1. Problem 1: Find two numbers where the sum of their squares decreased by 75 is 50, and their quotient is 2. 2. Let the two numbers be $x$ and $y$. The conditions are: $$x^2 + y^2 - 75 = 50$$ $$\frac{x}{y} = 2$$ 3. From the quotient, express $x$ in terms of $y$: $$x = 2y$$ 4. Substitute into the first equation: $$ (2y)^2 + y^2 - 75 = 50 $$ $$ 4y^2 + y^2 - 75 = 50 $$ $$ 5y^2 = 125 $$ $$ y^2 = 25 $$ $$ y = \pm 5 $$ 5. Find $x$: $$ x = 2y = \pm 10 $$ 6. So the numbers are $(10,5)$ or $(-10,-5)$. --- 7. Problem 2: Find two numbers where the difference of their squares is 16, and twice the square of the first plus the square of the second is 59. 8. Let the numbers be $a$ and $b$. The conditions: $$a^2 - b^2 = 16$$ $$2a^2 + b^2 = 59$$ 9. From the first equation: $$a^2 = b^2 + 16$$ 10. Substitute into the second: $$2(b^2 + 16) + b^2 = 59$$ $$2b^2 + 32 + b^2 = 59$$ $$3b^2 = 27$$ $$b^2 = 9$$ $$b = \pm 3$$ 11. Find $a^2$: $$a^2 = 9 + 16 = 25$$ $$a = \pm 5$$ 12. Numbers are $(5,3), (5,-3), (-5,3), (-5,-3)$. --- 13. Problem 3: Find length $L$ and width $W$ of a rectangle with perimeter 48 and area 95. 14. Formulas: $$P = 2(L + W) = 48$$ $$A = LW = 95$$ 15. From perimeter: $$L + W = 24$$ $$L = 24 - W$$ 16. Substitute into area: $$(24 - W)W = 95$$ $$24W - W^2 = 95$$ $$W^2 - 24W + 95 = 0$$ 17. Solve quadratic: $$W = \frac{24 \pm \sqrt{24^2 - 4 \times 95}}{2} = \frac{24 \pm \sqrt{576 - 380}}{2} = \frac{24 \pm \sqrt{196}}{2}$$ $$W = \frac{24 \pm 14}{2}$$ 18. Two solutions: $$W = 19, L = 5$$ $$W = 5, L = 19$$ --- 19. Problem 4: Find sides of a right triangle with area 60 and hypotenuse 17. 20. Let legs be $x$ and $y$. Area: $$\frac{1}{2}xy = 60 \Rightarrow xy = 120$$ 21. Pythagoras: $$x^2 + y^2 = 17^2 = 289$$ 22. Use identity: $$(x + y)^2 = x^2 + 2xy + y^2 = 289 + 2 \times 120 = 289 + 240 = 529$$ $$x + y = \sqrt{529} = 23$$ 23. Solve system: $$x + y = 23$$ $$xy = 120$$ 24. Quadratic for $x$: $$x^2 - 23x + 120 = 0$$ 25. Solve: $$x = \frac{23 \pm \sqrt{23^2 - 4 \times 120}}{2} = \frac{23 \pm \sqrt{529 - 480}}{2} = \frac{23 \pm \sqrt{49}}{2}$$ $$x = \frac{23 \pm 7}{2}$$ 26. Solutions: $$x = 15, y = 8$$ $$x = 8, y = 15$$ --- 27. Problem 5: Find length and width of a TV screen with diagonal 25 and area 300. 28. Let length $l$ and width $w$. Diagonal: $$l^2 + w^2 = 25^2 = 625$$ 29. Area: $$lw = 300$$ 30. Use identity: $$(l + w)^2 = l^2 + 2lw + w^2 = 625 + 2 \times 300 = 625 + 600 = 1225$$ $$l + w = \sqrt{1225} = 35$$ 31. Solve system: $$l + w = 35$$ $$lw = 300$$ 32. Quadratic for $l$: $$l^2 - 35l + 300 = 0$$ 33. Solve: $$l = \frac{35 \pm \sqrt{35^2 - 4 \times 300}}{2} = \frac{35 \pm \sqrt{1225 - 1200}}{2} = \frac{35 \pm \sqrt{25}}{2}$$ $$l = \frac{35 \pm 5}{2}$$ 34. Solutions: $$l = 20, w = 15$$ $$l = 15, w = 20$$ Final answers: 1. $(10,5)$ or $(-10,-5)$ 2. $(5,3), (5,-3), (-5,3), (-5,-3)$ 3. Length 19 m, Width 5 m (or vice versa) 4. Legs 15 m and 8 m 5. Length 20 in, Width 15 in (or vice versa)