Chapter 4, Problem 328RE

In the following exercises, determine if the following points are solutions to the given system of equations.

328. $\{\begin{array}{c}x+3y=-9\\ 2x-4y=12\end{array}$

(a) $\left(-3,-2\right)$

(b) $\left(0,-3\right)$

To determine

(a)

If the given point is a solution of given system of equations.

Answer to Problem 328RE

Point $(-3,-2)$ is not a solution of given system of equation.

Explanation of Solution

Given information:

A system of equation is given as $\begin{array}{l}x+3y=-9\\ 2x-4y=12\end{array}$.

A point is given as $(-3,-2)$.

Calculation:

As we know that a system of equation is consists two or more individual equation clubbed together.

The solution to the system of linear inequalities is the set of points whose coordinates satisfy all the equation in the system.

We have been given a system of equation as $\begin{array}{l}x+3y=-9\\ 2x-4y=12\end{array}$.

Also a point is given as $(-3,-2)$. i.e. $x=-3,y=-2$.

Let us take first equation from the given system of equation.

Putting $x=-3,y=-2$ in first equation,

$\begin{array}{l}LHS=x+3y\\ =-3+3\times (-2)\\ =-3-6\\ =-9=RHS\end{array}$

Let us take first equation from the given system of equation.

Putting $x=-3,y=-2$ in second equation,

$\begin{array}{l}LHS=2x-4y\\ =2\times (-3)-4\times (-2)\\ =-6+8\\ =2\ne RHS\end{array}$

Since, point $(-3,-2)$ do not satisfies both equation of given system of equation.

Therefore, we can say that point $(-3,-2)$ is not a solution of given system of equation.

To determine

(b)

If the given point is a solution of given system of equations.

Answer to Problem 328RE

Point $(0,-3)$ is a solution of given system of equation.

Explanation of Solution

Given information:

A system of equation is given as $\begin{array}{l}x+3y=-9\\ 2x-4y=12\end{array}$.

A point is given as $(0,-3)$.

Calculation:

As we know that a system of equation is consists two or more individual equation clubbed together.

The solution to the system of linear inequalities is the set of points whose coordinates satisfy all the equation in the system.

We have been given a system of equation as $\begin{array}{l}x+3y=-9\\ 2x-4y=12\end{array}$.

Also a point is given as $(0,-3)$. i.e. $x=0,y=-3$.

Let us take first equation from the given system of equation.

Putting $x=0,y=-3$ in first equation,

$\begin{array}{l}LHS=x+3y\\ =0+3\times (-3)\\ =-9=RHS\end{array}$

Let us take first equation from the given system of equation.

Putting $x=0,y=-3$ in second equation,

$\begin{array}{l}LHS=2x-4y\\ =2\times 0-4\times (-3)\\ =12\\ =2=RHS\end{array}$

Since, point $(0,-3)$ satisfies both equation of given system of equation.

Therefore, we can say that point $(0,-3)$ is a solution of given system of equation.