Question: An archaeologist finds the equation $ 4(x + 3) - 2x = 3(2x - 1) $ in an ancient tablet. Solve for $ x $. - Londonproperty

Understanding the Context

Decoding the tablet required unraveling the equation step by step, much like shedding light on a forgotten language. The equation reads:
4(x + 3) − 2x = 3(2x − 1)

On the surface, it’s a linear algebraic equation involving parentheses and variables. To solve for $ x $, archaeologists and mathematicians collaborated, applying familiar modern algebraic methods to verify the accuracy of the ancient scribe’s work.

Step-by-Step Solution: Solving the Equation

Let’s walk through how to solve this equation using standard algebraic procedures:

Key Insights

Step 1: Expand both sides
Distribute the terms on both sides of the equation.
$$
4(x + 3) - 2x = 3(2x - 1)
$$
Left side:
$$ 4x + 12 - 2x = 2x + 12 $$
Right side:
$$ 3(2x - 1) = 6x - 3 $$

Now the equation is:
$$ 2x + 12 = 6x - 3 $$

Step 2: Isolate variable terms
Bring all terms with $ x $ to one side and constant terms to the other. Subtract $ 2x $ from both sides:
$$ 12 = 4x - 3 $$

Step 3: Solve for $ x $
Add 3 to both sides:
$$ 15 = 4x $$
Now divide by 4:
$$ x = rac{15}{4} $$

Why This Discovery Matters

Final Thoughts

Beyond its mathematical value, this tablet offers a rare glimpse into the intellectual achievements of ancient scholars. The solving of equations like this suggests a developed understanding of algebra centuries before formal algebra was documented in the Islamic Golden Age or Renaissance Europe.

This find highlights the intersection of archaeology, history, and STEM education, inspiring public interest in mathematics and science through tangible historical evidence. It reminds us that problem-solving is a timeless human pursuit—deeply rooted in curiosity and logic.

Final Answer

$$
oxed{x = rac{15}{4}}
$$

So, the equation $ 4(x + 3) - 2x = 3(2x - 1) $, once etched in ancient stone, yields the elegant solution $ x = rac{15}{4} $—a testament to the enduring power of knowledge across millennia.