Problem 34 - Intermediate Algebra

Question

Solve each system of equations by elimination for real values of \(x\) and \(y .\) See Example 4 $$ \left\\{\begin{array}{l} x^{2}+y^{2}=25 \\ 2 x^{2}-3 y^{2}=5 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((4, 3), (-4, 3), (4, -3), (-4, -3)\).

Step 1: Identifying the System

We are given two equations in a system:\[ \begin{align*} x^2 + y^2 &= 25 \ 2x^2 - 3y^2 &= 5 \end{align*} \] We need to eliminate one variable to solve for the real values of \( x \) and \( y \).

Step 2: Multiplying to Facilitate Elimination

To eliminate one variable, we can adjust the equations by multiplying the first equation by 2, hoping to align the \( x^2 \) coefficients:\[ 2(x^2 + y^2) = 2 imes 25 \] Which simplifies to:\[ 2x^2 + 2y^2 = 50 \] This will help us cancel the \( x^2 \) terms when subtracted from the other equation.

Step 3: Subtracting the Equations

Subtract the modified first equation from the second equation to eliminate \( x^2 \):\[ \begin{align*} (2x^2 - 3y^2) - (2x^2 + 2y^2) &= 5 - 50 \ -5y^2 &= -45 \end{align*} \] Simplifying the equation gives us \( y^2 \):\[ y^2 = 9 \]

Step 4: Finding \( y \)

Solve \( y^2 = 9 \) for \( y \):\[ y = 3 \quad \text{or} \quad y = -3 \] We have two possible values for \( y \).

Step 5: Substituting \( y \) Back to Find \( x \)

Substitute \( y = 3 \) back into the first equation:\[ x^2 + 3^2 = 25 \] \[ x^2 + 9 = 25 \] \[ x^2 = 16 \] \[ x = 4 \quad \text{or} \quad x = -4 \]Substitute \( y = -3 \) back into the first equation:\[ x^2 + (-3)^2 = 25 \] \[ x^2 + 9 = 25 \] \[ x^2 = 16 \] \[ x = 4 \quad \text{or} \quad x = -4 \]

Step 6: Solution

We found the solutions for \( x \) and \( y \):\[ (x, y) = (4, 3), (-4, 3), (4, -3), (-4, -3) \] These are the real values for \( x \) and \( y \) that satisfy the system of equations.

Key Concepts

Elimination Method

When we want to solve a system of equations, such as the one given, the elimination method can be a very effective approach. The main goal of this method is to eliminate one variable by adding or subtracting the equations. This way, we can solve for one variable first and then find the others. In our specific problem, we have two equations that include both \( x^2 \) and \( y^2 \). The trick here lies in aligning these terms so that one can be eliminated when the equations are combined.
This often involves multiplying one or both equations by a number to create a situation where either the \( x^2 \) or \( y^2 \) terms have opposite coefficients. By achieving this alignment, you can subtract one equation from the other, effectively canceling out one variable. Here, multiplying the first equation by 2 allowed the \( 2x^2 \) terms to cancel out when subtracted, leaving us with an equation in just \( y^2 \).
Once a single variable is isolated, it becomes straightforward to solve for it, as demonstrated in the step-by-step solution.

Real Values

When dealing with systems of equations, we sometimes specify that solutions need to be real values. This distinguishes them from complex numbers, which have both real and imaginary parts. Real numbers include all the integers, fractions, and numbers with decimal points that you are accustomed to using in daily calculations. In our exercise, finding real values of \( x \) and \( y \) means ensuring that the answers we derive are within this real number set.
During this process, squaring a number can help simplify resolving equations. For instance, when you arrive at an equation such as \( y^2 = 9 \), solving for \( y \) involves finding each number whose square is 9, which are the real numbers \( y = 3 \) or \( y = -3 \). These solutions adhere to the real values requirement as they do not involve any imaginary units.

Quadratic Equations

Quadratic equations frequently pop up when solving systems that involve squared terms like \( x^2 \) or \( y^2 \), as in the given system. A quadratic equation is generally of the form \( ax^2 + bx + c = 0 \), where solving it may yield two solutions. For example, the equation \( x^2 = 16 \) we solved boils down to finding real numbers \( x \) such that its square equals 16.
This boils down to a standard quadratic solution where we recognize \( x = 4 \) and \( x = -4 \) as the solutions. To solve these types of equations, we often take the square root of both sides, remembering to account for both the positive and negative roots of the number.

• Ensure both solutions are checked against the original equations to affirm correctness.
• It's important to revisit the definitions of quadratic terms and recognize potential pitfalls like ignoring the negative solution.

This usually results in producing a pair of values that satisfy the equation, as seen in this problem.