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Turning a word problem into an algebraic equation is a fundamental skill in mathematics that allows us to solve real-world problems. When doing so, it's essential to understand the meaning of the phrases used. In the example given, 'Five times a number is 160', we identify the operation as multiplication and the constant as 160. To translate this, we represent 'a number' by a variable, let's say, 'n'.

The phrase 'Five times' indicates that the variable is to be multiplied by five. Thus, the sentence is expressed as the equation \(5n = 160\). Reading the sentence carefully and understanding the meaning of each word helps in translating correctly. This step is crucial for accuracy in forming the equation before attempting to solve it.

Example Translation

If the sentence was 'the sum of a number and four is equal to nine', the translation would be \(n + 4 = 9\), where 'n' is the unknown number, '+' denotes addition, and '9' is the constant.

Variables in algebra serve as placeholders for unknown quantities and are typically denoted by letters like 'n', 'x', 'y', etc. The process of identifying a variable is critical for setting up equations correctly. It allows us to solve for these unknowns systematically.

In our example, 'a number' is the phrase that indicates the variable. We chose 'n' to represent it, but any letter would work just as well. The key is consistency; once a variable is assigned, it should represent the same quantity throughout the problem.

Variable Consistency

For instance, if 'half of a number decreased by 2 is equal to 5', and we let 'x' be our variable, we would write \(\frac{x}{2} - 2 = 5\). Here, 'x' consistently stands for the unknown number across the entire equation.

Once we have translated a verbal sentence into an equation and identified the variable, the next step is solving it. Linear equations, like the one in our example \(5n = 160\), feature variables raised to the power of one and can usually be solved using simple algebraic manipulations.

The goal is to isolate the variable on one side of the equation, providing its value. In this case, we divide both sides by 5, the coefficient of 'n', giving us \(n = \frac{160}{5}\) and yielding the solution \(n = 32\). It's important to apply operations equally to both sides to maintain the equality.

Methodical Approach

A systematic approach involves performing permissible operations such as adding, subtracting, multiplying, or dividing both sides of the equation until the variable is isolated. This approach can be applied to any linear equation to find the unknown variable's value.