- Start with the given equation for \(w\): \(w = \frac{8 x^2}{y} - 2\).
- Find a common denominator \(y\) for both terms to combine them into a single expression.
- Rewrite \(2\) as \(\frac{2y}{y}\) and subtract the fractions: \(w = \frac{8 x^2 - 2y}{y}\).
- Identify the equivalent expression as \(\boxed{w = \frac{8 x^2 - 2 y}{y}}\).
Explanation
- Problem Analysis
We are given the equation \(\frac{2 x^2}{y} = \frac{w+2}{4}\) and are told that solving for \(w\) results in \(w = \frac{8 x^2}{y} - 2\). Our goal is to find an equivalent expression for \(w\) among the provided choices. We start by manipulating the given expression \(w = \frac{8 x^2}{y} - 2\) to see if it can be written as a single fraction.
- Finding a Common Denominator
To express \(w = \frac{8 x^2}{y} - 2\) as a single fraction, we need a common denominator for the two terms. The denominator of the first term is \(y\), so we can write the constant \(2\) as a fraction with a denominator of \(y\):
\[2 = \frac{2 \times y}{y} = \frac{2 y}{y}.\]
Substituting this back into our expression for \(w\), we get:
\[w = \frac{8 x^2}{y} - \frac{2 y}{y}.\]
- Simplifying and Concluding
Now that both terms have the same denominator, we can combine the numerators:
\[w = \frac{8 x^2 - 2 y}{y}.\]
Comparing this result with the given options, we see that it matches the choice \(w = \frac{8 x^2 - 2 y}{y}\).
Examples
Algebraic manipulation is useful in many real-world scenarios, such as in engineering or physics, where you might have formulas that define the relationship between variables. For instance, if you are calculating the power dissipated in an electrical circuit, you might start with one form of the power formula \(P = I^2 R\) and need to rearrange it to find current \(I = \sqrt{\frac{P}{R}}\) or resistance \(R = \frac{P}{I^2}\). Being able to manipulate these equations by finding common denominators or clearing terms helps you solve for the specific variable you need, just like we did with the variable \(w\) here.