Which equations have the variable term isolated to one side of the equals sign, and the constant isolated to the other side, for the equation 1

5
x + 1
3
= −1
2
x + 8
3
? Select all that apply.

One-fifth x minus StartFraction 7 Over 3 EndFraction = negative one-half
StartFraction 7 Over 10 EndFraction x = StartFraction 7 Over 3 EndFraction
Negative StartFraction 7 Over 3 EndFraction = Negative StartFraction 7 Over 10 EndFraction x
x = negative five-halves x + StartFraction 35 Over 3 EndFraction
mr001-5.jpg

1 answer

To isolate the variable term on one side and the constant term on the other for the given equation \( \frac{1}{5} x + \frac{1}{3} = -\frac{1}{2} x + \frac{8}{3} \), we can manipulate the equation step by step.

  1. Combine like terms: Start by getting all terms with \( x \) on one side and constants on the other.

    First, let's move \( -\frac{1}{2} x \) to the left side: \[ \frac{1}{5} x + \frac{1}{2} x + \frac{1}{3} = \frac{8}{3} \] This gives us: \[ \left( \frac{1}{5} + \frac{1}{2} \right)x + \frac{1}{3} = \frac{8}{3} \]

    Next, let's also move \( \frac{1}{3} \) to the right side: \[ \left( \frac{1}{5} + \frac{1}{2} \right)x = \frac{8}{3} - \frac{1}{3} \]

    Simplifying that: \[ \left( \frac{1}{5} + \frac{1}{2} \right)x = \frac{7}{3} \]

  2. Combine \( x \) coefficients: The least common multiple of 5 and 2 is 10. \[ \frac{1}{5} = \frac{2}{10}, \quad \frac{1}{2} = \frac{5}{10} \] Therefore, \[ \frac{2}{10} + \frac{5}{10} = \frac{7}{10} \] Now, substituting back: \[ \frac{7}{10}x = \frac{7}{3} \]

  3. Isolating \( x \): Dividing both sides by \( \frac{7}{10} \): \[ x = \frac{7}{3} \div \frac{7}{10} = \frac{7}{3} \cdot \frac{10}{7} = \frac{10}{3} \]

So, we can check the forms:

  • One-fifth \( x \) minus \( \frac{7}{3} \) = negative one-half: This rearranges correctly and is valid.

  • \( \frac{7}{10} x = \frac{7}{3} \): This resembles the correct formulation we arrived at.

  • Negative \( \frac{7}{3} = -\frac{7}{10} x \): This also is valid as a manipulation of the equation we derived.

  • \( x = -\frac{5}{2} x + \frac{35}{3} \): This does not match any transformations we applied readily to the original equation.

Given these evaluations, the correct equations to isolate would be:

  1. One-fifth \( x \) minus \( \frac{7}{3} \) = negative one-half
  2. \( \frac{7}{10} x = \frac{7}{3} \)
  3. Negative \( \frac{7}{3} = -\frac{7}{10} x \)

Thus, the valid options are first three statements:

  • One-fifth \( x \) minus \( \frac{7}{3} = -\frac{1}{2} \)
  • \( \frac{7}{10} x = \frac{7}{3} \)
  • Negative \( \frac{7}{3} = -\frac{7}{10} x \)