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(2n-3)(n^2+3n-2)=(n-1)(2n^2+5n-4)

2 min read Jun 16, 2024
(2n-3)(n^2+3n-2)=(n-1)(2n^2+5n-4)

Solving the Equation: (2n-3)(n^2+3n-2) = (n-1)(2n^2+5n-4)

This equation involves expanding and simplifying expressions to solve for the unknown variable 'n'. Here's a step-by-step solution:

1. Expand both sides of the equation:

  • Left side: (2n-3)(n^2+3n-2) = 2n(n^2+3n-2) - 3(n^2+3n-2) = 2n^3 + 6n^2 - 4n - 3n^2 - 9n + 6 = 2n^3 + 3n^2 - 13n + 6

  • Right side: (n-1)(2n^2+5n-4) = n(2n^2+5n-4) - 1(2n^2+5n-4) = 2n^3 + 5n^2 - 4n - 2n^2 - 5n + 4 = 2n^3 + 3n^2 - 9n + 4

2. Simplify by combining like terms:

Now our equation is: 2n^3 + 3n^2 - 13n + 6 = 2n^3 + 3n^2 - 9n + 4

3. Solve for 'n':

  • Subtract 2n^3 and 3n^2 from both sides: -13n + 6 = -9n + 4

  • Add 9n to both sides: -4n + 6 = 4

  • Subtract 6 from both sides: -4n = -2

  • Divide both sides by -4: n = 1/2

Therefore, the solution to the equation (2n-3)(n^2+3n-2) = (n-1)(2n^2+5n-4) is n = 1/2.

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