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.
Remaining Attempts : 3
Skip to navigation
page 12 of 13

Answer 1

whats the answer

Answer 2

To find a value that is a solution to two of the given equations, we will solve each equation one by one.

1. **First Equation:**
\[
p^2 + 4 = 10
\]
Subtract 4 from both sides:
\[
p^2 = 6
\]
Now take the square root of both sides:
\[
p = \sqrt{6} \quad \text{or} \quad p = -\sqrt{6}
\]

2. **Second Equation:**
\[
2 = 10 - 6h
\]
Subtract 10 from both sides:
\[
-8 = -6h
\]
Now divide by -6:
\[
h = \frac{8}{6} = \frac{4}{3}
\]

3. **Third Equation:**
\[
3y - 16 = 20
\]
Add 16 to both sides:
\[
3y = 36
\]
Now divide by 3:
\[
y = 12
\]

Next, we have the values:
- From the first equation, \( p = \sqrt{6}, -\sqrt{6} \)
- From the second equation, \( h = \frac{4}{3} \)
- From the third equation, \( y = 12 \)

Now, let's determine if any of these values are shared:
- The values are different across all three equations.

Thus, this means that there are no common solutions among two equations from the provided equations using the specific variable values derived from the equations.

The solution set is therefore:
\[
\{\}
\]

If you need further assistance or have additional requirements, please let me know!

Answer 3

Based on the calculations provided, the solutions to the equations are:

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

Since none of the values overlap, when asked for a solution that is common to two of the equations, the correct answer is:

**The solution set is empty**:

\[
\{\}
\]

If this is what you're looking for, then that is the answer. If there was a misunderstanding or if you meant specific equations, please clarify so I can assist you better!