- Analyze the equation \(A = xP + Py\) and recognize that \(P\) is a common factor on the right-hand side.
- Evaluate the options and determine that factoring out \(P\) allows for direct isolation of the variable.
- Execute the factoring: \(A = P(x + y)\).
- Final Answer: \(\boxed{\text{Factor out } P}\).
Explanation
- Analyzing the Equation
We are given the literal equation \(A = xP + Py\) and are asked to determine the most effective first step to solve for the variable \(P\). In this equation, \(A\), \(x\), and \(y\) are treated as constants, and we need to isolate \(P\). Notice that the variable \(P\) appears in two distinct terms on the right-hand side: \(xP\) and \(Py\). To solve for \(P\), we must first collect all terms containing \(P\) on one side, or better yet, condense these terms into a single instance of \(P\).
- Evaluating the Options
We examine the available options:
- Subtract \(xP\) from both sides: This results in \(A - xP = Py\). We still have \(P\) on both sides.
- Factor out \(P\): This results in \(A = P(x + y)\). Since \(P\) is now a factor in a single term, we can easily isolate it.
- Subtract \(Py\) from both sides: This results in \(A - Py = xP\). We still have \(P\) on both sides.
- Divide by \(xP\): This would result in \(\frac{A}{xP} = 1 + \frac{Py}{xP} = 1 + \frac{y}{x}\), which does not simplify the expression to solve for \(P\) effectively.
By factoring out \(P\), we transform the right side into the product \(P(x + y)\). This is the standard and most efficient algebraic technique for isolating a variable that appears as a common factor in multiple terms. Once factored, we simply divide both sides by the quantity \((x + y)\) to obtain \(P = \frac{A}{x+y}\).
- Conclusion
The most efficient first step to isolate \(P\) in the equation \(A = xP + Py\) is to factor it out. This directly converts the expression into \(A = P(x + y)\), setting the stage for a single final step of division. Therefore, the correct choice is to factor out \(P\). The final answer is \(\boxed{\text{Factor out } P}\).
Examples
In science and engineering, you often work with formulas that involve multiple parameters. For example, consider the Ohm's Law formula related to resistance in a circuit: \(V = IR_1 + IR_2\). If you want to find the current \(I\) given the total voltage \(V\) and two resistances \(R_1\) and \(R_2\), you would first factor out the current \(I\) to get \(V = I(R_1 + R_2)\) and then divide to find \(I = \frac{V}{R_1 + R_2}\). This method of factoring out common variables is a fundamental tool for solving literal equations across many fields, from physics to economics.