Problem 13 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} &x^{2}+2 y^{2}=11\\\ &x^{2}-y^{2}=8 \end{aligned}$$

Step-by-Step Solution

Answer

The solution set for the given system of equations is: \((x, y) = (3, 1), (-3, 1), (3, -1), (-3, -1)\).

Step 1: Solve one of the equations for x²

We don't need to solve for x or y directly in this case; instead, we will solve for x². From the second equation, we can write it as: \(x^2 = y^2 + 8\)

Step 2: Substitute the expression of x² into the first equation

Now, substitute the expression of x² from the second equation into the first equation: \((y^2 + 8) + 2y^2 = 11\)

Step 3: Solve for y

Combine the terms with y² and solve for y: \(3y^2 + 8 = 11\) \(3y^2 = 3\) \(y^2 = 1\) This gives us two possible values for y: \(y = 1\) or \(y = -1\)

Step 4: Substitute the values of y back into the expression of x²

Now, we need to find the corresponding values of x for each value of y. Substitute each y-value into the expression for x²: When \(y = 1\), \(x^2 = (1)^2 + 8 = 9\) \(x = \pm \sqrt{9} = \pm 3\) When \(y = -1\), \(x^2 = (-1)^2 + 8 = 9\) \(x = \pm \sqrt{9} = \pm 3\)

Step 5: Write the solution set

The solution set consists of all possible pairs of x and y values that satisfy the given system of equations. In this case, the solution set is: \((x, y) = (3, 1), (-3, 1), (3, -1), (-3, -1)\)

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra that arise when dealing with polynomial expressions of degree two. These equations are typically written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.

• The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of \( a \).
• Solving a quadratic can involve factoring, using the quadratic formula, or completing the square.

In our example, we work with the system:\[\begin{aligned}&x^{2}+2y^{2}=11\&x^{2}-y^{2}=8\end{aligned}\]Here, we notice that both equations have terms involving squares, making them a perfect fit for quadratic analysis.

Substitution Method

The substitution method is a widely used technique for solving systems of equations. It involves solving one equation for one variable and then substituting this expression into the other equation.

• It's particularly useful when one equation is easier to solve for one variable.
• Once substituted, the equation will contain only one variable, making it easier to solve.

In our solution, we first solve the second equation for \( x^2 \):\[ x^2 = y^2 + 8 \]We then substitute \( x^2 \) in the first equation:\[ (y^2 + 8) + 2y^2 = 11 \]This reduces the system to a single variable equation, simplifying the problem significantly.

Solutions to Equations

Finding solutions to equations is all about determining the values of variables that satisfy given mathematical statements. In the context of systems of equations, this means finding all possible sets of values for the variables.

• Each solution represents a point where the equations intersect when plotted on a graph.
• In systems of quadratic equations, solutions can be pairs of real numbers or sometimes complex numbers.

For our problem, after performing substitution, we find \( y^2 = 1 \), which provides two potential solutions: \( y = 1 \) and \( y = -1 \). We then derive \( x \) values for each \( y \), leading to the solution set:\[(x, y) = (3, 1), (-3, 1), (3, -1), (-3, -1)\]This showcases how theoretical analysis becomes applicable through algebraic operations to find actual numeric solutions.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations, allowing us to construct behaviors and relationships between different quantities. They are fundamental in creating equations.

• Understanding how to manipulate these expressions is crucial in solving equations and systems of equations.
• Expressions can be simplified, factored, or expanded to assist in solving for unknown values.

In this exercise, we simplified our expressions by combining like terms. By recognizing that similar terms could be combined, we reduced the first equation:\[ 3y^2 + 8 = 11 \]This step illustrates the power of algebraic manipulation to transform complex equations into more easily solvable forms. Recognizing and using algebraic expressions effectively is a key skill in all areas of mathematics.