To find a value that is a solution to two of the given equations, we need to solve each equation for their respective variables and identify any overlapping solutions.
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Equation 1: \( p^2 + 4 = 10 \)
Subtract 4 from both sides: \[ p^2 = 10 - 4 \] \[ p^2 = 6 \] Taking the square root of both sides gives: \[ p = \sqrt{6} \quad \text{or} \quad p = -\sqrt{6} \] So the solutions for \( p \) are \( p = \sqrt{6} \) and \( p = -\sqrt{6} \).
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Equation 2: \( 2 = 10 - 6h \)
Subtract 10 from both sides: \[ 2 - 10 = -6h \] \[ -8 = -6h \] Dividing both sides by -6: \[ h = \frac{8}{6} = \frac{4}{3} \] Thus, the solution for \( h \) is \( h = \frac{4}{3} \).
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Equation 3: \( 3y - 16 = 20 \)
Add 16 to both sides: \[ 3y = 20 + 16 \] \[ 3y = 36 \] Divide both sides by 3: \[ y = 12 \] Therefore, the solution for \( y \) is \( y = 12 \).
Now, we have:
- Solutions:
- From equation 1: \( p = \sqrt{6}, -\sqrt{6} \)
- From equation 2: \( h = \frac{4}{3} \)
- From equation 3: \( y = 12 \)
No common values are found from all three equations.
To check if there are any solutions to two equations:
- The solutions we have are:
- \( p = \sqrt{6}, -\sqrt{6} \) (from the first equation)
- \( h = \frac{4}{3} \) (from the second equation)
- \( y = 12 \) (from the third equation)
Since none of the variable solutions overlap, we conclude that there is no value solution that fits two of the equations simultaneously.
Thus, the solution set is:
\[ { } \]
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