1. **Problem (a): Factorise fully** $6y^2 - 5y - 4$ 2. To factorise a quadratic expression of the form $ay^2 + by + c$, we look for two numbers that multiply to $a \times c$ and add to $b$. 3. Here, $a=6$, $b=-5$, and $c=-4$. Calculate $a \times c = 6 \times (-4) = -24$. 4. Find two numbers that multiply to $-24$ and add to $-5$. These numbers are $3$ and $-8$ because $3 \times (-8) = -24$ and $3 + (-8) = -5$. 5. Rewrite the middle term using these numbers: $$6y^2 + 3y - 8y - 4$$ 6. Group terms: $$(6y^2 + 3y) + (-8y - 4)$$ 7. Factor each group: $$3y(2y + 1) - 4(2y + 1)$$ 8. Factor out the common binomial: $$(3y - 4)(2y + 1)$$ 9. **Answer (a):** $6y^2 - 5y - 4 = (3y - 4)(2y + 1)$ --- 10. **Problem (b): Express** $\frac{2x + 1}{4x} + \frac{7 - 5x}{3x}$ **as a single fraction in simplest form**. 11. The denominators are $4x$ and $3x$. The least common denominator (LCD) is $12x$. 12. Rewrite each fraction with denominator $12x$: $$\frac{2x + 1}{4x} = \frac{3(2x + 1)}{12x} = \frac{6x + 3}{12x}$$ $$\frac{7 - 5x}{3x} = \frac{4(7 - 5x)}{12x} = \frac{28 - 20x}{12x}$$ 13. Add the numerators: $$6x + 3 + 28 - 20x = (6x - 20x) + (3 + 28) = -14x + 31$$ 14. So the combined fraction is: $$\frac{-14x + 31}{12x}$$ 15. Check if numerator and denominator have common factors. They do not. 16. **Answer (b):** $\frac{-14x + 31}{12x}$ --- 17. The right triangle with legs 3 and 4 and hypotenuse 5 is a classic Pythagorean triple, confirming the triangle is right-angled.