1. **Problem 1:** Solve the systems: \(x + y = -2ع\) and \(\frac{x}{4} = -\frac{y}{3}\). From the second equation: \(\frac{x}{4} = -\frac{y}{3} \implies 3x = -4y \implies y = -\frac{3}{4}x\). Substitute into the first: \(x + (-\frac{3}{4}x) = -2ع \implies x - \frac{3}{4}x = -2ع \implies \frac{1}{4}x = -2ع \implies x = -8ع\). Then \(y = -\frac{3}{4}(-8ع) = 6ع\). 2. **Problem 2:** Solve the system: \(x - y = 7\) and \(\frac{x}{y} = -\frac{3}{11}\). From the ratio: \(x = -\frac{3}{11}y\). Substitute: \(-\frac{3}{11}y - y = 7 \implies -\frac{3}{11}y - \frac{11}{11}y = 7 \implies -\frac{14}{11}y = 7 \implies y = -\frac{77}{14} = -\frac{11}{2}\). Then \(x = -\frac{3}{11} \times -\frac{11}{2} = \frac{3}{2}\). 3. **Problem 3:** Solve: \(\frac{1}{a} + \frac{1}{b} = \frac{1}{3}\) and \(3a + 3b = ab\). Rewrite first: \(\frac{b+a}{ab} = \frac{1}{3} \implies 3(a+b) = ab\). From second: \(3a+3b=ab\), which matches, so the equations are consistent. Rewrite \(3(a+b) = ab\) as \(ab - 3a - 3b = 0\). Add 9 to both sides: \(ab - 3a - 3b +9 = 9\). Factor by grouping: \((a-3)(b-3) = 9\). Solutions satisfy \((a-3)(b-3) = 9\). 4. **Problem 4:** Given \(a - 5b + 7c = 90\), no further info is given, so this is a linear equation in three variables representing a plane. 5. **Problem 5:** Solve \(77x - 72y = 77\). Write \(77x = 77 + 72y \implies x = 1 + \frac{72}{77}y\). This is a linear relation between \(x\) and \(y\). **Final answers:** 1. \(x = -8ع, y = 6ع\) 2. \(x=\frac{3}{2}, y = -\frac{11}{2}\) 3. \( (a-3)(b-3) = 9 \) 4. \(a - 5b + 7c = 90\) (plane equation) 5. \(x = 1 + \frac{72}{77}y\)