Problem 34 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the nonlinear system. $$ \begin{aligned} 3 x^{2}-y^{2} &=12 \\ (x-1)^{2}+y^{2} &=4 \end{aligned} $$

Step-by-Step Solution

Answer

Solve the quadratic system by substitution, resulting in the values of \( x \) and \( y \) that satisfy both equations.

Step 1: Identify the System of Equations

We are given two equations to solve as a system: \( 3x^2 - y^2 = 12 \) (Equation 1) and \( (x-1)^2 + y^2 = 4 \) (Equation 2). These are nonlinear equations involving quadratic terms.

Step 2: Rearrange Equation 2

Rewrite Equation 2 by expanding \((x-1)^2\):\[ (x-1)^2 + y^2 = x^2 - 2x + 1 + y^2 = 4 \]This simplifies to:\[ x^2 - 2x + y^2 = 3 \]

Step 3: Express \( x^2 \) from Rearranged Equation 2

From \( x^2 - 2x + y^2 = 3 \), express \( x^2 \):\[ x^2 = 2x + 3 - y^2 \]Substitute \( x^2 \) in Equation 1.

Step 4: Substitute and Simplify in Equation 1

Now replace \( x^2 \) in Equation 1 \( 3x^2 - y^2 = 12 \) with the expression from Step 3:\[ 3(2x + 3 - y^2) - y^2 = 12 \]Simplify the equation:\[ 6x + 9 - 3y^2 - y^2 = 12 \]\[ 6x = 12 - 9 + 4y^2 \]\[ 6x = 3 + 4y^2 \]

Step 5: Solve for \( x \)

Solve the equation for \( x \):\[ 6x = 3 + 4y^2 \]\[ x = \frac{3 + 4y^2}{6} \]\[ x = \frac{1}{2} + \frac{2}{3}y^2 \]

Step 6: Substitute \( x \) in Rearranged Equation 2

Substitute \( x = \frac{1}{2} + \frac{2}{3}y^2 \) back into \( x^2 - 2x + y^2 = 3 \):Calculate \( x^2 = \left(\frac{1}{2} + \frac{2}{3}y^2\right)^2 \) and simplify the resulting equation.

Step 7: Solve for \( y \)

After substituting and simplifying, solve the resulting equation for \( y \). This may result in a quadratic equation, which can be solved using quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 8: Substitute Back to Find \( x \)

Use the value(s) of \( y \) found in Step 7 and substitute them back into \( x = \frac{1}{2} + \frac{2}{3}y^2 \) to find corresponding values of \( x \).

Key Concepts

Solving Quadratic Equations

Quadratic equations are fundamental in algebra and typically have the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations may have two, one, or no real solutions. Here's how we typically solve them:

• Factoring: Express the quadratic in a product of two binomials and solve for the roots. This is often possible when the coefficients are simple and yield integer solutions.
• Completing the square: Modify the equation so that the left-hand side forms a perfect square trinomial. Then solve for the variable by taking the square root of both sides.
• Quadratic formula: Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots. This method is straightforward as it works for all quadratics, even when factoring is challenging.

The approach chosen depends on the specific equation and its coefficients. In the context of solving nonlinear systems, like the ones in our example, solving a quadratic may involve substituting values or expressions derived from another equation.

Systems of Equations

A system of equations consists of two or more equations that share common variables. Solving a system requires finding values for each variable that satisfy all the equations simultaneously. In real-world terms, if each equation is viewed as a line or curve, solving the system equates to finding their intersection points.
Systems of equations can be classified as either:

• Linear systems: All equations involve variables to the first power, forming straight lines.
• Nonlinear systems: At least one equation includes variables raised to a power greater than one, typically forming curves or complex shapes.

Various methods to solve systems include graphing, substitution, elimination, and using matrices. In nonlinear equations, you'll often rely on substitution or graphical methods, given these systems need more careful consideration due to their complexity. The presence of curves adds interesting nuances, requiring specific techniques for an accurate solution.

Substitution Method

The substitution method is an algebraic technique used to solve systems of equations. This method involves solving one of the equations for a variable and then substituting that expression into the other equation(s). It is particularly effective when one equation is easily rearranged.
In practice, the method works as follows:

• Solve one equation for one variable: Rearrange an equation to express one variable in terms of others.
• Substitute the expression into the other equation(s): This replaces the chosen variable with its equivalent expression, reducing the number of variables in the remaining equations.
• Solve for the remaining variable: With only one variable left, you find its value.
• Back-substitute to find other variable(s): Replace back into the first expression to solve for the initial variable(s).

This method simplifies complex systems and is invaluable, especially when dealing with nonlinear equations where direct methods like graphing can be cumbersome.