answer:**Answer**: The relationship in this problem: The sum of the money already saved and the money saved per month should be at least 280 yuan. "At least" means greater than or equal to. Therefore, the correct answer is boxed{text{D}}.

answer:Let us solve the given system of equations: [ a^{2} b^{2}-a^{2}-a b+1=0 tag{1} ] [ a^{2} c - a b - a - c = 0 tag{2} ] [ a b c = -1 tag{3} ] 1. **Express ( c ) from Equation (3)**: From Equation (3), we have: [ c = -frac{1}{a b} ] 2. **Substitute ( c ) into Equation (2)**: Substitute ( c = -frac{1}{a b} ) into Equation (2): [ a^{2} left(-frac{1}{a b}right) - a b - a - left(-frac{1}{a b}right) = 0 ] 3. **Simplify the resulting equation**: Simplify the equation by combining like terms: [ -frac{a^2}{a b} - a b - a + frac{1}{a b} = 0 ] [ -frac{a}{b} - a b - a + frac{1}{a b} = 0 ] To eliminate the fractions, multiply through by (a b): [ -a^2 b - a^3 b^2 - a^2 b + 1 = 0 ] 4. **Combine with Equation (1)**: Let us revisit Equation (1): [ a^{2} b^{2} - a^{2} - a b + 1 = 0 tag{1} ] Now compare the simplified form from step 3: [ -a^2 - a^2 b^2 - a^2 b + 1 = 0 ] These two forms should yield consistent results, thus we add them: 5. **Simplify further**: Add the two equations: [ 1 - a^{2} - a^{2} b^{2} - a^{2} b = 0 ] Subtracting Equation (1): [ 2 a^{2} b^{2} + a^{2} b - a b = 0 ] 6. **Factorize the resulting equation**: Factorize the expression: [ a b (2 a b + a - 1) = 0 ] Since ( a b neq 0 ) from Equation (3), we must have: [ 2 a b + a - 1 = 0 ] Solving for ( a ): [ a = frac{1}{2 b + 1} ] Note that ( 2 b + 1 neq 0 implies b neq -frac{1}{2} ). 7. **Substitute into Equation (1)** and solve for ( b ): Substitute ( a = frac{1}{2 b + 1} ) into Equation (1): [ left(frac{1}{2 b + 1}right)^{2} b^{2} - left(frac{1}{2 b+1}right)^{2} - frac{1}{2 b + 1} b + 1 = 0 ] Multiply through by ((2 b + 1)^2): [ b^{2} - 1 - b(2 b + 1) + (2 b + 1)^2 = 0 ] Simplify and rearrange to form a standard quadratic equation: [ 3 b^{2} + 3 b = 0 ] Solution to the quadratic is: [ 3b(b+1) = 0 ] Given ( b neq 0 ), we conclude: [ b = -1 ] 8. **Determine ( a )** and ( c ): Using ( b = -1 ): [ a = frac{1}{2(-1) + 1} = frac{1}{-1} = -1 ] Using ( a = -1 ) and ( b = -1 ): [ c = -frac{1}{a b} = -frac{1}{(-1)(-1)} = -1 ] **Conclusion**: The solution to the system of equations is: [ boxed{a = -1, b = -1, c = -1} ]

answer:Let's call the number you put into the machine ( x ). According to the machine's operations, it first adds 15 to ( x ) and then subtracts 6 from the result. The final output is 35. So the equation representing the machine's operations is: ( x + 15 - 6 = 35 ) Now, let's solve for ( x ): ( x + 9 = 35 ) (combining the +15 and -6 into +9) ( x = 35 - 9 ) (subtracting 9 from both sides to isolate ( x )) ( x = 26 ) So the number you put into the machine was boxed{26} .

answer:We are given the equation: [ x y = x + y + A ] 1. **Substitute ( A = 22 ) into the equation:** [ x y = x + y + 22 ] 2. **Rearrange the equation:** [ x y - x - y = 22 ] 3. **Add 1 to both sides of the equation to facilitate factoring:** [ x y - x - y + 1 = 22 + 1 ] 4. **Simplify the right-hand side:** [ x y - x - y + 1 = 23 ] 5. **Recognize the left-hand side as a factored form:** [ (x - 1)(y - 1) = 23 ] 6. **Since 23 is a prime number, it can only be factored as ( 23 times 1 ) or ( 1 times 23 ). However, given ( x > y > 1 ), we must have ( x - 1 = 23 ) and ( y - 1 = 1 ):** [ x - 1 = 23 implies x = 24 ] [ y - 1 = 1 implies y = 2 ] 7. **Determine ( B ):** [ B = frac{x}{y} = frac{24}{2} = 12 ] # Conclusion: [ boxed{12} ]