Math Problem Statement

In the equation \(\frac{2}{3}(x + 2) - 2 = 3 + \frac{1}{3}(x - 5)\), what is the value of x?

Solution

Let's solve the given equation step by step.

The equation is:

23(x+2)2=3+13(x5)\frac{2}{3}(x + 2) - 2 = 3 + \frac{1}{3}(x - 5)

Step 1: Distribute the fractions.

Expand the terms that have fractions:

23(x+2)=23x+43\frac{2}{3}(x + 2) = \frac{2}{3}x + \frac{4}{3} 13(x5)=13x53\frac{1}{3}(x - 5) = \frac{1}{3}x - \frac{5}{3}

Now the equation becomes:

(23x+43)2=3+(13x53)\left(\frac{2}{3}x + \frac{4}{3}\right) - 2 = 3 + \left(\frac{1}{3}x - \frac{5}{3}\right)

Step 2: Simplify constants.

First, handle the subtraction of 2 on the left-hand side:

432=4363=23\frac{4}{3} - 2 = \frac{4}{3} - \frac{6}{3} = -\frac{2}{3}

Now the equation is:

23x23=3+13x53\frac{2}{3}x - \frac{2}{3} = 3 + \frac{1}{3}x - \frac{5}{3}

Step 3: Combine constants on the right-hand side.

Simplify the constants on the right-hand side:

353=9353=433 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}

Now the equation is:

23x23=43+13x\frac{2}{3}x - \frac{2}{3} = \frac{4}{3} + \frac{1}{3}x

Step 4: Eliminate the fractions by multiplying through by 3.

To simplify further, multiply both sides of the equation by 3 to eliminate the fractions:

3(23x23)=3(43+13x)3\left(\frac{2}{3}x - \frac{2}{3}\right) = 3\left(\frac{4}{3} + \frac{1}{3}x\right)

This gives:

2x2=4+x2x - 2 = 4 + x

Step 5: Solve for xx.

Now subtract xx from both sides:

2xx2=42x - x - 2 = 4

Simplify:

x2=4x - 2 = 4

Add 2 to both sides:

x=6x = 6

Thus, the value of xx is 66.

Final Answer: x=6x = 6

Would you like more details on any step? Here are some related questions you might want to explore:

  1. How do you solve equations involving fractions?
  2. How does distributing work with fractional coefficients?
  3. What techniques can simplify solving multi-step equations?
  4. Can you explain the logic behind multiplying both sides by the same value?
  5. How do you check if a solution to an equation is correct?

Tip: When working with equations that involve fractions, a good strategy is to eliminate fractions early by multiplying the entire equation by the least common denominator (LCD).

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Math Problem Analysis

Mathematical Concepts

Algebra
Solving Linear Equations
Fractions

Formulas

Distributive Property
Combining Like Terms
Solving for x

Theorems

Basic algebraic principles
Multiplication property of equality

Suitable Grade Level

Grades 8-10