Which System Is Equivalent To $\[
\left\{
\begin{array}{l}
y=-2x^2 \\
y=x-2
\end{array}
\right.
\\]?A. $\[
\left\{
\begin{array}{l}
y=-2y^2-2 \\
x=y-2
\end{array}
\right.
\\]B. $\[
\left\{
\begin{array}{l}
y=-2y^2+2
by ADMIN216 views
Introduction
In mathematics, a system of equations is a set of two or more equations that are solved simultaneously to find the values of the variables. Solving systems of equations is a fundamental concept in algebra and is used to model real-world problems. In this article, we will explore how to solve systems of equations and determine which system is equivalent to a given set of equations.
What is a System of Equations?
A system of equations is a set of two or more equations that are solved simultaneously to find the values of the variables. Each equation in the system is a statement that two expressions are equal. For example, the system of equations:
{y=β2x2y=xβ2β
consists of two equations: y=β2x2 and y=xβ2. The goal is to find the values of x and y that satisfy both equations simultaneously.
Methods for Solving Systems of Equations
There are several methods for solving systems of equations, including:
Substitution Method: This method involves solving one equation for one variable and then substituting that expression into the other equation.
Elimination Method: This method involves adding or subtracting the equations to eliminate one variable.
Graphical Method: This method involves graphing the equations on a coordinate plane and finding the point of intersection.
Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. For example, let's use the system of equations:
{y=β2x2y=xβ2β
To solve this system using the substitution method, we can solve the first equation for y:
y=β2x2
Then, we can substitute this expression for y into the second equation:
xβ2=β2x2
Simplifying this equation, we get:
2x2+xβ2=0
This is a quadratic equation, which can be solved using the quadratic formula:
x=2aβbΒ±b2β4acββ
In this case, a=2, b=1, and c=β2. Plugging these values into the quadratic formula, we get:
x=4β1Β±1+16ββ
Simplifying this expression, we get:
x=4β1Β±17ββ
Now that we have found the value of x, we can substitute this value back into one of the original equations to find the value of y. Let's use the first equation:
y=β2x2
Substituting x=4β1Β±17ββ into this equation, we get:
y=β2(4β1Β±17ββ)2
Simplifying this expression, we get:
y=81β17ββ
Therefore, the solution to the system of equations is:
x=4β1Β±17ββ
y=81β17ββ
Elimination Method
The elimination method involves adding or subtracting the equations to eliminate one variable. For example, let's use the system of equations:
{y=β2x2y=xβ2β
To solve this system using the elimination method, we can add the two equations together:
β2x2+xβ2=0
Simplifying this equation, we get:
2x2βx+2=0
This is a quadratic equation, which can be solved using the quadratic formula:
x=2aβbΒ±b2β4acββ
In this case, a=2, b=β1, and c=2. Plugging these values into the quadratic formula, we get:
x=41Β±1β16ββ
Simplifying this expression, we get:
x=41Β±β15ββ
Since the square root of a negative number is not a real number, this equation has no real solutions.
Graphical Method
The graphical method involves graphing the equations on a coordinate plane and finding the point of intersection. For example, let's use the system of equations:
{y=β2x2y=xβ2β
To graph these equations, we can use a graphing calculator or a computer program. The graph of the first equation is a parabola that opens downward, while the graph of the second equation is a line.
The point of intersection of these two graphs is the solution to the system of equations. To find this point, we can set the two equations equal to each other:
β2x2=xβ2
Simplifying this equation, we get:
2x2+xβ2=0
This is a quadratic equation, which can be solved using the quadratic formula:
x=2aβbΒ±b2β4acββ
In this case, a=2, b=1, and c=β2. Plugging these values into the quadratic formula, we get:
x=4β1Β±1+16ββ
Simplifying this expression, we get:
x=4β1Β±17ββ
Now that we have found the value of x, we can substitute this value back into one of the original equations to find the value of y. Let's use the first equation:
y=β2x2
Substituting x=4β1Β±17ββ into this equation, we get:
y=β2(4β1Β±17ββ)2
Simplifying this expression, we get:
y=81β17ββ
Therefore, the solution to the system of equations is:
x=4β1Β±17ββ
y=81β17ββ
Which System is Equivalent?
Now that we have solved the system of equations, we can determine which system is equivalent. Let's compare the two systems:
A. ${
\left{
\begin{array}{l}
y=-2y^2-2 \
x=y-2
\end{array}
\right.
}$
B. ${
\left{
\begin{array}{l}
y=-2y^2+2
\end{array}
\right.
}$
To determine which system is equivalent, we can substitute the expression for y from the first equation into the second equation:
x=yβ2
x=(β2y2β2)β2
Simplifying this expression, we get:
x=β2y2β4
Now, we can substitute this expression for x into the first equation:
y=β2y2β2
β2y2β4=β2y2β2
Simplifying this equation, we get:
β4=β2
This equation is not true, so the system A is not equivalent to the original system.
To determine which system is equivalent, we can substitute the expression for y from the first equation into the second equation:
x=yβ2
x=(β2y2+2)β2
Simplifying this expression, we get:
x=β2y2
Now, we can substitute this expression for x into the first equation:
y=β2y2+2
β2y2=β2y2+2
Simplifying this equation, we get:
0=2
This equation is not true, so the system B is not equivalent to the original system.
However, if we substitute the expression for y from the first equation into the second equation:
x=yβ2
x=(β2y2+2)β2
Simplifying this expression, we get:
x=β2y2
Now, we can substitute this expression for x into the first equation:
y=β2y2+2
β2y2=β2y2+2
Simplifying this equation, we get:
0=2
This equation is not true, so the system B is not equivalent to the original system.
However, if we substitute the expression for y from the first equation into the second equation: