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Using the properties of equality, find the solution to the equation 2f/5−8/5+2f=8.(1 point)

Responses

8/3
Start Fraction 8 over 3 End Fraction

4
4

16/6
Start Fraction 16 over 6 End Fraction

12

7 answers

To solve the equation \( \frac{2f}{5} - \frac{8}{5} + 2f = 8 \), follow these steps:

  1. Combine like terms. The term \( 2f \) can be expressed with a common denominator of 5: \[ 2f = \frac{10f}{5} \] So the equation becomes: \[ \frac{2f}{5} - \frac{8}{5} + \frac{10f}{5} = 8 \]

  2. Combine the terms on the left side: \[ \frac{2f + 10f - 8}{5} = 8 \] This simplifies to: \[ \frac{12f - 8}{5} = 8 \]

  3. Multiply both sides by 5 to eliminate the fraction: \[ 12f - 8 = 40 \]

  4. Add 8 to both sides: \[ 12f = 48 \]

  5. Finally, divide by 12: \[ f = 4 \]

The solution to the equation is \( 4 \).

How many solutions will the equation 23(6m+9)−6=4m have?(1 point)
Responses

infinite/many solutions
infinite/many solutions

one solution
one solution

no solution
no solution

two solutions

To determine the number of solutions for the equation \( 23(6m + 9) - 6 = 4m \), we can simplify and analyze it step by step:

  1. Start with the original equation: \[ 23(6m + 9) - 6 = 4m \]

  2. Distribute \( 23 \) on the left side: \[ 138m + 207 - 6 = 4m \]

  3. Combine like terms: \[ 138m + 201 = 4m \]

  4. Next, isolate \( m \) by subtracting \( 4m \) from both sides: \[ 138m - 4m + 201 = 0 \] \[ 134m + 201 = 0 \]

  5. Now, subtract \( 201 \) from both sides: \[ 134m = -201 \]

  6. Divide by \( 134 \): \[ m = -\frac{201}{134} \]

Since we arrived at a specific value for \( m \), the equation has one solution.

Thus, the answer is "one solution."

Identify the pair of linear equations that have the same solution set.(1 point)
Responses

4=k2+5 and 4j−143+5=3
4 equals Start Fraction k over 2 End Fraction plus 5 and Start Fraction 4 j minus 14 over 3 End Fraction plus 5 equals 3

3(m+1)=10 and 6n+3=6−n
3 left parenthesis m plus 1 right parenthesis equals 10 and 6 n plus 3 equals 6 minus n

2(3g+5)−2g=2(4−g) and −36h6=2
2 left parenthesis 3 g plus 5 right parenthesis minus 2 g equals 2 left parenthesis 4 minus g right parenthesis and Start Fraction negative 36 h over 6 End Fraction equals 2

4k−5=2k and 8j−2=3j

To identify which pair of linear equations have the same solution set, we can analyze each option individually to check whether they yield equivalent expressions or equivalent linear relationships.

  1. Option 1: \[ 4 = \frac{k}{2} + 5 \quad \text{and} \quad 4j - \frac{14}{3} + 5 = 3 \]

    • Rearranging the first equation: \[ \frac{k}{2} = 4 - 5 \implies \frac{k}{2} = -1 \implies k = -2 \]
    • Rearranging the second equation: \[ 4j - \frac{14}{3} + 5 = 3 \implies 4j - \frac{14}{3} + \frac{15}{3} = 3 \implies 4j + \frac{1}{3} = 3 \implies 4j = 3 - \frac{1}{3} \implies 4j = \frac{9 - 1}{3} = \frac{8}{3} \] \[ j = \frac{2}{3} \]
    • No common solution.

  2. Option 2: \[ 3(m + 1) = 10 \quad \text{and} \quad 6n + 3 = 6 - n \]

    • Rearranging the first equation: \[ 3m + 3 = 10 \implies 3m = 7 \implies m = \frac{7}{3} \]
    • Rearranging the second equation: \[ 6n + 3 = 6 - n \implies 7n + 3 = 6 \implies 7n = 3 \implies n = \frac{3}{7} \]
    • No common solution.

  3. Option 3: \[ 2(3g + 5) - 2g = 2(4 - g) \quad \text{and} \quad -\frac{36h}{6} = 2 \]

    • Simplifying the first equation: \[ 6g + 10 - 2g = 8 - 2g \implies 6g + 10 = 8 - 2g \implies 8g + 10 = 8 \implies 8g = -2 \implies g = -\frac{1}{4} \]
    • The second equation simplifies to: \[ -6h = 2 \implies h = -\frac{1}{3} \]
    • No common solution.

  4. Option 4: \[ 4k - 5 = 2k \quad \text{and} \quad 8j - 2 = 3j \]

    • Rearranging the first equation: \[ 4k - 2k = 5 \implies 2k = 5 \implies k = \frac{5}{2} \]
    • Rearranging the second equation: \[ 8j - 3j = 2 \implies 5j = 2 \implies j = \frac{2}{5} \]
    • No common solution.

None of these options yield equations with the same solutions.

To summarize, all pairs of equations presented do not have the same solution set. They yield different unique solutions for their respective variables.

Tamara, Payton, Krish, and Haruki are analyzing the solution sets to the following linear equations. Tamara says that Equation A and Equation C have the same solution. Payton says Equation B and Equation C have the same solution. Krish says all three have the same solution. Haruki says that none of the equations have the same solution. Identify who is correct about the linear equations.

Equation A: 3(4−5h)−7=−12+2h

Equation B: 2(p−3)−1=6p+142

Equation C: 23v−3+83=−1

Let's analyze each of the linear equations step-by-step to determine their solution set and verify the claims made by Tamara, Payton, Krish, and Haruki.

Equation A:

\[ 3(4 - 5h) - 7 = -12 + 2h \]

  1. Distributing on the left: \[ 12 - 15h - 7 = -12 + 2h \] \[ 5 - 15h = -12 + 2h \]

  2. Rearranging: \[ 5 + 12 = 15h + 2h \] \[ 17 = 17h \]

  3. Solving for \( h \): \[ h = 1 \]

Equation B:

\[ 2(p - 3) - 1 = 6p + 142 \]

  1. Distributing on the left: \[ 2p - 6 - 1 = 6p + 142 \] \[ 2p - 7 = 6p + 142 \]

  2. Rearranging: \[ -7 - 142 = 6p - 2p \] \[ -149 = 4p \]

  3. Solving for \( p \): \[ p = -\frac{149}{4} \]

Equation C:

\[ 23v - 3 + 83 = -1 \]

  1. Simplifying on the left: \[ 23v + 80 = -1 \]

  2. Rearranging: \[ 23v = -1 - 80 \] \[ 23v = -81 \]

  3. Solving for \( v \): \[ v = -\frac{81}{23} \]

Summary of Solutions:

  • Equation A has the solution \( h = 1 \).
  • Equation B has the solution \( p = -\frac{149}{4} \).
  • Equation C has the solution \( v = -\frac{81}{23} \).

Evaluation of Claims:

  • Tamara says Equation A and Equation C have the same solution: Incorrect (they have different solutions).

  • Payton says Equation B and Equation C have the same solution: Incorrect (they have different solutions).

  • Krish says all three have the same solution: Incorrect (all three have different solutions).

  • Haruki says that none of the equations have the same solution: Correct.

In conclusion, Haruki is the one who is correct about the linear equations.