To solve the equation \(0.12z + \left(0.5 + \frac{z}{2}\right) = \frac{z}{3} + 1.5\), we need to simplify and isolate \(z\).

First, distribute and combine like terms:

\[ 0.12z + 0.5 + \frac{z}{2} = \frac{z}{3} + 1.5 \]

Next, express all terms with \(z\) on one side and constants on the other:

1. Convert \(\frac{z}{2}\) and \(\frac{z}{3}\) to a common denominator. The least common multiple of 2 and 3 is 6.
   \[ \frac{z}{2} = \frac{3z}{6} \]
   \[ \frac{z}{3} = \frac{2z}{6} \]

So the equation becomes:
\[ 0.12z + 0.5 + \frac{3z}{6} = \frac{2z}{6} + 1.5 \]

2. Combine the terms involving \(z\):
   \[ 0.12z + \frac{3z}{6} = \frac{0.12 \cdot 6z + 3z}{6} = \frac{0.72z + 3z}{6} = \frac{3.72z}{6} \]

Thus, the equation simplifies to:
\[ \frac{3.72z}{6} + 0.5 = \frac{2z}{6} + 1.5 \]

3. Eliminate the fractions by multiplying the entire equation by 6:
   \[ 3.72z + 3 = 2z + 9 \]

4. Isolate \(z\) on one side of the equation:
   \[ 3.72z - 2z = 9 - 3 \]
   \[ 1.72z = 6 \]

5. Solve for \(z\):
   \[ z = \frac{6}{1.72} \]
   \[ z \approx 3.49 \]

So, the solution is approximately \( z \approx 3.49 \).

Would you like more details on any step or have any questions? Here are five more questions you might want to explore:

1. How do you solve quadratic equations using the quadratic formula?
2. What are the steps to factorize polynomials?
3. Can you explain the method of completing the square?
4. How do you solve a system of linear equations using matrix methods?
5. What are the key properties of logarithmic functions?

**Tip:** When solving equations, always remember to check your solution by substituting it back into the original equation to ensure it satisfies the equation.

Learn how to solve the equation 0.12z + (0.5 + z/2) = z/3 + 1.5 with a detailed step-by-step solution. This problem involves equations with fractions and algebraic simplification. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation