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

The solution to the equation is g=

5 answers

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: \[ \boxed{\frac{8}{15}} \]

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}

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:

\[ \boxed{2} \]

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.

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.