Chapter 6, Problem 1RE

To determine

To solve: The system of linear equations $4x-3y=-1$ and $3x+5y=50$ and also identify the system with dependent equations and inconsistent system.

Answer to Problem 1RE

The solution set of the system of linear equations $4x-3y=-1$ and $3x+5y=50$ is $\underset{\_}{\left\{\left(5,7\right)\right\}}$.

Explanation of Solution

Given:

The system of linear equations is $\{\begin{array}{c}4x-3y=-1\\ 3x+5y=50\end{array}$.

Procedure used:

“To solve the system of linear equations 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, you can skip this step.)

Step 2: Substitute the expression found in step 1 into the other equation.

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 obtained ordered pair by putting the corresponding value to the variables of the system of linear equations.”

Calculation:

The first equation is given below:

$4x-3y=-1\left(1\right)$

The second equation is given below:

$3x+5y=50\left(2\right)$

Step 1:

In equation (1), isolate the variable $x$ as follows:

$\begin{array}{l}4x-3y=-1\\ 4x=3y-1\\ x=\frac{3y-1}{4}\end{array}$

Step 2:

Substitute $\frac{3y-1}{4}$ for $x$  in equation (2) as follows:

$3\left(\frac{3y-1}{4}\right)+5y=50$

Here, the variable $x$ is eliminated.

Step 3:

Solve the equation obtained in Step 2 to find the value of $y$ as follows:

$\begin{array}{l}3\left(\frac{3y-1}{4}\right)+5y=50\\ \frac{3\left(3y-1\right)+20y}{4}=50\\ 9y-3+20y=50\left(4\right)\\ 29y-3=200\end{array}$

Simplify the above equation as follows:

$\begin{array}{l}29y=200+3\\ 29y=203\\ y=\frac{203}{29}\\ y=7\end{array}$

Therefore, the value of $y$ is $7$.

Step 4:

Substitute $7$ for $y$ in equation (1) to obtain the value of $x$ as follows:

$\begin{array}{l}4x-3\left(7\right)=-1\\ 4x-21=-1\\ 4x=-1+21\\ 4x=20\end{array}$

Simplify the above equation as follows:

$\begin{array}{c}x=\frac{20}{4}\\ x=5\end{array}$

Therefore, the value of $x$ is $5$.

Step 5:

Check whether the obtained solution $\left(5,7\right)$ satisfies the given equations or not.

Substitute $5$ for $x$ and $7$ for $y$  in equation (1) to verify the obtained solution.

$\begin{array}{l}4\left(5\right)-3\left(7\right)\stackrel{?}{=}-1\\ 20-21\stackrel{?}{=}-1\\ -1=-1\left(\text{True}\right)\end{array}$

Substitute $5$ for $x$ and $7$ for $y$  in equation (2) to verify the obtained solution.

$\begin{array}{l}3\left(5\right)+5\left(7\right)\stackrel{?}{=}50\\ 15+35\stackrel{?}{=}50\\ 50=50\left(\text{True}\right)\end{array}$

Therefore, the obtained solution $\left(5,7\right)$ is verified.

Thus, the solution set of the system of linear equations $4x-3y=-1$ and $3x+5y=50$ is $\underset{\_}{\left\{\left(5,7\right)\right\}}$.