Simplify the following expression:
$\frac{2 y+9}{y^2-7 y+12}+rac{2}{y-3}$
The objective is to simplify the rational expression \(\frac{2y+9}{y^2-7y+12} + \frac{2}{y-3}\).
First, we analyze the denominator of the first term: \(y^2-7y+12\). This is a quadratic trinomial. To factor it, we seek two numbers that multiply to \(12\) and add up to \(-7\). These numbers are \(-3\) and \(-4\). Thus, \(y^2-7y+12 = (y-3)(y-4)\).
The expression can then be written as:
\[\frac{2y+9}{(y-3)(y-4)} + \frac{2}{y-3}.\]
For the expression to be defined, the denominators must be non-zero. This implies \(y-3
eq 0\) and \(y-4
eq 0\), so \(y
eq 3\) and \(y
eq 4\).
To add these rational expressions, we need a common denominator. The least common denominator (LCD) is \((y-3)(y-4)\).
The first term already has this denominator: \(\frac{2y+9}{(y-3)(y-4)}\).
For the second term, we multiply the numerator and denominator by \((y-4)\):
\[\frac{2}{y-3} = \frac{2(y-4)}{(y-3)(y-4)}.\]
Now, combine the two fractions:
\[\frac{2y+9}{(y-3)(y-4)} + \frac{2(y-4)}{(y-3)(y-4)} = \frac{2y+9 + 2(y-4)}{(y-3)(y-4)}.\]
Now, we simplify the numerator:
\[2y + 9 + 2(y-4) = 2y + 9 + 2y - 8 = 4y + 1.\]
So the entire expression simplifies to:
\[\frac{4y+1}{(y-3)(y-4)}.\]
The simplified expression is \(\frac{4y+1}{y^2-7y+12}\) for \(y
eq 3\) and \(y
eq 4\).
Rational expressions like this appear frequently in physics and engineering when calculating total resistance in circuits or average velocities over segments of a path. For example, if the time taken to travel two segments of a trip is represented by different rational functions of speed, finding the total time requires adding these fractions. Mastering this algebraic simplification allows you to combine such complex dependencies into a single, manageable formula, making it easier to analyze overall system behavior.