1. Stating the problems: Solve the systems of equations given for each case. **Part 1:** Given $$x^2 + y^2 = 13$$ and $$xy = 6$$. Step 1: Recall the identity $$(x + y)^2 = x^2 + y^2 + 2xy$$. Step 2: Substitute known values: $$ (x + y)^2 = 13 + 2 \cdot 6 = 13 + 12 = 25 $$ Step 3: Taking square root gives: $$ x + y = \pm 5 $$ Step 4: Now, $x$ and $y$ satisfy: $$ t^2 - (x+y)t + xy = 0 \implies t^2 - (\pm 5)t + 6 = 0 $$ Solve for $t$ in both cases: - For $x + y = 5$: $$ t^2 - 5t + 6 = 0 \Rightarrow (t - 2)(t - 3) = 0 \Rightarrow t = 2, 3 $$ - For $x + y = -5$: $$ t^2 + 5t + 6 = 0 \Rightarrow (t + 2)(t + 3) = 0 \Rightarrow t = -2, -3 $$ Therefore, solutions are $(x,y) = (2,3), (3,2), (-2,-3), (-3,-2)$. --- **Part 2:** Given $$x^2 + y^2 = 41$$ and $$xy = 20$$. Step 1: Using identity: $$ (x + y)^2 = 41 + 2 \cdot 20 = 41 + 40 = 81 $$ Step 2: So: $$ x + y = \pm 9 $$ Step 3: Quadratic for $t$: $$ t^2 - (x+y) t + xy = 0 $$ For $x + y = 9$: $$ t^2 - 9t + 20 = 0 \Rightarrow (t - 4)(t - 5) = 0 \Rightarrow t = 4, 5 $$ For $x + y = -9$: $$ t^2 + 9t + 20 = 0 \Rightarrow (t + 4)(t + 5) = 0 \Rightarrow t = -4, -5 $$ Solutions: $(4,5), (5,4), (-4,-5), (-5,-4)$. --- **Part 3:** Given $$x^2 + y^2 = 10$$ and $$xy = 3$$. Step 1: $$ (x + y)^2 = 10 + 2 \cdot 3 = 16 $$ Step 2: $$ x + y = \pm 4 $$ Step 3: Quadratic equations: For $x + y = 4$: $$ t^2 - 4t + 3 = 0 \Rightarrow (t - 3)(t - 1) = 0 $$ For $x + y = -4$: $$ t^2 + 4t + 3 = 0 \Rightarrow (t + 3)(t + 1) = 0 $$ Solutions: $(3,1), (1,3), (-3,-1), (-1,-3)$. --- **Part 4:** Given $$x^2 + y^2 = 34$$ and $$xy = 15$$. Step 1: $$ (x + y)^2 = 34 + 2 \cdot 15 = 64 $$ Step 2: $$ x + y = \pm 8 $$ Step 3: For $x + y = 8$: $$ t^2 - 8t + 15 = 0 \Rightarrow (t - 5)(t - 3) = 0 $$ For $x + y = -8$: $$ t^2 + 8t + 15 = 0 \Rightarrow (t + 5)(t + 3) = 0 $$ Solutions: $(5,3), (3,5), (-5,-3), (-3,-5)$. --- **Part 5:** Given $$x^3 + y^3 = 7$$ and $$x^2 y^3 = -8$$. Step 1: Recall that $$x^3 + y^3 = (x + y)^3 - 3xy(x + y)$$. Step 2: Need values of $x+y$ and $xy$, but not given explicitly. Step 3: Let $s = x + y$ and $p = xy$. Then: $$ x^3 + y^3 = s^3 - 3ps = 7 $$ Step 4: Given $$x^2 y^3 = x^2 (y^3)$$ but no explicit way to separate. This is tricky; try assumptions or methods better suited to specific values. Since $x^2 y^3 = x^2 y^3 = (xy)^2 y = p^2 \cdot y$. Without $y$, this is a complex system needing substitution and possibly more info. Without further info, no unique solution is available. --- **Part 6:** Given $$\frac{x + y}{y} = 20$$ and $$(x + y) + \frac{x}{y} = 9$$. Step 1: From first equation: $$ \frac{x + y}{y} = 20 \implies x + y = 20y $$ Step 2: Substitute into second: $$ (x + y) + \frac{x}{y} = 9 \implies 20y + \frac{x}{y} = 9 $$ Step 3: Replace $x$ using $x = 20y - y = 19y$ from step 1: $$ 20y + \frac{19y}{y} = 9 \implies 20y + 19 = 9 $$ Step 4: Solve for $y$: $$ 20y = 9 - 19 = -10 \implies y = -\frac{1}{2} $$ Step 5: Substitute back to find $x$: $$ x = 20y - y = 19y = 19 \cdot -\frac{1}{2} = -\frac{19}{2} $$ Solution: $$\left(-\frac{19}{2}, -\frac{1}{2}\right)$$. --- **Part 7:** Given $$x^2 + y^2 + xy = 6$$ and $$xy + x + y = 5$$. Step 1: Let $s = x + y$ and $p = xy$. Step 2: Note that: $$ x^2 + y^2 = (x + y)^2 - 2xy = s^2 - 2p $$ The first equation becomes: $$ s^2 - 2p + p = 6 \implies s^2 - p = 6 $$ Step 3: The second equation is: $$ p + s = 5 $$ Step 4: From the second equation: $$ p = 5 - s $$ Substitute into the first: $$ s^2 - (5 - s) = 6 \implies s^2 - 5 + s = 6 $$ Step 5: Rearrange: $$ s^2 + s - 11 = 0 $$ Step 6: Solve quadratic: $$ s = \frac{-1 \pm \sqrt{1 + 44}}{2} = \frac{-1 \pm \sqrt{45}}{2} = \frac{-1 \pm 3\sqrt{5}}{2} $$ Step 7: Find corresponding $p$ values: $$ p = 5 - s $$ So two cases: - $s_1 = \frac{-1 + 3\sqrt{5}}{2} \Rightarrow p_1 = 5 - s_1 = \frac{11 - 3\sqrt{5}}{2}$ - $s_2 = \frac{-1 - 3\sqrt{5}}{2} \Rightarrow p_2 = 5 - s_2 = \frac{11 + 3\sqrt{5}}{2}$ Step 8: Find $x,y$ by solving: $$ t^2 - s t + p = 0 $$ For both sets. Therefore, solutions correspond to roots of quadratics: $$ t^2 - \frac{-1 + 3\sqrt{5}}{2} t + \frac{11 - 3\sqrt{5}}{2} = 0 $$ and $$ t^2 - \frac{-1 - 3\sqrt{5}}{2} t + \frac{11 + 3\sqrt{5}}{2} = 0 $$. --- **Summary:** - Problems 1-4 solved completely using sum and product of roots. - Problem 6 solved. - Problem 7 reduced to quadratic form solutions for $s$ and $p$. - Problem 5 is underdetermined without further info.