Problem 19 - Intermediate Algebra

Question

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} 2 x^{2}+3 y^{2} &=14 \\ -x^{2}+y^{2} &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions are \((1, 2), (1, -2), (-1, 2), (-1, -2)\).

Step 1: Solve Equation 2 for One Variable

Start with Equation 2: \(-x^2 + y^2 = 3\).Rearrange it to find \(y^2\) in terms of \(x^2\):\[y^2 = x^2 + 3.\]

Step 2: Substitute into Equation 1

Substitute the expression for \(y^2\) from Step 1 into Equation 1:\[2x^2 + 3(x^2 + 3) = 14.\]Simplify this to obtain:\[2x^2 + 3x^2 + 9 = 14.\]

Step 3: Simplify and Solve for \(x^2\)

Combine like terms and isolate \(x^2\):\[5x^2 + 9 = 14.\]Subtract 9 from both sides:\[5x^2 = 5.\]Divide both sides by 5:\[x^2 = 1.\]

Step 4: Determine the Values of \(x\)

Solve for \(x\) by taking the square root of both sides:\[x = \pm 1.\]

Step 5: Substitute \(x\) Values to Find Corresponding \(y\) Values

For each \(x\) value, use the equation from Step 1:\[y^2 = x^2 + 3.\]Substitute \(x = 1\) to get:\[y^2 = 1 + 3 = 4,\]so \(y = \pm 2.\)Substitute \(x = -1\) to get:\[y^2 = 1 + 3 = 4,\]so \(y = \pm 2.\)

Step 6: List All Possible Solutions

Combine the \(x\) and \(y\) values found:- For \(x = 1\), \(y = 2\) or \(y = -2\), the solutions are \((1, 2)\) and \((1, -2)\).- For \(x = -1\), \(y = 2\) or \(y = -2\), the solutions are \((-1, 2)\) and \((-1, -2)\).

Key Concepts

Understanding Real Solutions

A real solution in the context of a system of equations refers to solutions involving real numbers rather than complex numbers. This means that the solutions have values that you might encounter in day-to-day life, like integers or fractions. In our problem, we aim to find values of \(x\) and \(y\) which satisfy both equations simultaneously with real numbers only.

For example, if an equation's solution was a square root of a negative number, it would be a complex solution involving imaginary numbers. However, our goal here is strictly real solutions, which naturally occur on the familiar number line.

When solving systems of equations, identifying real solutions is crucial:

• They provide meaningful results applicable to real-world situations.
• Real solutions can be directly interpreted without transformations.
• These solutions often offer direct insight into the relationship between variables.

Solving Nonlinear Equations

Solving equations, especially nonlinear ones, involves finding values of variables that satisfy the given relationships. Nonlinear equations can include powers, exponents, or other components that create curves rather than straight lines on graphs.

In our exercise, we have a system of equations combining quadratic expressions, which describe curves. The key to solving such systems is to work methodically:

• Start by simplifying the system by isolating one variable in one of the equations.
• Use algebraic substitution to simplify the process, as it's often easier to handle one equation at a time.
• Once the system is reduced, solve for one variable first, then backtrack to find the complementary variable.

Nonlinear systems can often have multiple solutions due to their curved nature, allowing for more than one intersection point between the plotted equations.

Algebraic Substitution Technique

Algebraic substitution is a powerful tool in solving systems of equations, particularly nonlinear systems. This method involves expressing one variable in terms of another and substituting it into the other equation.

For instance, in our problem, we started with the variable \(y^2\) expressed from the second equation:

• Rearranging \(-x^2 + y^2 = 3\) into \(y^2 = x^2 + 3\).

This expression is then substituted in the first equation to eliminate \(y^2\), simplifying the system:

• The substitution transformed the complex system into a simpler quadratic equation in terms of \(x^2\).

The method eases the process by reducing variables, making it simpler to solve one equation at a time. Thus, this technique allows us to focus on one variable and simplifies our approach to finding real solutions.

Mastering Square Roots in Equations

The concept of square roots is an important part of solving quadratic equations, which are common in nonlinear systems. When we have an equation like \(x^2 = 1\), finding the square root helps us determine the possible values of \(x\).

Calculating square roots is straightforward:

• If \(x^2 = 1\), we take the square root of both sides to find \(x = \pm 1\).

Square roots can yield both positive and negative solutions because squaring either value results in the same positive number. This leads to multiple possible solutions when dealing with quadratic equations.

In the exercise, after substituting and solving for \(x^2\), the square root process reveals both possible values of \(x\).

• Similarly, solving for \(y^2\) translates to \(y = \pm 2\) once the square root is applied.

Understanding and accurately applying square roots is crucial in identifying all potential solutions and their related variable values.