1. The problem involves simplifying algebraic expressions and understanding the floor plan dimensions with tile prices. 2. First, simplify the fraction $$\frac{14x^2 - 63x}{7x}$$. 3. Factor the numerator: $$14x^2 - 63x = 7x(2x - 9)$$. 4. Substitute back: $$\frac{7x(2x - 9)}{7x}$$. 5. Cancel the common factor $$7x$$: $$\frac{\cancel{7x}(2x - 9)}{\cancel{7x}} = 2x - 9$$. 6. Next, simplify the fraction $$\frac{-10x^2 - 40x^2}{2x^2 - 8x}$$. 7. Combine like terms in numerator: $$-10x^2 - 40x^2 = -50x^2$$. 8. Factor denominator: $$2x^2 - 8x = 2x(x - 4)$$. 9. Write the fraction: $$\frac{-50x^2}{2x(x - 4)}$$. 10. Cancel common factor $$2x$$: $$\frac{-50x^2}{2x(x - 4)} = \frac{-\cancel{50}x^{\cancel{2}}}{\cancel{2}x(x - 4)} = \frac{-25x}{x - 4}$$. 11. The problem states this equals $$-5x$$, so check if $$\frac{-25x}{x - 4} = -5x$$. 12. Multiply both sides by $$x - 4$$: $$-25x = -5x(x - 4)$$. 13. Expand right side: $$-25x = -5x^2 + 20x$$. 14. Rearrange: $$0 = -5x^2 + 20x + 25x = -5x^2 + 45x$$. 15. Divide both sides by $$-5$$: $$0 = x^2 - 9x$$. 16. Factor: $$x(x - 9) = 0$$. 17. Solutions: $$x = 0$$ or $$x = 9$$. 18. Since $$x=0$$ is not valid for dimensions, $$x=9$$. 19. Tile prices and dimensions: - Tile A: size $$x \times 2$$, price 10 per tile. - Tile B: not available. - Tile C: size $$x \times 0.5$$, price 3.5 per tile. - Tile D: size $$1.5 \times x \times 0.5$$, price 13 per tile. - Tile E: size 2, price 1.5 per tile. 20. The floor plan dimensions: - Main floor: width $$3x + 4$$, height $$x - 9$$. - Smaller block: $$x + 6$$. 21. Since Tile B is not included, only consider Tiles A, C, D, and E for calculations. Final simplified expressions: $$\frac{14x^2 - 63x}{7x} = 2x - 9$$ $$\frac{-10x^2 - 40x^2}{2x^2 - 8x} = -5x$$ with $$x=9$$.