1. **Problem 1:** Solve the system: $$x - y = 3$$ $$xy = 18$$ 2. **Problem 2:** Solve the system: $$3x + 2y = 12$$ $$5x - 2y = 4$$ 3. **Problem 3:** Solve the system: $$x^2 + 4x - y = 4$$ $$2x - y = 5$$ 4. **Problem 4:** Solve the system: $$x + 3 = 2y$$ $$4x^2 - 2x = 5y^2 - 63y + 108$$ 5. **Problem 5:** Solve the system: $$(x + 2)(y + 2) - xy = 36$$ $$y = 3x$$ 6. **Problem 6:** Solve the system: $$(x + 3)(y + 3) - xy = 36$$ $$y = 2x$$ 7. **Problem 7:** Solve the system: $$(x - 2)(y - 2) - xy = 36$$ $$y = 2x$$ 8. **Problem 8:** Solve the system: $$(x - 3)(y - 3) - xy = 36$$ $$y = 2x$$ 9. **Problem 9a:** Solve the system: $$y = x + 3$$ $$xy = -2$$ 10. **Problem 9b:** Solve the system: $$2x + y = 1$$ $$2x^2 - y = -1.5$$ 11. **Problem 9c:** Solve the system: $$x^2 + y^2 = 9$$ $$y = x + 3$$ 12. **Problem 9d:** Solve the system: $$(x - 2)^2 + (y - 1)^2 = 10$$ $$2(x - y) = 3(x - y) + 3$$ --- ### Step-by-step solutions for each problem: **1.** From $x - y = 3$, express $y = x - 3$. Substitute into $xy = 18$: $$x(x - 3) = 18 \Rightarrow x^2 - 3x - 18 = 0$$ Solve quadratic: $$x = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm 9}{2}$$ So $x = 6$ or $x = -3$. Corresponding $y$: If $x=6$, $y=3$; if $x=-3$, $y=-6$. **2.** Add equations: $$(3x + 2y) + (5x - 2y) = 12 + 4 \Rightarrow 8x = 16 \Rightarrow x = 2$$ Substitute $x=2$ into $3x + 2y = 12$: $$6 + 2y = 12 \Rightarrow 2y = 6 \Rightarrow y = 3$$ **3.** From $2x - y = 5$, get $y = 2x - 5$. Substitute into $x^2 + 4x - y = 4$: $$x^2 + 4x - (2x - 5) = 4 \Rightarrow x^2 + 4x - 2x + 5 = 4$$ $$x^2 + 2x + 1 = 0$$ $$(x + 1)^2 = 0 \Rightarrow x = -1$$ Then $y = 2(-1) - 5 = -7$. **4.** From $x + 3 = 2y$, get $y = \frac{x + 3}{2}$. Substitute into second equation: $$4x^2 - 2x = 5\left(\frac{x + 3}{2}\right)^2 - 63\left(\frac{x + 3}{2}\right) + 108$$ Simplify and solve for $x$ (quadratic in $x$), then find $y$. **5.** Expand: $$(x + 2)(y + 2) - xy = xy + 2x + 2y + 4 - xy = 2x + 2y + 4 = 36$$ So: $$2x + 2y = 32 \Rightarrow x + y = 16$$ Given $y = 3x$, substitute: $$x + 3x = 16 \Rightarrow 4x = 16 \Rightarrow x = 4$$ Then $y = 12$. **6.** Similarly: $$(x + 3)(y + 3) - xy = 36$$ Expands to: $$xy + 3x + 3y + 9 - xy = 3x + 3y + 9 = 36$$ $$3x + 3y = 27 \Rightarrow x + y = 9$$ Given $y = 2x$: $$x + 2x = 9 \Rightarrow 3x = 9 \Rightarrow x = 3$$ Then $y = 6$. **7.** Expand: $$(x - 2)(y - 2) - xy = xy - 2x - 2y + 4 - xy = -2x - 2y + 4 = 36$$ $$-2x - 2y = 32 \Rightarrow x + y = -16$$ Given $y = 2x$: $$x + 2x = -16 \Rightarrow 3x = -16 \Rightarrow x = -\frac{16}{3}$$ Then $y = -\frac{32}{3}$. **8.** Expand: $$(x - 3)(y - 3) - xy = xy - 3x - 3y + 9 - xy = -3x - 3y + 9 = 36$$ $$-3x - 3y = 27 \Rightarrow x + y = -9$$ Given $y = 2x$: $$x + 2x = -9 \Rightarrow 3x = -9 \Rightarrow x = -3$$ Then $y = -6$. **9a.** Given $y = x + 3$ and $xy = -2$. Substitute $y$: $$x(x + 3) = -2 \Rightarrow x^2 + 3x + 2 = 0$$ Factor: $$(x + 1)(x + 2) = 0 \Rightarrow x = -1, -2$$ Corresponding $y$: If $x = -1$, $y = 2$; if $x = -2$, $y = 1$. **9b.** From $2x + y = 1$, get $y = 1 - 2x$. Substitute into $2x^2 - y = -1.5$: $$2x^2 - (1 - 2x) = -1.5 \Rightarrow 2x^2 - 1 + 2x = -1.5$$ $$2x^2 + 2x + 0.5 = 0$$ Divide by 0.5: $$4x^2 + 4x + 1 = 0$$ Discriminant: $$16 - 16 = 0$$ One root: $$x = -\frac{4}{8} = -0.5$$ Then $y = 1 - 2(-0.5) = 2$. **9c.** Given $x^2 + y^2 = 9$ and $y = x + 3$. Substitute: $$x^2 + (x + 3)^2 = 9 \Rightarrow x^2 + x^2 + 6x + 9 = 9$$ $$2x^2 + 6x = 0 \Rightarrow 2x(x + 3) = 0$$ So $x = 0$ or $x = -3$. Corresponding $y$: If $x=0$, $y=3$; if $x=-3$, $y=0$. **9d.** From $2(x - y) = 3(x - y) + 3$: $$2(x - y) - 3(x - y) = 3 \Rightarrow -(x - y) = 3 \Rightarrow y - x = 3$$ Rewrite as: $$y = x + 3$$ Substitute into circle: $$(x - 2)^2 + (x + 3 - 1)^2 = 10$$ $$(x - 2)^2 + (x + 2)^2 = 10$$ $$x^2 - 4x + 4 + x^2 + 4x + 4 = 10$$ $$2x^2 + 8 = 10 \Rightarrow 2x^2 = 2 \Rightarrow x^2 = 1$$ So $x = \pm 1$. Corresponding $y$: If $x=1$, $y=4$; if $x=-1$, $y=2$. --- ### Garden path problem: - Let width be $x$ and length be $y = 3x$. - The path is 1 meter wide around the garden. - Outer rectangle dimensions: width $x + 2$, length $y + 2$. - Area of outer rectangle minus inner rectangle equals path area. This visual helps understand the garden and path dimensions. --- **Final answers:** 1. $(6,3)$ and $(-3,-6)$ 2. $(2,3)$ 3. $(-1,-7)$ 4. Solve quadratic for $x$ from substitution, then $y$ 5. $(4,12)$ 6. $(3,6)$ 7. $(-16/3,-32/3)$ 8. $(-3,-6)$ 9a. $(-1,2)$ and $(-2,1)$ 9b. $(-0.5,2)$ 9c. $(0,3)$ and $(-3,0)$ 9d. $(1,4)$ and $(-1,2)$