Chapter 3, Problem 1CT

Solution

To identify: The system without graphing. Then find the solution to each system using a graph.

  $\left\{\begin{array}{c}y=5x-2\\ y=x+4\end{array}\right.$

The resultant solution is $\left(\frac{3}{2},\frac{11}{2}\right)$ .

Given information:

The given system is:

  $\left\{\begin{array}{c}y=5x-2\\ y=x+4\end{array}\right.$

Explanation:

For two equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ .

The system of equation will be consistent but independent with one solution if: $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

The system of equation will be consistent but dependent with infinitely many solutions if: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The system of equation will be inconsistent with no solution if: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Now consider the given system:

  $\left\{\begin{array}{c}y=5x-2\\ y=x+4\end{array}\right.$

The equations can be written as:

  $\begin{array}{c}5x-y-2=0\\ x-y+4=0\end{array}$

Now check if the system of equations is consistent but independent with one solution, consistent but independent with one solution, or inconsistent for no solution:

  $\frac{5}{1}\ne \frac{-1}{-1}$

Therefore, the system of equation will be consistent but independent with one solution

Therefore, the lines will intersect once at a point or the lines are independent.

Now consider the equation $y=5x-2$ :

For $x=0$ ,

  $\begin{array}{c}y=5\left(0\right)-2\\ y=-2\end{array}$

For $y=0$ ,

  $\begin{array}{c}0=5x-2\\ 2=5x\\ x=\frac{2}{5}\end{array}$

Now consider the equation $y=x+4$ :

For $x=0$ ,

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

For $y=0$ ,

  $\begin{array}{c}0=x+4\\ x=-4\end{array}$

Now graph each equation:

  

From the graph the solution is $\left(\frac{3}{2},\frac{11}{2}\right)$ .