Problem 24 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} x=-y^{2}-3 \\ x=y^{2}-5 \end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are \((-4, 1)\) and \((-4, -1)\).

Step 1: Set up the equations

We have the system of equations: \( x = -y^2 - 3 \) and \( x = y^2 - 5 \). Our objective is to find values of \( x \) and \( y \) that satisfy both equations simultaneously.

Step 2: Equate the two expressions for x

Since both equations set \( x \) equal to expressions involving \( y \), we can equate these expressions: \( -y^2 - 3 = y^2 - 5 \).

Step 3: Simplify the equation

Add \( y^2 \) to both sides to get \( -3 = 2y^2 - 5 \).

Step 4: Solve for y^2

Add 5 to both sides of the equation: \( 2 = 2y^2 \). Divide both sides by 2 to get \( y^2 = 1 \).

Step 5: Find the values of y

Take the square root of both sides: \( y = 1 \) or \( y = -1 \).

Step 6: Substitute y back to find x

Substitute \( y = 1 \) into the first equation: \( x = -(1)^2 - 3 = -4 \). Now, substitute \( y = -1 \) into the first equation: \( x = -(-1)^2 - 3 = -4 \).

Step 7: Verify solutions

Substitute both \( (x, y) = (-4, 1) \) and \( (x, y) = (-4, -1) \) into the second equation to verify: both will satisfy \( x = y^2 - 5 \) as \( -4 = 1 - 5 \). Hence, both solutions are correct.

Key Concepts

Solving Equations

When dealing with a nonlinear system of equations, our main task is to find values for the variables that make each equation true simultaneously. In this exercise, we have two equations where \( x \) is expressed in terms of \( y^2 \). This means that both equations are linked together through their relationship with \( x \). To start solving them, we can use equivalent expressions. Since both are set equal to \( x \), we can equate the right-hand sides of the equations:

• First equation: \( x = -y^2 - 3 \)
• Second equation: \( x = y^2 - 5 \)

To find a solution, we set \( -y^2 - 3 \) equal to \( y^2 - 5 \). This step is crucial as it simplifies the problem into a single equation with just one variable, \( y \). From here, solving becomes much more straightforward.

Algebraic Manipulation

Once the equations are equated, algebraic manipulation begins. The equation we have now is:\[ -y^2 - 3 = y^2 - 5 \]The standard approach is to move all terms involving \( y \) to one side and constant terms to the other side. Start by adding \( y^2 \) to both sides to eliminate it from one side:\[ -3 = 2y^2 - 5 \]Next, rearrange to isolate \( y^2 \). This involves adding 5 to both sides, thus:\[ 2 = 2y^2 \]Finally, divide both sides by 2 to solve for \( y^2 \):\[ y^2 = 1 \]Now, take the square root of both sides to find the values of \( y \). Consequently, \( y \) can be \( 1 \) or \( -1 \). These steps highlight the power of manipulating equations to simplify and solve complex problems.

Solution Verification

Once potential solutions for \( y \) are found, it’s time to verify and find the corresponding \( x \) values, ensuring they satisfy both original equations. For each \( y \) value:

• Substitute \( y = 1 \) in the first equation:
\( x = -(1)^2 - 3 = -4 \)
• Substitute \( y = -1 \):
\( x = -(-1)^2 - 3 = -4 \)

Both give \( x = -4 \). Next, verify both solutions \((x, y) = (-4, 1) \) and \((x, y) = (-4, -1) \) in the second equation:For \( y = 1 \): \( x = 1 - 5 = -4 \), and for \( y = -1 \): \( x = 1 - 5 = -4 \).Each satisfies \( x = y^2 - 5 \). This verification step confirms that both pairs \((x, y) \) are indeed the correct solutions, demonstrating the importance of checking our work to ensure accuracy.