- Define the domain of the expression by setting all denominators to be non-zero (\(y
eq -1, 0, 3\)).
- Combine the terms in the numerator to get \(\frac{2y+4}{y+1}\).
- Combine the terms in the denominator to get \(\frac{y-3}{y}\).
- Multiply the numerator by the reciprocal of the denominator to obtain the final simplified result: \(\boxed{\frac{2y(y+2)}{(y+1)(y-3)}}\).
Explanation
- Analyzing the Expression and Domain Restrictions
To simplify the expression \(\frac{2+\frac{2}{y+1}}{1-\frac{3}{y}}\), we first identify the domain restrictions. The expression is defined when the denominators are non-zero:
- \(y+1
eq 0 \implies y
eq -1\)
- \(y
eq 0\)
- \(1 - \frac{3}{y}
eq 0 \implies \frac{y-3}{y}
eq 0 \implies y
eq 3\)
Thus, the expression is defined for \(y \in \mathbb{R} \setminus \{-1, 0, 3\}\).
- Simplifying the Numerator and Denominator
Next, we simplify the numerator, \(2 + \frac{2}{y+1}\). To do this, we find a common denominator:
\[ 2 + \frac{2}{y+1} = \frac{2(y+1) + 2}{y+1} = \frac{2y + 2 + 2}{y+1} = \frac{2y+4}{y+1} = \frac{2(y+2)}{y+1} \]
Now, we simplify the denominator, \(1 - \frac{3}{y}\):
\[ 1 - \frac{3}{y} = \frac{y - 3}{y} \]
Combining these results, the original expression becomes:
\[ \frac{\frac{2(y+2)}{y+1}}{\frac{y-3}{y}} \]
- Finalizing the Simplification
To divide by a fraction, we multiply the numerator by the reciprocal of the denominator:
\[ \frac{2(y+2)}{y+1} \cdot \frac{y}{y-3} = \frac{2y(y+2)}{(y+1)(y-3)} \]
Expanding the expression, we get:
\[ \frac{2y^2 + 4y}{y^2 - 2y - 3} \]
This is the simplified form of the given expression.
Examples
Rational expressions appear frequently in physics and engineering, such as when calculating the combined resistance of parallel electrical circuits or modeling velocity in fluid dynamics. Simplifying these complex fractions helps in analyzing how a system behaves as a specific variable changes. For example, if the expression above represented a control system gain, simplifying it allows engineers to quickly determine the system's poles and zeros, which are essential for ensuring stable performance.