3×(6/7-3/4) (6x-2)×(3-x)=0

Let's solve both problems step by step. ### Problem 1: \( 3 \times \left( \frac{6}{7} - \frac{3}{4} \right) \) First, we need to find a common denominator for the fractions \(\frac{6}{7}\) and \(\frac{3}{4}\). The least common denominator (LCD) of 7 and 4 is 28. Now, we convert each fraction: \[ \frac{6}{7} = \frac{6 \times 4}{7 \times 4} = \frac{24}{28} \] \[ \frac{3}{4} = \frac{3 \times 7}{4 \times 7} = \frac{21}{28} \] Now, we can subtract the two fractions: \[ \frac{6}{7} - \frac{3}{4} = \frac{24}{28} - \frac{21}{28} = \frac{24 - 21}{28} = \frac{3}{28} \] Next, we multiply by 3: \[ 3 \times \left( \frac{3}{28} \right) = \frac{3 \times 3}{28} = \frac{9}{28} \] So, the result of the first problem is: \[ \boxed{\frac{9}{28}} \] --- ### Problem 2: \( (6x - 2)(3 - x) = 0 \) To solve this equation, we can use the zero product property, which states that if the product of two factors equals zero, then at least one of the factors must be zero. **Step 1: Set each factor to zero.** 1. \( 6x - 2 = 0 \) 2. \( 3 - x = 0 \) **Step 2: Solve each equation.** 1. For \( 6x - 2 = 0 \): \[ 6x = 2 \\ x = \frac{2}{6} = \frac{1}{3} \] 2. For \( 3 - x = 0 \): \[ x = 3 \] **Step 3: Combine the solutions.** The solutions to the equation \( (6x - 2)(3 - x) = 0 \) are: \[ x = \frac{1}{3} \quad \text{and} \quad x = 3 \] So, the final result for the second problem is: \[ \boxed{\frac{1}{3}, 3} \]