Solve $\frac{1}{y}-\frac{1}{y^2}=-\frac{y-1}{y}$
Answer (1)
Multiply both sides by y 2 to eliminate fractions: y − 1 = − y ( y − 1 ) .
Simplify and rearrange to get a quadratic equation: y 2 − 1 = 0 .
Factor the quadratic: ( y − 1 ) ( y + 1 ) = 0 .
Solve for y and check for extraneous solutions: y = − 1 or y = 1 . Both are valid.
The solutions are y = − 1" , " y = 1 .
Explanation
Analyze the problem We are given the equation y 1 − y 2 1 = − y y − 1 and we want to solve for y .
Eliminate fractions First, we multiply both sides of the equation by y 2 to eliminate the fractions. This gives us y 2 ( y 1 − y 2 1 ) = y 2 ( − y y − 1 ) y − 1 = − y ( y − 1 ) y − 1 = − y 2 + y
Simplify the equation Next, we move all terms to one side to obtain a quadratic equation: y 2 + y − y − 1 = 0 y 2 − 1 = 0
Factor the quadratic We can factor the quadratic equation as ( y − 1 ) ( y + 1 ) = 0
Check for extraneous solutions Thus, the solutions are y = 1 and y = − 1 . However, we must check for extraneous solutions. If y = 1 , the original equation becomes 1 1 − 1 2 1 = − 1 1 − 1 1 − 1 = 0 0 = 0 which is true. However, if y = 1 , the term y 1 − y 2 1 becomes 1 1 − 1 2 1 = 1 − 1 = 0 , and the term − y y − 1 becomes − 1 1 − 1 = 0 . So y = 1 is a valid solution. If y = − 1 , the original equation becomes − 1 1 − ( − 1 ) 2 1 = − − 1 − 1 − 1 − 1 − 1 = − − 1 − 2 − 2 = − 2 which is also true. So y = − 1 is also a valid solution.
State the solutions Therefore, the solutions are y = 1 and y = − 1 .
Examples
When designing electrical circuits, engineers often encounter equations involving reciprocals and squares, similar to the one we solved. For instance, analyzing the impedance of parallel components might lead to such equations. Solving these equations accurately ensures the circuit functions as intended, preventing failures and optimizing performance. This type of algebraic manipulation is crucial for ensuring the reliability and efficiency of electronic devices.
