91Ó°ÊÓ

Understanding algebraic manipulation is crucial for solving linear equations. This involves rearranging the terms of the equation to make the variable of interest, most often denoted as 'x', more accessible. In the exercise \( 7.45x - 8.81 = 5.29 + 9.47x \), we start with the rearrangement of terms that contain 'x' to one side and numeric terms to the other.

For this, we perform operations uniformly on both sides of the equation. The practice of doing the same operation on each side ensures we keep the equation balanced or equal. To combine 'x' terms on one side, we subtract the term involving 'x' that is on the opposite side of our desired variable from both sides. Here, subtracting \( 7.45x \) from each side starts our process of algebraic manipulation, leading us to move toward equation simplification.

After rearranging the terms, we enter the equation simplification stage. It’s all about combining like terms and simplifying both sides of the equation into a more manageable form. This makes the calculation and the final solution much clearer.

In the given problem, once we consolidated like terms, we simplified the algebraic expression on the left (\( 9.47x - 7.45x \)) to \( 2.02x \) by subtracting the coefficients of 'x'. On the right side, we combined the constants (\( 5.29 + 8.81 \) to get \( 14.1 \) and arrive at a simplified equation \( 2.02x = 14.1 \).

While simplification, it is critical to keep track of your mathematical operations to avoid any mistakes. Using a structured, step-by-step approach can prevent confusion and ensure that each term is simplified accurately.

With our simplified equation \( 2.02x = 14.1 \), the final core concept to apply is to isolate the variable. Isolation means to have the variable 'x' on one side and the numerical answer on the other.

To achieve this, we consider the inverse operation to the one that is applied to 'x'. Since 'x' is being multiplied by \( 2.02 \), we use division as the inverse operation. Dividing both sides by \( 2.02 \) effectively isolates 'x', yielding \( x = \frac{14.1}{2.02} \).

It is essential to understand that isolating the variable is a crucial step in solving for 'x' because it gives us the value that 'x' must be for the original equation to hold true. With 'x' isolated, you should be able to clearly see the solution and, in this case, can round off to the nearest hundredth as needed to find the final answer.