To determine which equation Hill is solving when his second step is to divide both sides by 5, we need to identify an equation where multiplying or having a coefficient of 5 in front of \( x \) would necessitate this step.
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For the equation \( 7 + 5x = 22 \):
- First, you would subtract 7 from both sides to get \( 5x = 15 \). Then you would divide both sides by 5, which fits the scenario.
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For the equation \( 7 + \frac{x}{5} = 22 \):
- First, you would subtract 7 from both sides to get \( \frac{x}{5} = 15 \). Then you would multiply both sides by 5 or divide both sides by \( \frac{1}{5} \) (which involves a different process), so this doesn't fit.
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For the equation \( 7 - 5x = 22 \):
- First, you would subtract 7 from both sides to get \( -5x = 15 \). You would then divide by -5, not 5, so this doesn't fit.
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For the equation \( 7 - \frac{x}{5} = 22 \):
- First, you would subtract 7 from both sides to get \( -\frac{x}{5} = 15 \). Then you would multiply by -5, so this also doesn't fit.
The only equation that leads to dividing by 5 in the second step is: \( 7 + 5x = 22 \).