Chapter 1.3, Problem 42PPSE

Solution

To state: The expression $\frac{5\left(2k-3\right)-3\left(k+4\right)}{3k+2}$ for the values $k=-2$ .

The resultant answer is $\frac{41}{4}$ .

Given information:

The given expression is $\frac{5\left(2k-3\right)-3\left(k+4\right)}{3k+2}$ .

Explanation:

Consider the given expression $\frac{5\left(2k-3\right)-3\left(k+4\right)}{3k+2}$ :

  $\frac{5\left(2k-3\right)-3\left(k+4\right)}{3k+2}$

Substitute the values of variable $k\text{as}-2$ into the given expression,

  $\frac{5\left(2\left(-2\right)-3\right)-3\left(\left(-2\right)+4\right)}{3\left(-2\right)+2}$

Multiply,

  $\frac{5\left(-4-3\right)-3\left(-2+4\right)}{-6+2}$

Simplify within the grouping symbol,

  $\frac{5\left(-7\right)-3\left(2\right)}{-4}$

Multiply,

  $\frac{-35-6}{-4}$

Add and subtract from left to right,

  $\frac{-41}{-4}=\frac{41}{4}$

Therefore, $\frac{41}{4}$ is the resultant value of the expression $\frac{5\left(2k-3\right)-3\left(k+4\right)}{3k+2}$ .