Chapter 5.4, Problem 13Q

(a)

To determine

To find:The system of linear of equation if spruce tree grows 4 inches each year and hemlock tree grows 6 inches each year.

Answer to Problem 13Q

The system of linear of equation is $y=4x+14,y=6x+8$ .

Explanation of Solution

Given information:

1. The initial height of spruce tree is $14\text{ inches}$ and the height of hemlock tree is $8\text{ inches}$ .
2. The rate of growth per year of spruce tree is $4\text{ inches}$ and rate of growth per year of hemlock tree is $\text{6 inches}$ .

Consider, the heights of given trees be $y$ and the be number of years be $x$ .

According to the given information the equation that represents the situation of spruce tree is as follows:

  $y=4x+14$ .

According to the given information the equation that represents the situation of hemlock tree is as follows:

  $y=6x+8$ .

Therefore, the system of linear of equation is $y=4x+14,y=6x+8$ .

Conclusion:

Thus, the system of linear of equation is $y=4x+14,y=6x+8$ .

(b)

To determine

To find: The solution set of the system of linear equations $y=4x+14$ and $y=6x+8$ by method of graphing.

Answer to Problem 13Q

The solution set of the systems of linear equations $y=4x+14$ and $y=6x+8$ is $\left(3,26\right)$ .

Explanation of Solution

Given information:

1. The initial height of spruce tree is $14\text{ inches}$ and the height of hemlock tree is $8\text{ inches}$ .
2. The rate of growth per year of spruce tree is $4\text{ inches}$ and rate of growth per year of hemlock tree is $\text{6 inches}$ .

Calculation:

As theobtained system of linear equations from part (a) $y=4x+14$ and $y=6x+8$ .

Consider equation $y=4x+14$ as equation 1 and equation $y=6x+8$ as equation 2.

Equation 1:

Substitute $\text{0 for }x$ in equation $y=4x+14$ to obtain the value of $y$ .

  $\begin{array}{c}y=4\left(0\right)+14\\ =14\end{array}$

Substitute $\text{1 for }x$ in equation $y=4x+14$ to obtain the value of $y$ .

  $\begin{array}{c}y=4\left(1\right)+14\\ =4+14\\ =18\end{array}$

Substitute $\text{2 for }x$ in equation $y=4x+14$ to obtain the value of $y$ .

  $\begin{array}{c}y=4\left(2\right)+14\\ =8+14\\ =22\end{array}$

Therefore, the ordered pair of the equation $y=4x+14$ is as follows:

  $\left(0,14\right),\left(1,18\right)\text{ and }\left(2,22\right)$ .

Equation 2:

Substitute $\text{0 for }x$ in equation $y=6x+8$ to obtain the value of $y$ .

  $\begin{array}{c}y=6\left(0\right)+8\\ =8\end{array}$

Substitute $\text{1 for }x$ in equation $y=6x+8$ to obtain the value of $y$ .

  $\begin{array}{c}y=6\left(1\right)+8\\ =6+8\\ =14\end{array}$

Substitute $\text{2 for }x$ in equation $y=6x+8$ to obtain the value of $y$ .

  $\begin{array}{c}y=6\left(2\right)+8\\ =12+8\\ =20\end{array}$

Therefore, the ordered pairs of the equation $y=6x+8$ is as follows:

  $\left(0,8\right),\left(1,14\right)\text{ and }\left(2,20\right)$ .

So, according to the ordered pairs of equation 1 and equation 2 the graph of equation 1 and equation 2 is as follows:

  

The solution of the system of equations is the intersection point of the system of equations.

It can be observed that from figure 1 that the both equations intersect each other at points at point $\left(3,26\right)$ ,

Therefore, the solution set of the systems of linear equations $y=4x+14$ and $y=6x+8$ is $\left(3,26\right)$ .

The interpretation of the given problem is the height of both the trees is same after 3 years from that is $26\text{ inches}$ .

Conclusion:

Thus, the solution set of the systems of linear equations $y=4x+14$ and $y=6x+8$ is $\left(3,26\right)$ .