1. Problem 224: Find how many possible values of $x$ satisfy the equation $5 - \frac{6}{x} = x$. 2. Start by rewriting the equation: $$5 - \frac{6}{x} = x$$ 3. Multiply both sides by $x$ (assuming $x \neq 0$) to clear the denominator: $$5x - 6 = x^2$$ 4. Rearrange to standard quadratic form: $$x^2 - 5x + 6 = 0$$ 5. Factor the quadratic: $$ (x - 2)(x - 3) = 0 $$ 6. So, the solutions are: $$x = 2 \quad \text{or} \quad x = 3$$ 7. Both values are valid since neither is zero (which would make the original denominator zero). 8. Therefore, there are two possible values of $x$. --- 9. Problem 225: Given seed mixtures X and Y with ryegrass percentages 40% and 25% respectively, find the percent weight of X in a mixture that contains 30% ryegrass. 10. Let $w$ be the fraction of mixture X in the total mixture (so $1-w$ is the fraction of mixture Y). 11. The ryegrass percentage in the mixture is given by: $$0.40w + 0.25(1 - w) = 0.30$$ 12. Simplify the equation: $$0.40w + 0.25 - 0.25w = 0.30$$ 13. Combine like terms: $$0.15w + 0.25 = 0.30$$ 14. Subtract 0.25 from both sides: $$0.15w = 0.05$$ 15. Solve for $w$: $$w = \frac{0.05}{0.15} = \frac{1}{3} \approx 0.3333$$ 16. Convert to percentage: $$33 \frac{1}{3} \%$$ 17. Therefore, the mixture contains 33 1/3% of seed mixture X. Final answers: - Problem 224: Two possible values. - Problem 225: 33 1/3% of the mixture is X.