To determine which equation has exactly one solution, let's solve each equation one by one.
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Equation 1: \( 3 + 10x^2 = 4x + 2 \)
Rearranging gives: \[ 10x^2 - 4x + 1 = 0 \] We can find the discriminant to check the number of solutions: \[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 10 \cdot 1 = 16 - 40 = -24 \] Since the discriminant is negative, this equation has no real solutions.
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Equation 2: \( 15x - \frac{5}{3} = 5x + 4 \)
Rearranging gives: \[ 15x - 5x = 4 + \frac{5}{3} \] \[ 10x = 4 + \frac{5}{3} \] Converting 4 to have a common denominator: \[ 4 = \frac{12}{3} \] \[ 10x = \frac{12}{3} + \frac{5}{3} = \frac{17}{3} \] Dividing by 10: \[ x = \frac{17}{30} \] This equation has exactly one solution.
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Equation 3: \( \frac{2 + 6x}{2} = 3x + 1 \)
Multiplying both sides by 2: \[ 2 + 6x = 6x + 2 \] Subtracting \( 6x \) and 2 from both sides results in: \[ 0 = 0 \] This equation has infinite solutions.
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Equation 4: \( \frac{4x - 10}{2} = 2x - 5 \)
Multiplying both sides by 2: \[ 4x - 10 = 4x - 10 \] Subtracting \( 4x - 10 \) gives: \[ 0 = 0 \] This equation also has infinite solutions.
After analyzing all the equations, the equation that has exactly one solution is:
Equation 2: \( 15x - \frac{5}{3} = 5x + 4 \) which has the solution \( x = \frac{17}{30} \).
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