[tex]\frac{4 n-3}{6 n+1}=\frac{2 n-1}{3 n+4}[/tex]
Answer (1)
Cross-multiply the given equation to eliminate fractions.
Expand and simplify the resulting equation.
Solve for n .
Verify the solution by substituting it back into the original equation: 1 .
Explanation
Understanding the Problem We are given the equation 6 n + 1 4 n − 3 = 3 n + 4 2 n − 1 and we want to find the value(s) of n that satisfy this equation.
Cross-Multiplying To solve this equation, we first cross-multiply to eliminate the fractions. This gives us ( 4 n − 3 ) ( 3 n + 4 ) = ( 2 n − 1 ) ( 6 n + 1 ) .
Expanding Both Sides Next, we expand both sides of the equation: ( 4 n − 3 ) ( 3 n + 4 ) = 12 n 2 + 16 n − 9 n − 12 = 12 n 2 + 7 n − 12 ( 2 n − 1 ) ( 6 n + 1 ) = 12 n 2 + 2 n − 6 n − 1 = 12 n 2 − 4 n − 1
Equating the Expressions Now we set the expanded expressions equal to each other: 12 n 2 + 7 n − 12 = 12 n 2 − 4 n − 1
Simplifying the Equation We simplify the equation by subtracting 12 n 2 from both sides: 7 n − 12 = − 4 n − 1
Adding 4n to Both Sides Next, we add 4 n to both sides: 11 n − 12 = − 1
Adding 12 to Both Sides Then, we add 12 to both sides: 11 n = 11
Solving for n Finally, we divide both sides by 11 to solve for n :
n = 11 11 = 1
Checking for Extraneous Solutions We should check if n = 1 is an extraneous solution by plugging it back into the original equation: 6 ( 1 ) + 1 4 ( 1 ) − 3 = 6 + 1 4 − 3 = 7 1 3 ( 1 ) + 4 2 ( 1 ) − 1 = 3 + 4 2 − 1 = 7 1 Since both sides are equal and the denominators are not zero, n = 1 is a valid solution.
Final Answer Therefore, the solution to the equation is n = 1 .
Examples
In electrical engineering, when analyzing circuits, you might encounter equations involving rational expressions. Solving these equations helps determine the values of circuit components, such as resistors or capacitors, needed to achieve a desired circuit behavior. For example, the value 'n' could represent a scaling factor in a circuit design, and finding its value ensures the circuit operates as intended.
