Chapter 4.6, Problem 42WE

To determine

To find: The solution set of the equation $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ .

Answer to Problem 42WE

The solution set for the equation $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ is $\left\{\frac{1}{2}\right\}$ .

Explanation of Solution

Given information:

The equation is $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ .

Calculation:

The given equation is $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ .

Solve the given equation $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ to obtain the value of solution set.

  $\begin{array}{c}\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)\\ 2n^3+6n^2-4n-3n^2-9n+6=2n^3+5n^2-4n-2n^2-5n+4\\ 2n^3+3n^2-13n+6=2n^3+3n^2-9n+4\left(\text{merge the like terms}\right)\\ 3n^2-13n+6=3n^2-9n+4\left(\text{subtract 2}{n}^3\text{from both sides}\right)\end{array}$

Further solve the above calculation,

  $\begin{array}{c}-13n+6=-9n+4\left(\text{subtract 3}{n}^2\text{from both sides}\right)\\ 6=4n+4\left(\text{add 13}n\text{ on both sides}\right)\\ 2=4n\left(\text{subtract 4 on both sides}\right)\\ \frac{1}{2}=n\left(\text{divide both side by 4}\right)\end{array}$

Therefore, the solution set for the equation $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ is $\left\{\frac{1}{2}\right\}$ .

Conclusion:

Thus, the solution set for the equation $\left(2n-3\right)\left(n^2+3n-2\right)=\left(n-1\right)\left(2n^2+5n-4\right)$ is $\left\{\frac{1}{2}\right\}$ .