1. **State the problem:** Factorise the expression $7y(5x+10)-(x+2)(3x+5)$. 2. **Expand each term:** $$7y(5x+10) = 7y \times 5x + 7y \times 10 = 35xy + 70y$$ $$(x+2)(3x+5) = x \times 3x + x \times 5 + 2 \times 3x + 2 \times 5 = 3x^2 + 5x + 6x + 10 = 3x^2 + 11x + 10$$ 3. **Rewrite the expression:** $$35xy + 70y - (3x^2 + 11x + 10) = 35xy + 70y - 3x^2 - 11x - 10$$ 4. **Group terms to factor:** Group terms with $y$ and without $y$: $$(35xy + 70y) - (3x^2 + 11x + 10)$$ 5. **Factor out common factors in each group:** $$35y(x + 2) - (3x^2 + 11x + 10)$$ 6. **Factor the quadratic $3x^2 + 11x + 10$:** Find two numbers that multiply to $3 \times 10 = 30$ and add to $11$: these are $5$ and $6$. Rewrite: $$3x^2 + 5x + 6x + 10 = (3x^2 + 5x) + (6x + 10)$$ Factor each group: $$x(3x + 5) + 2(3x + 5) = (x + 2)(3x + 5)$$ 7. **Rewrite the expression:** $$35y(x + 2) - (x + 2)(3x + 5)$$ 8. **Factor out the common binomial $(x + 2)$:** $$\cancel{(x + 2)}(35y) - \cancel{(x + 2)}(3x + 5) = (x + 2)(35y - (3x + 5))$$ 9. **Simplify inside the parentheses:** $$35y - 3x - 5$$ 10. **Final factorised form:** $$\boxed{(x + 2)(35y - 3x - 5)}$$