1. Given the system \(\begin{cases} xy = 7, \\ xy = 18; \end{cases}\) Since both equations say \(xy\) equals different values (7 and 18), there is no solution because one product cannot equal two different numbers simultaneously. 2. Given the system \(\begin{cases} x - y = 2, \\ xy = 15; \end{cases}\) 1. Solve for \(x\) from the first equation: \(x = y + 2\). 2. Substitute into the second equation: \((y + 2)y = 15 \Rightarrow y^2 + 2y - 15 = 0\). 3. Solve the quadratic equation for \(y\) using the quadratic formula: $$y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 60}}{2} = \frac{-2 \pm \sqrt{64}}{2} $$ $$ = \frac{-2 \pm 8}{2}$$ 4. The two values for \(y\) are: - \(y = \frac{-2 + 8}{2} = 3\) - \(y = \frac{-2 - 8}{2} = -5\) 5. Calculate corresponding \(x\) values: - For \(y = 3\), \(x = 3 + 2 = 5\) - For \(y = -5\), \(x = -5 + 2 = -3\) 6. Solutions: \((x, y) = (5, 3)\) or \((-3, -5)\). 3. Given the system \(\begin{cases} x - y = 16, \\ xy = -48; \end{cases}\) 1. Express \(x\) as \(x = y + 16\). 2. Substitute into the product: \((y + 16) y = -48 \Rightarrow y^2 + 16 y + 48 = 0\). 3. Solve the quadratic: $$y = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 48}}{2} = \frac{-16 \pm \sqrt{256 - 192}}{2} = \frac{-16 \pm \sqrt{64}}{2} $$ $$ = \frac{-16 \pm 8}{2}$$ 4. The roots for \(y\) are: - \(y = \frac{-16 + 8}{2} = -4\) - \(y = \frac{-16 - 8}{2} = -12\) 5. Find corresponding \(x\): - For \(y = -4\), \(x = -4 + 16 = 12\) - For \(y = -12\), \(x = -12 + 16 = 4\) 6. Solutions: \((x, y) = (12, -4)\) or \((4, -12)\). 4. Given the system \(\begin{cases} x - y = 3, \\ xy = -2; \end{cases}\) 1. Express \(x = y + 3\). 2. Substitute into product: $$(y + 3) y = -2 \Rightarrow y^2 + 3 y + 2 = 0$$ 3. Factor the quadratic: $$(y + 1)(y + 2) = 0$$ 4. Solutions for \(y\): \(-1\) or \(-2\) 5. Corresponding \(x\): - For \(y = -1\), \(x = -1 + 3 = 2\) - For \(y = -2\), \(x = -2 + 3 = 1\) 6. Solutions: \((x, y) = (2, -1)\) or \((1, -2)\). Final answers: 1. No solution 2. \((5,3), (-3,-5)\) 3. \((12,-4), (4,-12)\) 4. \((2,-1), (1,-2)\)