2 = 9y^2 + 6xy + x^2 $$ $$ (y - x)^2 = y^2 - 2xy + x^2 $$ $$ (x + 5)^2 = x^2 + 10x + 25 $$ $$ (y + 5)^2 = y^2 + 10y + 25 $$ 7. **Sum left side:** $$ 9y^2 + 6xy + x^2 + y^2 - 2xy + x^2 = 10y^2 + 4xy + 2x^2 $$ 8. **Sum right side:** $$ x^2 + 10x + 25 + y^2 + 10y + 25 = x^2 + y^2 + 10x + 10y + 50 $$ 9. **Set equal and simplify:** $$ 10y^2 + 4xy + 2x^2 = x^2 + y^2 + 10x + 10y + 50 $$ Move all terms to one side: $$ 10y^2 + 4xy + 2x^2 - x^2 - y^2 - 10x - 10y - 50 = 0 $$ Simplify: $$ 9y^2 + 4xy + x^2 - 10x - 10y - 50 = 0 $$ 10. **Leg equality:** For HL, one leg must be equal. We can try: - $3y + x = x + 5$ or - $3y + x = y + 5$ or - $y - x = x + 5$ or - $y - x = y + 5$ Check each: - $3y + x = x + 5 \Rightarrow 3y = 5 \Rightarrow y = \frac{5}{3}$ - $3y + x = y + 5 \Rightarrow 2y + x = 5$ - $y - x = x + 5 \Rightarrow y - 2x = 5$ - $y - x = y + 5 \Rightarrow -x = 5 \Rightarrow x = -5$ 11. **Try $y = \frac{5}{3}$:** Substitute into the hypotenuse equation: $$ 9\left(\frac{5}{3}\right)^2 + 4x\left(\frac{5}{3}\right) + x^2 - 10x - 10\left(\frac{5}{3}\right) - 50 = 0 $$ Calculate: $$ 9 \times \frac{25}{9} + \frac{20}{3}x + x^2 - 10x - \frac{50}{3} - 50 = 0 $$ Simplify constants: $$ 25 + \frac{20}{3}x + x^2 - 10x - \frac{50}{3} - 50 = 0 $$ Combine constants: $$ 25 - 50 = -25 $$ $$ -25 - \frac{50}{3} = -\frac{75}{3} - \frac{50}{3} = -\frac{125}{3} $$ Equation: $$ x^2 + \frac{20}{3}x - 10x - \frac{125}{3} = 0 $$ Simplify $x$ terms: $$ x^2 + \left(\frac{20}{3} - 10\right)x - \frac{125}{3} = 0 $$ $$ x^2 - \frac{10}{3}x - \frac{125}{3} = 0 $$ 12. **Solve quadratic for $x$:** $$ x = \frac{\cancel{-}(-\frac{10}{3}) \pm \sqrt{\left(-\frac{10}{3}\right)^2 - 4 \times 1 \times \left(-\frac{125}{3}\right)}}{2} $$ $$ = \frac{\frac{10}{3} \pm \sqrt{\frac{100}{9} + \frac{500}{3}}}{2} $$ Convert to common denominator inside sqrt: $$ \frac{100}{9} + \frac{1500}{9} = \frac{1600}{9} $$ $$ x = \frac{\frac{10}{3} \pm \frac{40}{3}}{2} $$ Two solutions: $$ x = \frac{\frac{10}{3} + \frac{40}{3}}{2} = \frac{50/3}{2} = \frac{50}{6} = \frac{25}{3} $$ $$ x = \frac{\frac{10}{3} - \frac{40}{3}}{2} = \frac{-30/3}{2} = \frac{-10}{2} = -5 $$ 13. **Check leg equality for $x = -5$ and $y = \frac{5}{3}$:** $$ 3y + x = 3 \times \frac{5}{3} + (-5) = 5 - 5 = 0 $$ $$ x + 5 = -5 + 5 = 0 $$ Legs equal, valid. 14. **Final answer:** $$ \boxed{x = -5, \quad y = \frac{5}{3}} $$ These values satisfy the HL theorem for congruence of the two right triangles in item 6.