1. **Problem statement:** There are two numbers. When the second number is subtracted from twice the first number, the result is 18. When 9 is added to the first number, the result is equal to twice the second number. 2. **Form the simultaneous equations:** Let $x$ be the first number and $y$ be the second number. From the problem: - Twice the first number minus the second number equals 18: $$2x - y = 18$$ - Adding 9 to the first number equals twice the second number: $$x + 9 = 2y$$ 3. **Rewrite the second equation for substitution:** $$x + 9 = 2y \implies 2y = x + 9$$ Divide both sides by 2: $$\cancel{2}y = \frac{x + 9}{\cancel{2}} \implies y = \frac{x + 9}{2}$$ 4. **Substitute $y$ into the first equation:** $$2x - \frac{x + 9}{2} = 18$$ Multiply both sides by 2 to clear the denominator: $$2 \times 2x - 2 \times \frac{x + 9}{2} = 2 \times 18$$ $$4x - (x + 9) = 36$$ 5. **Simplify and solve for $x$:** $$4x - x - 9 = 36$$ $$3x - 9 = 36$$ Add 9 to both sides: $$3x = 45$$ Divide both sides by 3: $$\cancel{3}x = \frac{45}{\cancel{3}} \implies x = 15$$ 6. **Find $y$ using $x=15$:** $$y = \frac{15 + 9}{2} = \frac{24}{2} = 12$$ 7. **Final answer:** The two numbers are $x = 15$ and $y = 12$. --- **Slug:** two numbers **Subject:** algebra **Desmos:** {"latex":"y=\frac{x+9}{2}","features":{"intercepts":true,"extrema":true}} **q_count:** 2