Problem 32 - College Algebra

Question

For the following exercises, use any method to solve the nonlinear system. $$\begin{aligned} 16 x^{2}-9 y^{2}+144 &=0 \\ y^{2}+x^{2} &=16 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \((0, 4)\) and \((0, -4)\).

Step 1: Identify the Type of Equations

The first equation is a form of a hyperbola given by \(16x^2 - 9y^2 + 144 = 0\). The second equation is a circle with equation \(y^2 + x^2 = 16\).

Step 2: Simplify the First Equation

Rearrange the first equation \(16x^2 - 9y^2 + 144 = 0\) to \(16x^2 - 9y^2 = -144\). Divide the whole equation by 144 to get \(\frac{x^2}{36} - \frac{y^2}{16} = -1\). This confirms it as a hyperbola equation.

Step 3: Express x in Terms of y from the Circle Equation

The equation of the circle is \(y^2 + x^2 = 16\). Rearrange to express \(x^2\) as \(x^2 = 16 - y^2\).

Step 4: Substitute into the Hyperbola Equation

Substitute \(x^2 = 16 - y^2\) into \(16x^2 - 9y^2 = -144\), yielding \(16(16 - y^2) - 9y^2 = -144\).

Step 5: Simplify the Expression

The equation becomes \(256 - 16y^2 - 9y^2 = -144\). Combine like terms to get \(-25y^2 = -400\).

Step 6: Solve for y

Solve \(-25y^2 = -400\) by dividing both sides by \(-25\), resulting in \(y^2 = 16\). Taking the square root gives \(y = 4\) or \(y = -4\).

Step 7: Find Corresponding x Values

Substitute \(y = 4\) into \(x^2 = 16 - y^2\), resulting in \(x^2 = 0\). Thus, \(x = 0\). Do the same for \(y = -4\), also resulting in \(x = 0\).

Step 8: Verify Solutions

Verify the solutions \((0, 4)\) and \((0, -4)\) satisfy both original equations. Substituting into both equations confirms that both solutions satisfy the hyperbola and the circle equations.

Key Concepts

Hyperbola Equations

A hyperbola is a type of conic section that can be defined by its standard equation \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] in its simplified form. In our exercise, the first equation given is \( 16x^2 - 9y^2 + 144 = 0 \), which is a hyperbola.
To recognize a hyperbola equation:

• Look for the subtraction of squared terms.
• Involve coefficients leading to a difference, reflecting the 'hyperbolic' nature.

Hyperbolas consist of two separate curves opening either horizontally or vertically, depending on the terms in the equation. In this case, by simplifying to \( \frac{x^2}{36} - \frac{y^2}{16} = -1 \), we confirm it represents a hyperbola.

Circle Equations

Equations of a circle are simplified to \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius.
The equation in the problem \( y^2 + x^2 = 16 \) indicates a circle centered at the origin \((0, 0)\) with a radius of 4 (since \( 16 = 4^2 \)).

• The equation is already in the standard form, making it easy to identify these parameters.
• Equations describing circles have the feature of having equal coefficients for \(x^2\) and \(y^2\) with a positive sum involved.

Substitution Method

The substitution method is particularly useful for solving systems of nonlinear equations. It involves isolating one variable in one equation and substituting it into another.
In the exercise, we isolate \(x^2\) from the circle equation \( y^2 + x^2 = 16 \) and get \( x^2 = 16 - y^2 \). This expression for \(x^2\) can then be substituted into the hyperbola equation, \( 16x^2 - 9y^2 = -144 \).

• It reduces the system to one equation in terms of one variable.
• Allows for easier solving and elimination of one variable.

Solution Verification

After finding potential solutions, it's crucial to verify them within the original equations to ensure they're correct. For the points \((0, 4)\) and \((0, -4)\):

• Substitute back into the hyperbola \(16x^2 - 9y^2 + 144 = 0\).
• Check the circle equation \(y^2 + x^2 = 16\).

This dual process ensures no miscalculations obscure the solution.
Verification ensures that both solutions not only mathematically satisfy but make logical sense considering both graphs.

Algebraic Manipulation

Critical to solving equations, especially in systems involving multiple variables, is skillful algebraic manipulation. In this task:

• Rearrange the equation \(16x^2 - 9y^2 + 144 = 0\) to its canonical form \(\frac{x^2}{36} - \frac{y^2}{16} = -1\) by dividing the entire equation by 144.
• Extract \(x^2\) from the circle as \(x^2 = 16 - y^2\).

These steps not only simplify your equations but make subsequent solutions more straightforward.
Algebraic manipulation supports solving complex nonlinear systems by rearranging them into more manageable forms.