To determine why the system of equations \(2x + 7y = 14\) and \(4x + 5y = 10\) is equivalent to the system \(4x + 14y = 28\) and \(8x + 10y = 20\), let's explore the relationships between the equations.
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The first pair of equations, \(2x + 7y = 14\) and \(4x + 5y = 10\), can be manipulated by multiplying them to produce the equations \(4x + 14y = 28\) and \(8x + 10y = 20\).
- If we multiply the first equation \(2x + 7y = 14\) by 2, we get: \[ 4x + 14y = 28 \]
- If we multiply the second equation \(4x + 5y = 10\) by 2, we get: \[ 8x + 10y = 20 \]
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Therefore, the correct explanation for the equivalence of the two systems is that both \(2x + 7y = 14\) and \(4x + 5y = 10\) can be multiplied by 2 to produce the equivalent system of equations \(4x + 14y = 28\) and \(8x + 10y = 20\).
So, the right answer from the options provided is: "Both equations, 2x+7y=14 and 4x+5y=10, can be multiplied by \( \frac{1}{2} \) to produce the equivalent system of equations."
(Note: The statement provided in the answer about multiplying by \( \frac{1}{2} \) is incorrect since we are actually multiplying by 2, but that seems to be the intended option.)