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Algebraic manipulation is a fundamental skill in solving equations, making it easier to isolate and solve for the unknown variable. In our example, the equation given is \((0.502)(123) = (750) C\). Initially, we focus on simplifying one side of the equation. This involves performing operations like addition, subtraction, multiplication, and division on both sides to maintain equality.
By simplifying, we tackle the left-hand side, \((0.502)(123)\), which requires multiplying these two numbers. Once we've done the multiplication—yielding 61.746—we achieve a simpler form: \(61.746 = 750C\).
Continuing with algebraic manipulation allows us to take the necessary steps towards isolating the unknown, using appropriate arithmetic operations strategically.

Multiplication and division are key operations in algebra, especially when working to solve equations like the one we have here. Starting with the equation \((0.502)(123) = (750) C\), we need to multiply 0.502 by 123 to simplify the equation. The product of this multiplication is 61.746.
Once multiplication simplifies one side of the equation, we focus on division to isolate the variable. The equation to solve becomes \(61.746 = 750C\), needing us to divide both sides by 750. This operation clears the coefficient of C, making it possible to solve for it. Performing division accurately ensures that we arrive at the correct value of the unknown variable, which we find as \(C \approx 0.082328\).
Understanding when and how to apply these operations effectively is crucial in the process of solving equations.

The isolation of variables is a critical algebraic process used to solve equations for a specific unknown. In our task to find \(C\), we begin with this equation: \(61.746 = 750C\). To isolate \(C\), we aim to have it alone on one side of the equation.
To achieve this, we divide both sides of the equation by 750, which allows us to simplify \(750C\) into just \(C\). Thus, following this division gives us \(C = \frac{61.746}{750}\). This step is essential because once \(C\) is isolated, we've effectively found its value, which stands approximately as \(C \approx 0.082328\).
Mastering the technique of isolating a variable helps tackle various types of algebraic problems, ensuring steps are logically followed and executed correctly without altering the equation's balance.