Simplifying each side of the equation results in $x^2-3 x-4=x^2-5 x+6$. Find the solution: $\frac{x+2}{3 x}-\frac{1}{x-2}=\frac{x-3}{3 x}$
Answer (2)
Subtract 3 x x + 2 from both sides: − x − 2 1 = 3 x x − 3 − 3 x x + 2 .
Combine the terms: − x − 2 1 = 3 x − 5 .
Multiply both sides by -1: x − 2 1 = 3 x 5 .
Cross-multiply and solve for x: 3 x = 5 ( x − 2 ) ⇒ x = 5 .
The solution is 5 .
Explanation
Understanding the Problem We are given the equation 3 x x + 2 − x − 2 1 = 3 x x − 3 and we need to find the value of x that satisfies it. First, we note that x cannot be 0 or 2, because that would make the denominators zero, which is not allowed.
Isolating the Term with x To solve the equation, we want to isolate x . Let's start by subtracting 3 x x + 2 from both sides of the equation: 3 x x + 2 − x − 2 1 − 3 x x + 2 = 3 x x − 3 − 3 x x + 2 − x − 2 1 = 3 x x − 3 − ( x + 2 ) − x − 2 1 = 3 x x − 3 − x − 2 − x − 2 1 = 3 x − 5
Simplifying the Equation Now, multiply both sides by -1: x − 2 1 = 3 x 5
Cross-Multiplication Next, we cross-multiply: 3 x = 5 ( x − 2 )
Expanding the Equation Expand the right side: 3 x = 5 x − 10
Isolating x Subtract 5 x from both sides: 3 x − 5 x = − 10 − 2 x = − 10
Solving for x Finally, divide both sides by -2: x = − 2 − 10 x = 5
Checking the Solution Now, we need to check if our solution is valid. Since x cannot be 0 or 2, and our solution is x = 5 , it is a valid solution. Therefore, the solution to the equation is x = 5 .
Examples
Imagine you are distributing flyers for a local event. You want to divide the flyers equally among several locations, but you also need to account for some flyers getting damaged or lost in transit. Solving equations like this helps you determine the exact number of flyers needed to ensure each location receives its fair share, even with some potential losses. This kind of problem-solving is useful in resource allocation, logistics, and many other real-world scenarios where precision is important.
The solution to the equation 3 x x + 2 − x − 2 1 = 3 x x − 3 is x = 5 . This value does not violate the conditions that x cannot be 0 or 2. Therefore, the solution is valid.
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