Problem 22 - Precalculus: Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l} y=4-x^{2} \\ y=x^{2}-4 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are \((2, 0)\) and \((-2, 0)\).

Step 1: Set Equations Equal

In this step, since both equations are equal to \(y\), equate them: \[4 - x^2 = x^2 - 4\] Simplifying, we get: \[4 + 4 = x^2 + x^2\] \[8 = 2x^2\] Divide by 2: \[4 = x^2\]

Step 2: Solve for x

Continue solving for \(x\) from \(4 = x^2\):Take the square root of both sides:\[x = \pm 2\] This gives two possible values for \(x\): \(x = 2\) and \(x = -2\).

Step 3: Solve for y When x = 2

Use \(x = 2\) in either original equation to find \(y\). Using the equation \(y = 4 - x^2\): \[y = 4 - (2)^2 = 4 - 4 = 0\] Thus, when \(x = 2\), \(y = 0\).

Step 4: Solve for y When x = -2

Use \(x = -2\) in either original equation to find \(y\). Using the same equation \(y = 4 - x^2\): \[y = 4 - (-2)^2 = 4 - 4 = 0\] Thus, when \(x = -2\), \(y = 0\).

Step 5: Verify Solutions

Verify that both solutions satisfy the second equation \(y = x^2 - 4\):\((x,y) = (2, 0)\): \[0 = 2^2 - 4 = 4 - 4 = 0\] \((x,y) = (-2, 0)\): \[0 = (-2)^2 - 4 = 4 - 4 = 0\] Both solutions satisfy both equations.

Key Concepts

Solving Systems

When you come across a set of equations listed together, like those in the exercise above, it means you're dealing with a system of equations. Systems of equations can have one solution, many solutions, or sometimes no solution at all. Here, the goal is to find values for the variables, usually denoted as \(x\) and \(y\), that will satisfy all given equations simultaneously.

Think of it like finding a common bridge or link between two different paths. The solution to a system provides the point or points where these paths intersect. Solving systems can be done using several methods, including substitution, elimination, and graphing.

• Substitution: This involves replacing one of the variables in one equation with an expression from another, as will be demonstrated in more detail in a moment.
• Elimination: By adding or subtracting equations, you can eliminate one variable to solve for the other.
• Graphing: This is useful to visually see where the equations may intersect on a graph.

In our current exercise, we focus on substitution since both equations are already solved for \(y\), making it a straightforward choice.

Quadratic Equations

Quadratic equations are a special type of equation that involve the square of a variable. A typical quadratic equation looks like this: \(ax^2 + bx + c = 0\). The key feature is the \(x^2\) term, which makes the graph of the equation a parabola.

In the given exercise, we have two quadratic equations set up in a system:

• \(y = 4 - x^2\)
• \(y = x^2 - 4\)

Both equations express \(y\) in terms of \(x^2\). The parabolas formed by these equations open in opposite directions (one opens downward and the other upward). The exercise involves finding where these parabolas intersect.

Quadratic equations can be solved by:

• Factoring
• Using the quadratic formula
• Completing the square

In our context, solving involved equating the two expressions for \(y\) to find values for \(x\). This step simplifies the problem to a more basic form, allowing for straightforward solutions through basic algebra.

Substitution Method

The substitution method is particularly useful for solving systems of equations when one variable is already isolated. The core idea is to substitute one equation into the other, reducing the number of variables and making it simpler to solve.

In our example, both equations express \(y\) in terms of \(x\). This makes substitution a natural choice. Since both equations equal \(y\), you set them equal to each other, like this: \[4 - x^2 = x^2 - 4\].

This step effectively eliminates \(y\) from the equations, allowing us to focus solely on \(x\). This transforms our system into a basic quadratic equation, \(8 = 2x^2\), that can be solved easily for \(x\). Once we find \(x\)-values, substitute them back into either original equation to solve for \(y\).

Benefits of the substitution method include:

• Simplifies complex problems by reducing the number of variables
• When set up correctly, it can provide solutions clearly and quickly

The substitution method is a powerful tool in the toolbox for solving systems of equations, helping to bridge simple concepts and complex solutions seamlessly.