Let's define the two numbers as \( x \) and \( y \). According to the problem, we can outline the following system of equations:
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The product of the two numbers is 10: \[ xy = 10 \]
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The first number is 3 more than the second: \[ x = y + 3 \]
Now, we can substitute the second equation into the first to find solutions for \( x \) and \( y \).
Substituting \( x = y + 3 \) into \( xy = 10 \), we get: \[ (y + 3)y = 10 \] Expanding this: \[ y^2 + 3y - 10 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 3 \), and \( c = -10 \).
Calculating the discriminant: \[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49 \] Now applying the quadratic formula: \[ y = \frac{-3 \pm \sqrt{49}}{2 \cdot 1} = \frac{-3 \pm 7}{2} \]
Calculating the two possible values for \( y \):
- \( y = \frac{-3 + 7}{2} = \frac{4}{2} = 2 \)
- \( y = \frac{-3 - 7}{2} = \frac{-10}{2} = -5 \)
Now, we can find \( x \) for both values of \( y \):
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If \( y = 2 \): \[ x = 2 + 3 = 5 \] So, one solution is \( (x, y) = (5, 2) \).
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If \( y = -5 \): \[ x = -5 + 3 = -2 \] So, another solution is \( (x, y) = (-2, -5) \).
Thus, the pairs that satisfy the conditions are:
- \( (5, 2) \)
- \( (-2, -5) \)
To verify:
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For \( (5, 2) \):
- Product: \( 5 \times 2 = 10 \)
- The first number is 3 more: \( 5 = 2 + 3 \) (true).
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For \( (-2, -5) \):
- Product: \( -2 \times -5 = 10 \)
- The first number is 3 more: \( -2 = -5 + 3 \) (true).
Both pairs satisfy the system of equations.