Solve each equation. Check your solution.

$69.\left|k+7\right|=3k-11$

On real number line, the distance of a number from 0 is the absolute value. It is always nonnegative.

The absolute value of a function is expressed as, if x is a real number then absolute value of xis defined as.

$\left|x\right|=x$ when x is greater than or equal to 0 . In other words, the absolute value of x is x when x is either positive or zero.

$\left|x\right|=-x$ when x is less than 0 . In other words, the absolute value of x is opposite of x when x is negative.

The absolute value of 5 is expressed as $\left|5\right|=5$ and of $-5$ is expressed as $\left|-5\right|=5$

Consider the provided equation.

$\left|k+7\right|=3k-11$

Recall the concept of absolute value and apply it.

Case 1. $k+7=3k-11$

Subtract k from both the sides.

$\begin{array}{l}k+7-k=3k-11-k\\ 7=2k-11\end{array}$

Subtract 7 from both the sides.

$\begin{array}{l}7-7=2k-11-7\\ 0=2k-18\\ 2k=18\\ k=9\end{array}$

Case 2. $k+7=-\left(3k-11\right)$

Subtract k from both the sides.

$\begin{array}{l}k+7-k=-3k+11-k\\ 7=-2k+11\end{array}$

Subtract 7 from both the sides.

$\begin{array}{l}7-7=-2k+11-7\\ 0=-2k+4\\ 2k=4\\ k=2\end{array}$

Therefore, there are two possible solutions for the equation $\left|k+7\right|=3k-11$ that are $k=9$ and$k=2$.

Substitute the value $k=9$in the equation $\left|k+7\right|=3k-11$.

$\begin{array}{l}|9+7|=3\left(9\right)-11\\ \left|16\right|=27-11\\ 16=16\end{array}$

Since, this is true so the value $k=9$satisfy the equation $\left|k+7\right|=3k-11$.

Substitute the value $k=2$in the equation $\left|k+7\right|=3k-11$

$\begin{array}{l}|9+2|=3\left(2\right)-11\\ \left|11\right|=6-11\\ 11=-5\end{array}$

Since, this is not true so the value $k=2$ does not satisfy the equation$\left|k+7\right|=3k-11$.

Thus, the solution set is $\left\{9\right\}$.

Hence, the solution of the equation is$k=9$.