Question

What value is a solution to two of the equations? p2+4=10 2=10−6h 3y−16=20 (1 point) The solution set {} is the solution to two equations.

Answer 1

what is the answer

Answer 2

To find a value that satisfies at least two of the given equations, we can solve each of them one by one.

1. **First equation: \( p^2 + 4 = 10 \)**
\[
p^2 = 10 - 4 \\
p^2 = 6 \\
p = \pm\sqrt{6}
\]
So, \( p = \sqrt{6} \) or \( p = -\sqrt{6} \).

2. **Second equation: \( 2 = 10 - 6h \)**
\[
6h = 10 - 2 \\
6h = 8 \\
h = \frac{8}{6} = \frac{4}{3}
\]
So, \( h = \frac{4}{3} \).

3. **Third equation: \( 3y - 16 = 20 \)**
\[
3y = 20 + 16 \\
3y = 36 \\
y = \frac{36}{3} = 12
\]
So, \( y = 12 \).

Now we have:

- From the first equation, the solutions are \( p = \sqrt{6} \) and \( p = -\sqrt{6} \).
- From the second equation, the solution is \( h = \frac{4}{3} \).
- From the third equation, the solution is \( y = 12 \).

Now, we can see that there are no common values among the solutions of the three equations:

- \( \sqrt{6} \) and \(-\sqrt{6}\) do not match with \( \frac{4}{3} \) or \( 12 \)
- \( \frac{4}{3} \) does not match with \( \sqrt{6} \) or \( 12 \)
- \( 12 \) does not match with \( \sqrt{6} \) or \( \frac{4}{3} \)

Since there are no common values that satisfy two equations, the result would be:

The solution set {} is the solution to two equations.

Answer 3

The answer is that there is no single value that satisfies two of the given equations simultaneously.

The equations provided are:

1. \( p^2 + 4 = 10 \) (solutions: \( p = \sqrt{6} \) and \( p = -\sqrt{6} \))
2. \( 2 = 10 - 6h \) (solution: \( h = \frac{4}{3} \))
3. \( 3y - 16 = 20 \) (solution: \( y = 12 \))

Because none of the obtained solutions \( \sqrt{6}, -\sqrt{6}, \frac{4}{3}, \) and \( 12 \) are the same, the final conclusion is that the solution set is empty:

**The solution set is \(\{\}\)**.

If you are looking for a specific form, it can be stated as "**no common solutions**" or "**no value satisfies two equations simultaneously**."