- Start with the equation \(12y + d = -19y + t\).
- Add \(19y\) to both sides to group the \(y\) terms: \(31y + d = t\).
- Subtract \(d\) from both sides to isolate the \(y\) term: \(31y = t - d\).
- Divide by \(31\) to find \(y = \frac{t - d}{31}\).
- The correct option is \(\boxed{y = \frac{t - d}{31}}\).
Explanation
- Analyzing the Equation
The objective is to isolate the variable \(y\) in the given linear equation:
\[12y + d = -19y + t\]
To isolate \(y\), we must collect all terms involving \(y\) on one side of the equality and all other terms (in this case, \(d\) and \(t\)) on the opposite side. This is a standard algebraic procedure.
- Grouping the \(y\) Terms
To gather the \(y\) terms, add \(19y\) to both sides of the equation:
\[12y + 19y + d = -19y + 19y + t\]
\[(12 + 19)y + d = t\]
\[31y + d = t\]
This step effectively groups the coefficients of \(y\) together.
- Isolating the \(y\) Term
Next, move the constant \(d\) to the right side by subtracting it from both sides of the equation:
\[31y + d - d = t - d\]
\[31y = t - d\]
Now, the variable \(y\) is part of a simple multiplication expression on the left side.
- Solving for \(y\) and Conclusion
Finally, solve for \(y\) by dividing both sides of the equation by the coefficient \(31\):
\[\frac{31y}{31} = \frac{t - d}{31}\]
\[y = \frac{t - d}{31}\]
This gives us the explicit expression for \(y\) in terms of \(d\) and \(t\). Comparing this result with the provided options, we see that it matches the first option.
Examples
Rearranging equations is a fundamental skill in science and engineering. For example, consider the kinematic equation \(v = u + at\), where \(v\) is final velocity, \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is time. If you want to solve for time \(t\), you would subtract \(u\) from both sides (\(v - u = at\)) and then divide by acceleration \(a\), giving \(t = \frac{v - u}{a}\). This simple algebraic manipulation allows us to determine the duration of an object's motion given its velocity changes and acceleration.