Chapter 4, Problem 1PCT

Determine whether the given ordered pairs are solutions of the given system of equations. Answer Yes or No.

$\{\begin{array}{ccc}4x+y& =& -20\\ y& =& -\frac{1}{6}x+3\end{array}$

1. a. (3, $\frac{5}{2}$)
2. b. (−6, 4)
3. c. (−4, −4)

To determine

Whether the given ordered pair $\left(3,\frac{5}{2}\right)$ is a solution of the system of equations $4x+y=-20\text{and}y=-\frac{1}{6}x+3$ or not.

Answer to Problem 1PCT

No, the ordered pair $\left(3,\frac{5}{2}\right)$ is not a solution of the given system of linear equations.

Explanation of Solution

Procedure used:

Step 1: Substitute the values of $x$ and $y$ in first equation and verify whether the equality holds.

Step 2: Repeat the procedure in step 1 for the second equation as well.

Step 3: If the equality holds in both step 1 and step 2, then conclude that the given ordered pair is a solution of the given system.

Calculation:

Consider the given equations as,

$4x+y=-20$                                                                                     (1)

$y=-\frac{1}{6}x+3$                                                                                    (2)

The given ordered pair is $\left(3,\frac{5}{2}\right)$.

Substitute $x=3$ and $y=\frac{5}{2}$ in equation (1).

$\begin{array}{l}4x+y=-20\\ 4\left(3\right)+\left(\frac{5}{2}\right)\stackrel{?}{=}-20\\ 12+\frac{5}{2}\stackrel{?}{=}-20\\ \frac{29}{2}=6\text{False}\end{array}$

Here, the ordered pair $\left(3,\frac{5}{2}\right)$ does not satisfy the first equation.

Since the given ordered pair $\left(3,\frac{5}{2}\right)$ does not satisfy the equation (1) of the system, the ordered pair is not a solution of the given system.

To determine

Whether the given ordered pair $\left(-6,4\right)$ is a solution of the system of equations $4x+y=-20\text{and}y=-\frac{1}{6}x+3$ or not.

Answer to Problem 1PCT

Yes, the given ordered pair $\left(-6,4\right)$ is a solution of the given system of linear equations.

Explanation of Solution

The given ordered pair is $\left(-6,4\right)$.

Substitute $x=-6$ and $y=4$ in equation (1).

$\begin{array}{l}4x+y=-20\\ 4\left(-6\right)+\left(4\right)\stackrel{?}{=}-20\\ -24+4\stackrel{?}{=}-20\\ -20=-20\text{True}\end{array}$

Here, the ordered pair $\left(-6,4\right)$ satisfies the first equation.

Substitute $x=-6$ and $y=4$ in equation (2).

$\begin{array}{l}y=-\frac{1}{6}x+3\\ \left(4\right)\stackrel{?}{=}-\frac{1}{6}\left(-6\right)+3\\ 4\stackrel{?}{=}1+3\\ 4=4\text{True}\end{array}$

Here, the ordered pair $\left(-6,4\right)$ satisfies the second equation.

Since both the equations are true, the given ordered pair $\left(-6,4\right)$ is a solution of the given system of linear equations.

To determine

Whether the given ordered pair $\left(-4,-4\right)$ is a solution of system of the equations $4x+y=-20\text{and}y=-\frac{1}{6}x+3$ or not.

Answer to Problem 1PCT

No, the given ordered pair $\left(-4,-4\right)$ is not a solution of the given system of linear equations.

Explanation of Solution

The given ordered pair is $\left(-4,-4\right)$.

Substitute $x=-4$ and $y=-4$ in equation (1).

$\begin{array}{l}4x+y=-20\\ 4\left(-4\right)+\left(-4\right)\stackrel{?}{=}-20\\ -16-4\stackrel{?}{=}-20\\ -20=-20\text{True}\end{array}$

Here, the ordered pair $\left(-4,-4\right)$ satisfies the first equation.

Substitute $x=-4$ and $y=-4$ in equation (2).

$\begin{array}{l}y=-\frac{1}{6}x+3\\ \left(-4\right)\stackrel{?}{=}-\frac{1}{6}\left(-4\right)+3\\ -4\stackrel{?}{=}\frac{2}{3}+3\\ -4=\frac{11}{3}\text{False}\end{array}$

Here, the ordered pair $\left(-4,-4\right)$ does not satisfy the second equation.

Since the ordered pair fails to satisfy both the equations, the given ordered pair $\left(-4,-4\right)$ is not a solution of the system of linear equation.