Chapter 6.4, Problem 14P

Solve each system. Explain why you chose the method you used.

$-x-3y=-3$

$2x+3y=5$

To determine

To find: the solution and identify which method is used to solve it.

  $\begin{array}{l}-x-3y=-3\hfill \\ 2x+3y=5\hfill \end{array}$

Answer to Problem 14P

In order to solve two systems we used substitution method, as in first equation x has − 1 as coefficient .Therefore the solution is $(x,y)=\left(2,\frac{1}{3}\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}-x-3y=-3\hfill \\ 2x+3y=5\hfill \end{array}$

Calculation:

In order to solve two systems we used substitution method, as in first equation x has -1.

Coefficient as:

  $\begin{array}{l}-x-3y=-3\hfill \\ \Rightarrow x=3-3y\hfill \end{array}$

Replace x in the second equation we get,

  $\begin{array}{l}2x+3y=5\hfill \\ \Rightarrow 2(3-3y)+3y=5\hfill \\ \Rightarrow -3y=-1\hfill \\ \Rightarrow y=\frac{1}{3}\hfill \end{array}$

Replace y in the first equation we get,

  $\begin{array}{l}x=3-3y\hfill \\ \Rightarrow x=3-3\left(\frac{1}{3}\right)\hfill \\ \Rightarrow x=2\hfill \end{array}$

We have one solution, $(x,y)=\left(2,\frac{1}{3}\right)$ which are consistent and independent.