Chapter 11, Problem 1CRT

a.

To determine

To state: that whether the system linear or nonlinear.

Answer to Problem 1CRT

The system is nonlinear.

Explanation of Solution

Given information:

The system is $\left\{\begin{array}{l}{x}^2+y^2=4y\\ x^2-2y=0\end{array}\right\}$

the equation system is nonlinear because the equations graph are different.

b.

To determine

To find: the all solutions of the system.

Answer to Problem 1CRT

The solutions of the system is $\left(0,0\right),\left(2,2\right),\left(-2,2\right)$ .

Explanation of Solution

Given information:

The system is $\left\{\begin{array}{l}{x}^2+y^2=4y\\ x^2-2y=0\end{array}\right\}$

Calculation: to find the solutions of the system,

  $\begin{array}{l}{x}^2+y^2=4y\text{ [eq}\text{.1]}\\ x^2-2y=0\text{ [eq}\text{.2]}\\ 2y+y^2=4y\text{ [substitute from eq}\text{.]}\\ y^2-2y=0\text{ [simplify]}\\ y\left(y-2\right)=0\text{ [factorise]}\\ y=0,2\\ x^2-2\left(0\right)=0\text{ [substitute }y=0]\\ x=0\\ x^2-2\left(2\right)=0\text{ [substitute }y=2]\\ x=\pm 2\text{ [simplify]}\end{array}$

Thus, the all solution of the system is $\left(0,0\right),\left(2,2\right),\left(-2,2\right)$ .

c.

To determine

To state: the type of conic section.

Answer to Problem 1CRT

The first equation is a circle and another equation is parabola.

Explanation of Solution

Given information:

The system is $\left\{\begin{array}{l}{x}^2+y^2=4y\\ x^2-2y=0\end{array}\right\}$

from the above graph it can be observed that the first equation makes a circle and the another one makes the parabola.

d.

To determine

To graph: the both equation.

Explanation of Solution

Given information:

The system is $\left\{\begin{array}{l}{x}^2+y^2=4y\\ x^2-2y=0\end{array}\right\}$

graph: to draw the graph, first use the plotting point method,

$x$	$x^2+y^2=4y$	$x^2-2y=0$
$0$	$0$	$0$
$2$	$2$	$2$
$-2$	$2$	$2$

By using the above table and join the points,

The graph can be obtained as:

  

Interpretation: from the above graph it can be observed that the first equation graph is circle and another one is parabola.

e.

To determine

To graph: the shaded region of the solutions of the system of inequalitites.

Explanation of Solution

Given information:

The system is $\left\{\begin{array}{l}{x}^2+y^2\le 4y\\ x^2-2y\le 0\end{array}\right\}$

graph: to draw the graph, first use the plotting point method,

$x$	$x^2+y^2=4y$	$x^2-2y=0$
$0$	$0$	$0$
$2$	$2$	$2$
$-2$	$2$	$2$

By using the above table and join the points,

The graph can be obtained as:

  

Interpretation: from the above graph it can be observed that the first equation graph is circle and another one is parabola.