Chapter 8, Problem 1RE

To determine

The solution of the provided system of equation by using any method of your choice and determine whether the system with no solution and system with infinitely many solution.

Answer to Problem 1RE

The solution set is $\{-1,2\}$.

Explanation of Solution

Given information:

The provided system of equation is shown below:

$\begin{array}{cc}\left\{\begin{array}{l}3x-y=-5\\ x+2y=3\end{array}\right\}& \begin{array}{l}(1)\\ (2)\end{array}\end{array}$

Formula used:

Solving Linear System by Substitution:

Step 1: Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, we can skip this step).

Step 2: Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable.

Step 3: Solve the equation containing one variable.

Step 4: Back-substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable.

Step 5: Check the proposed solution in both of the system’s give equations.

Calculation:

First, solve one of the equation for any of the variable $x\text{or}y$. Since the coefficient of $x$ in equation (2) is 1. So, its easy to solve for $x$ in equation (2).

$x=3-2y$

Now, substitute the value of $x$ in equation (1), we get

$\begin{array}{c}3(3-2y)-y=-5\\ 9-6y-y=-5\left(\text{Distributive}\text{Property}\right)\\ 9-7y=-5\\ 9-7y-9=-5-9(\text{Subtract}\text{9}\text{from}\text{both}\text{sides})\\ -7y=-14\\ \frac{-7y}{-7}=\frac{-14}{-7}(\text{Divide}\text{both}\text{sides}\text{by}\text{-7})\\ y=2\end{array}$

Substitute the value of $y$ in equation (2), we get

$\begin{array}{c}x+2(2)=3\\ x+4=3\\ x=3-4\\ x=-1\end{array}$

Now, we have to check the above solution.

Substitute the value of $x=-1\text{and}y=2$ in an equation (1), we get

$\begin{array}{l}3(-1)-(2)\stackrel{?}{=}-5\\ -3-2\stackrel{?}{=}-5\\ -5=-5(\text{True})\end{array}$

Substitute the value of $x=-1\text{and}y=2$ in an equation (2), we get

$\begin{array}{l}(-1)+2(2)\stackrel{?}{=}3\\ -1+4\stackrel{?}{=}3\\ 3=3(\text{True})\end{array}$

Therefore, the required solution set is $\{-1,2\}$.