Question

Using the properties of equality, solve the equation 14(8−16g)+8=g .(1 point)
The solution to the equation is g=

Answer 1

How many of the following equations have the solution set {5}?

2(x−3)+1=5

3b15=1

2h−30=40

(1 point)
The number of equations that have the solution set {5}

Answer 2

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 3

To solve the equation \( 14(8 - 16g) + 8 = g \), we will first distribute and then combine like terms.

1. Distribute \( 14 \) in the left side of the equation:
\[
14 \cdot 8 - 14 \cdot 16g + 8 = g
\]

This simplifies to:
\[
112 - 224g + 8 = g
\]

2. Combine the constant terms on the left side:
\[
120 - 224g = g
\]

3. To isolate \( g \), add \( 224g \) to both sides:
\[
120 = g + 224g
\]
\[
120 = 225g
\]

4. Now, divide both sides by \( 225 \):
\[
g = \frac{120}{225}
\]

5. Simplify the fraction:
\[
g = \frac{120 \div 15}{225 \div 15} = \frac{8}{15}
\]

Thus, the solution to the equation is:
\[
\boxed{\frac{8}{15}}
\]

Answer 4

Let's solve each equation one by one to determine if the solution set is {5}.

1. **Equation:** \( 2(x - 3) + 1 = 5 \)

\[
2(x - 3) + 1 = 5
\]
Subtract 1 from both sides:
\[
2(x - 3) = 4
\]
Divide by 2:
\[
x - 3 = 2
\]
Add 3:
\[
x = 5
\]

This equation has the solution {5}.

2. **Equation:** \( 3b = 15 \)

Dividing both sides by 3:
\[
b = 5
\]

This equation also has the solution {5}.

3. **Equation:** \( 2h - 30 = 40 \)

Add 30 to both sides:
\[
2h = 70
\]
Divide by 2:
\[
h = 35
\]

This equation does **not** have the solution {5}.

Now, we summarize the results:

- The first equation has the solution {5}.
- The second equation has the solution {5}.
- The third equation does not have the solution {5}.

Thus, the total number of equations having the solution set {5} is:

\[
\boxed{2}
\]

Answer 5

Let's solve each equation one by one to see if we can find a common solution.

1. **Equation:** \( p^2 + 4 = 10 \)

Subtract 4 from both sides:
\[
p^2 = 6
\]
Taking the square root of both sides:
\[
p = \sqrt{6} \quad \text{or} \quad p = -\sqrt{6}
\]

So the solutions are \( p = \sqrt{6} \) and \( p = -\sqrt{6} \).

2. **Equation:** \( 2 = 10 - 6h \)

Subtract 10 from both sides:
\[
-8 = -6h
\]
Divide by -6:
\[
h = \frac{4}{3}
\]

There are no solutions in common yet.

3. **Equation:** \( 3y - 16 = 20 \)

Add 16 to both sides:
\[
3y = 36
\]
Divide by 3:
\[
y = 12
\]

Since none of the solutions from the equations \( p^2 + 4 = 10 \) (which gives \( \sqrt{6} \) and \( -\sqrt{6} \)), \( 2 = 10 - 6h \) (which gives \( \frac{4}{3} \)), or \( 3y - 16 = 20 \) (which gives 12) are common, we need to check if any specific value repeats in at least two equations.

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

Since we did not find any common solutions that appear in at least two equations, **there is no value that is a solution to two of the equations.**

Thus, we conclude:
\[
\text{The solution set is } \{\}
\] (empty set) because no single value satisfies two equations from the given set.