Problem 36 - College Algebra

Question

Use any method to solve the nonlinear system. $$ \begin{aligned} x^{2}-y^{2}-6 x-4 y-11 &=0 \\ -x^{2}+y^{2} &=5 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (\pm \sqrt{\frac{44}{5}}, -\frac{3}{2}x - 4)\).

Step 1: Simplify the System

Add the two equations to eliminate the squared terms. \[(x^2 - y^2 - 6x - 4y - 11) + (-x^2 + y^2) = 0 + 5\]This simplifies to: \[-6x - 4y - 11 = 5\]

Step 2: Solve for y

Rearrange the equation from Step 1 to isolate \(y\): \[-4y = 6x + 11 + 5\]\[-4y = 6x + 16\]\[y = -\frac{3}{2}x - 4\]

Step 3: Substitute into the First Equation

Substitute \(y = -\frac{3}{2}x - 4\) into the first equation: \[ x^2 - \left(-\frac{3}{2}x - 4\right)^2 - 6x - 4\left(-\frac{3}{2}x - 4\right) - 11 = 0 \] Expand and simplify: \[ x^2 - \left(\frac{9}{4}x^2 + 12x + 16\right) - 6x + 6x + 16 - 11 = 0 \] Continue simplifying: \[ -\frac{5}{4}x^2 -11 = 0 \]

Step 4: Solve for x

Rearrange the equation: \[ \frac{5}{4}x^2 = 11 \] Multiply by 4/5:\[ x^2 = \frac{44}{5} \] Take the square root of both sides:\[ x = \pm \sqrt{\frac{44}{5}} \]

Step 5: Solve for y

Using the expression \(y = -\frac{3}{2}x - 4\), find \(y\) for each value of \(x\):- If \( x = \sqrt{\frac{44}{5}} \), then \[ y = -\frac{3}{2}\sqrt{\frac{44}{5}} - 4 \] - If \( x = -\sqrt{\frac{44}{5}} \), then \[ y = \frac{3}{2}\sqrt{\frac{44}{5}} - 4 \]

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. An equation in the form \( ax^2 + bx + c = 0 \) is a quadratic equation where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These equations form a parabolic graph, and their solutions can be found using various methods like factoring, completing the square, or applying the quadratic formula.
In the context of nonlinear systems, quadratic equations come in handy for representing complex curves that can intersect with other curves, such as linear equations, at multiple points.

Elimination Method

The elimination method is a key technique used in solving systems of equations. The goal is to eliminate one of the variables, allowing you to solve for the other. This is often achieved by adding or subtracting the equations so that one variable cancels out.

• Start by aligning two equations to target one variable for elimination.
• Add or subtract the equations to remove this variable.
• The resulting equation can be solved like an ordinary equation to find the value of the remaining variable.

The most challenging part is aligning the coefficients for successful elimination, especially in nonlinear systems where squaring and expanding terms might be needed.

Square Root

The square root is a mathematical function that returns a number which, when multiplied by itself, gives the original value. It plays a crucial role in solving quadratic equations.
When solving for the roots of a quadratic equation in the form \( x^2 = d \), you take the square root on both sides to find \( x = \pm \sqrt{d} \). This gives two solutions because both the positive and negative values can be squared to get the same result.
The presence of a square root often introduces the necessity to consider both positive and negative potential solutions, especially when solving systems of equations.

Solution of Equations

The solution of equations involves finding the values for the unknown variables that satisfy all given equations simultaneously. This is particularly interesting in nonlinear systems because the solutions are usually where the graphs of the equations intersect.
Here's a step-by-step outline:

• Formulate each equation clearly.
• Use methods like substitution or elimination to reduce the system down to a simpler form, if necessary.
• Solve these simplified equations to find the values of the unknowns.

In nonlinear systems with polynomials, you often arrive at multiple solutions, reflecting the multiple points at which the curves intersect.