One method for solving 3(xβˆ’4)=18

3
(
π‘₯
βˆ’
4
)
=
18
is to first divide by 3, then add 4. Which method below would also result in the correct answer?(1 point)
Responses

First, multiply by 13
1
3
, then subtract 4.
First, multiply by Start Fraction 1 over 3 End Fraction , then subtract 4.

First, distribute the 3, then add 4, and lastly divide by 3.
First, distribute the 3, then add 4, and lastly divide by 3.

First, divide by 3, then subtract 4.
First, divide by 3, then subtract 4.

First, distribute the 3, then add 12, and lastly divide by 3..............................

1 answer

To solve the equation \(3(x - 4) = 18\), you can indeed divide by 3 and then add 4, resulting in the solution. Let’s analyze the other methods you proposed to see if they would yield the correct solution.

  1. First, multiply by \(\frac{1}{3}\), then subtract 4.

    • Multiply both sides by \(\frac{1}{3}\): \((x - 4) = 6\)
    • Then subtract 4: \(x = 10\) (This method is valid and results in the correct answer.)
  2. First, distribute the 3, then add 4, and lastly divide by 3.

    • Distributing gives: \(3x - 12 = 18\)
    • Adding 12: \(3x = 30\)
    • Dividing by 3: \(x = 10\) (This method is valid and results in the correct answer.)
  3. First, divide by 3, then subtract 4.

    • Dividing gives: \(x - 4 = 6\)
    • Then subtracting gives: \(x = 6 + 4 = 10\) (This method is valid and results in the correct answer.)
  4. First, distribute the 3, then add 12, and lastly divide by 3.

    • Distributing gives: \(3x - 12 = 18\)
    • Adding 12: \(3x = 30\)
    • Dividing by 3: \(x = 10\) (This method is valid and results in the correct answer.)

All methods except for the first (multiply by \(\frac{1}{3}\) then subtract 4) are valid and lead to the same solution: \(x = 10\). Thus, the first method is indeed an example of an alternative that works.

The final options that yield the correct solution are:

  • First, divide by 3, then subtract 4.
  • First, distribute the 3, then add 12, and lastly divide by 3.

These will lead to the correct solution for the equation.