Let the first figure be \( x \) and the second figure be \( y \).
From the problem, we can derive two equations:
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Quarter of the first figure is equal to three times the second figure: \[ \frac{x}{4} = 3y \] Therefore, multiplying both sides by 4: \[ x = 12y \quad \text{(Equation 1)} \]
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If 12 is added to four times the second figure, the resultant figure equals three times the first figure: \[ 4y + 12 = 3x \] Substituting \( x \) from Equation 1: \[ 4y + 12 = 3(12y) \] Simplifying this gives: \[ 4y + 12 = 36y \] Rearranging the equation gives: \[ 12 = 36y - 4y \] \[ 12 = 32y \] Dividing by 32: \[ y = \frac{12}{32} = \frac{3}{8} \]
Now, substituting \( y \) back into Equation 1 to find \( x \): \[ x = 12y = 12 \times \frac{3}{8} = \frac{36}{8} = \frac{9}{2} = 4.5 \]
Thus, the first figure \( x \) is \( 4.5 \) and the second figure \( y \) is \( \frac{3}{8} \).
To summarize:
- First figure \( x = 4.5 \)
- Second figure \( y = \frac{3}{8} \)
In sequence: \( 4.5, \frac{3}{8} \).
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