This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the step-by-step solutions for the problems: 1) Evaluate 2e - 3 + 6e + 2 in mod 5 Step 1: Combine like terms. 2e - 3 + 6e + 2 = (2e + 6e) + (-3 + 2) = 8e - 1 Step 2: Reduce the coefficients modulo 5. Since 8 3 5 and -1 4 5: 8e - 1 3e + 4 5 The expression in mod 5 is 3e + 4. 2) Evaluate (x+1) - (4x-10) in mod 7 Step 1: Distribute the negative sign and combine like terms. (x+1) - (4x-10) = x+1 - 4x + 10 = (x - 4x) + (1 + 10) = -3x + 11 Step 2: Reduce the coefficients modulo 7. Since -3 4 7 and 11 4 7: -3x + 11 4x + 4 7 The expression in mod 7 is 4x + 4. 3) Evaluate 4f + e - 7f - 5e in mod 7 Step 1: Combine like terms. 4f + e - 7f - 5e = (4f - 7f) + (e - 5e) = -3f - 4e Step 2: Reduce the coefficients modulo 7. Since -3 4 7 and -4 3 7: -3f - 4e 4f + 3e 7 The expression in mod 7 is 4f + 3e. 4) Evaluate (2x^2 + 4)(2x - x + 1) ÷ (x-3) Step 1: Simplify the second factor. 2x - x + 1 = x + 1 The expression becomes (2x^2 + 4)(x + 1) ÷ (x-3). Step 2: Perform the multiplication. (2x^2 + 4)(x + 1) = 2x^2(x) + 2x^2(1) + 4(x) + 4(1) = 2x^3 + 2x^2 + 4x + 4 Step 3: Perform polynomial division (2x^3 + 2x^2 + 4x + 4) ÷ (x-3). Using polynomial long division: ` 2x^2 + 8x + 28 ________________ x - 3 | 2x^3 + 2x^2 + 4x + 4 -(2x^3 - 6x^2) ________________ 8x^2 + 4x -(8x^2 - 24x) ________________ 28x + 4 -(28x - 84) ___________ 88 ` The quotient is 2x^2 + 8x + 28 with a remainder of 88. So, the result is 2x^2 + 8x + 28 + (88)/(x-3). The evaluated expression is 2x^2 + 8x + 28 + (88)/(x-3). 5) Evaluate (2t^2+1) ÷ (2t-1) in mod 3 Step 1: Reduce the coefficients modulo 3 in the expressions. For 2t^2+1: 2 2 3, 1 1 3. So, 2t^2+1 3. For 2t-1: 2 2 3, -1 2 3. So, 2t+2 3. The problem becomes (2t^2+1) ÷ (2t+2) in mod 3. Step 2: Perform polynomial division modulo 3. We need to find Q(t) and R such that 2t^2+1 = (2t+2)Q(t) + R 3. We can multiply the divisor by the inverse of its leading coefficient to simplify. The inverse of 2 3 is 2 (since 2 × 2 = 4 1 3). So, 2t+2 2(t+1) 3. Let's divide 2t^2+1 by 2t+2 directly. ` t + 2 _________ 2t+2 | 2t^2 + 0t + 1 -(2t^2 + 2t) ___________ -2t + 1 -(2t + 2) _________ -1 ` All calculations are modulo 3. -2t t 3 -1 2 3 So, the division steps are: ` t + 2 _________ 2t+2 | 2t^2 + 0t + 1 -(2t^2 + 2t) (since 2t * (2t+2) = 4t^2 + 4t = t^2 + t mod 3, this is wrong. We need to find a term to multiply (2t+2) to get 2t^2. (2t+2) * t = 2t^2 + 2t. So, subtract (2t^2 + 2t)) ___________ -2t + 1 (which is t + 1 mod 3) -(2t + 2) (We need to find a term to multiply (2t+2) to get t. (2t+2) * 2 = 4t + 4 = t + 1 mod 3. This is not t. Let's try to multiply by 2. (2t+2) * 2 = t+1. We need to get t. Let's restart the division carefully modulo 3. ` Let P(t) = 2t^2+1 and D(t) = 2t+2. To get 2t^2 from 2t, we multiply by t. t(2t+2) = 2t^2 + 2t. Subtract this from P(t): (2t^2+1) - (2t^2+2t) = -2t+1 t+1 3. Now we need to divide t+1 by 2t+2. To get t from 2t, we multiply by 2 (since 2 × 2t = 4t t 3). 2(2t+2) = 4t+4 t+1 3. Subtract this from t+1: (t+1) - (t+1) = 0 3. So the quotient is t+2 and the remainder is 0. The evaluated expression in mod 3 is t+2. 6) Solve 3a = 6 in mod 5 Step 1: Reduce the constant term modulo 5. 6 1 5. So the congruence is 3a 1 5. Step 2: Find the multiplicative inverse of 3 modulo 5. We need a number x such that 3x 1 5. By testing values: 3 × 1 = 3 5 3 × 2 = 6 1 5 So, the inverse of 3 modulo 5 is 2. Step 3: Multiply both sides of the congruence by the inverse. 2 × (3a) 2 × 1 5 6a 2 5 a 2 5 The solution is a 2 5. 7) Solve 5r = 4 in mod 8 Step 1: The congruence is 5r 4 8. Step 2: Find the multiplicative inverse of 5 modulo 8. We need a number x such that 5x 1 8. By testing values: 5 × 1 = 5 8 5 × 2 = 10 2 8 5 × 3 = 15 7 8 5 × 4 = 20 4 8 5 × 5 = 25 1 8 So, the inverse of 5 modulo 8 is 5. Step 3: Multiply both sides of the congruence by the inverse. 5 × (5r) 5 × 4 8 25r 20 8 r 4 8 (since 25 1 8 and 20 4 8) The solution is r 4 8. 8) Solve 35x+3 = 6x-1 Step 1: Collect x terms on one side and constant terms on the other side. Subtract 6x from both sides: 35x - 6x + 3 = -1 29x + 3 = -1 Step 2: Subtract 3 from both sides. 29x = -1 - 3 29x = -4 Step 3: Divide by 29 to solve for x. x = -(4)/(29) The solution is x = -(4)/(29). 9) Evaluate (3x+6)(3x-2) in mod 9 Step 1: Expand the product. (3x+6)(3x-2) = (3x)(3x) + (3x)(-2) + (6)(3x) + (6)(-2) = 9x^2 - 6x + 18x - 12 = 9x^2 + 12x - 12 Step 2: Reduce the coefficients modulo 9. Since 9 0 9: 9x^2 0x^2 0 9 Since 12 3 9: 12x 3x 9 Since -12 6 9 (because -12 = -2 × 9 + 6 or -12+18=6): -12 6 9 Substitute these back into the expression: 9x^