Chapter 8, Problem 1RE

Determine if the ordered pair is a solution to the system.

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1\end{array}$

a. $\left(1,\frac{2}{3}\right)$

b. $\left(6,4\right)$

To determine

If the following ordered pair is the solution to the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$

a) $\left(1,\frac{2}{3}\right)$

b) $\left(6,4\right)$

Answer to Problem 1RE

a) The ordered pair $\left(1,\frac{2}{3}\right)$ is the solution to the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$

b) The ordered pair $\left(6,4\right)$ is not the solution to the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$

Explanation of Solution

(a).

An ordered pair is the solution to the system of equations, if it satisfies all the equations in that system.

The aim is to check whether the given ordered pair $\left(1,\frac{2}{3}\right)$ satisfies the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$

Consider $2x-3y=0$ .

Put $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ in the above equation.

$\begin{array}{c}2\left(1\right)-3\left(\frac{2}{3}\right)=0\\ 0=0\text{ true}\end{array}$

So, $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ satisfies $2x-3y=0$ .

So, $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ is the solution of $2x-3y=0$ .

Consider $-5x+6y=-1$ .

Put $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ in the above equation.

$\begin{array}{c}-5\left(1\right)+6\left(\frac{2}{3}\right)=-1\\ -1=-1\text{ true}\end{array}$

So, $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ satisfies $-5x+6y=-1$ .

So, $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ is the solution of $-5x+6y=-1$ .

As $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ satisfies both the equations in the given system, the ordered pair, $\left(x,y\right)=\left(1,\frac{2}{3}\right)$ , is the solution to the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$

(b)

An ordered pair is the solution to the system of equations if it satisfies all the equations in that system.

The aim is to check whether the given ordered pair $\left(6,4\right)$ satisfies the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$

Consider $2x-3y=0$ .

Put $\left(x,y\right)=\left(6,4\right)$ in the above equation.

$\begin{array}{c}2\left(6\right)-3\left(4\right)=0\\ 0=0\text{ true}\end{array}$

So, $\left(x,y\right)=\left(6,4\right)$ satisfies $2x-3y=0$ .

So, $\left(x,y\right)=\left(6,4\right)$ is the solution of $2x-3y=0$ .

Consider $-5x+6y=-1$ .

Put $\left(x,y\right)=\left(6,4\right)$ in the above equation.

$\begin{array}{c}-5\left(6\right)+6\left(4\right)=-1\\ -6=-1\text{ false}\end{array}$

So, $\left(x,y\right)=\left(6,4\right)$ does not satisfy $-5x+6y=-1$ .

So, $\left(x,y\right)=\left(6,4\right)$ is not the solution of $-5x+6y=-1$ .

As $\left(x,y\right)=\left(6,4\right)$ does not satisfy $-5x+6y=-1$ , the ordered pair, $\left(6,4\right)$ , is not the solution to the system,

$\begin{array}{c}2x-3y=0\\ -5x+6y=-1.\end{array}$