1. **State the problem:** Solve the simultaneous equations: $$2xy=45$$ and $$\frac{x}{2} + \frac{y}{3} = 4$$ 2. **Rewrite the second equation:** Multiply both sides by 6 (the least common multiple of 2 and 3) to clear denominators: $$6 \times \left( \frac{x}{2} + \frac{y}{3} \right) = 6 \times 4$$ $$3x + 2y = 24$$ 3. **Express one variable in terms of the other:** From the linear equation, $$2y = 24 - 3x \implies y = \frac{24 - 3x}{2}$$ 4. **Substitute into the first equation:** $$2x \times y = 45 \implies 2x \times \frac{24 - 3x}{2} = 45$$ Simplify: $$x(24 - 3x) = 45$$ $$24x - 3x^2 = 45$$ 5. **Rearrange into standard quadratic form:** $$-3x^2 + 24x - 45 = 0$$ Divide both sides by -3: $$x^2 - 8x + 15 = 0$$ 6. **Factor the quadratic:** $$x^2 - 8x + 15 = (x - 3)(x - 5) = 0$$ So, $$x = 3 \quad \text{or} \quad x = 5$$ 7. **Find corresponding y values:** - For $x=3$: $$y = \frac{24 - 3(3)}{2} = \frac{24 - 9}{2} = \frac{15}{2} = 7.5$$ - For $x=5$: $$y = \frac{24 - 3(5)}{2} = \frac{24 - 15}{2} = \frac{9}{2} = 4.5$$ **Final solutions:** $$(x,y) = (3, 7.5) \quad \text{or} \quad (5, 4.5)$$