https://file.aiquickdraw.com/mathbot/file/172909420422398dx0hmn.jpg

To solve the equation:

\[
\frac{2x - 4}{3} - \frac{x + 1}{6} = t + 1 \quad \text{and} \quad t = 3
\]

First, substitute \( t = 3 \) into the equation:

\[
\frac{2x - 4}{3} - \frac{x + 1}{6} = 3 + 1
\]
\[
\frac{2x - 4}{3} - \frac{x + 1}{6} = 4
\]

Now, clear the fractions by multiplying the entire equation by 6, which is the least common denominator of 3 and 6:

\[
6 \times \left(\frac{2x - 4}{3}\right) - 6 \times \left(\frac{x + 1}{6}\right) = 6 \times 4
\]
\[
2 \times (2x - 4) - (x + 1) = 24
\]
\[
4x - 8 - x - 1 = 24
\]
\[
3x - 9 = 24
\]
\[
3x = 33
\]
\[
x = 11
\]

Thus, the value of \( x \) is:

\[
\boxed{11}
\]

If \( \frac{2x - 4}{3} - \frac{x + 1}{6} = t + 1 \) and \( t = 3 \), what is the value of \( x \)?

In this algebra problem, solve the linear equation \( \frac{2x - 4}{3} - \frac{x + 1}{6} = t + 1 \) with \( t = 3 \) by clearing fractions and isolating x. Suitable for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation