Problem 14 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{r} 2 x^{2}-y^{2}=7 \\ 2 y^{2}-3 x^{2}=2 \end{array}$$

Step-by-Step Solution

Answer

The solutions for the given system of equations are: Solution 1: \(x = 4\), \(y = 5\) Solution 2: \(x = 4\), \(y = -5\) Solution 3: \(x = -4\), \(y = 5\) Solution 4: \(x = -4\), \(y = -5\)

Step 1: Solve the first equation for y

First, let's solve the first equation for y in terms of x: \(2x^2 - y^2 = 7\) Add \(y^2\) to both sides: \(2x^2 = y^2 + 7\) Subtract 7 from both sides: \(y^2 = 2x^2 - 7\)

Step 2: Substitute the expression for y into the second equation

Next, substitute the expression \(y^2 = 2x^2 - 7\) into the second equation: \(2(2x^2 - 7) - 3x^2 = 2\)

Step 3: Solve the equation for x

Now, we will solve the equation for x: \(4x^2 - 14 - 3x^2 = 2\) Combine like terms: \(x^2 - 14 = 2\) Add 14 to both sides: \(x^2 = 16\) Take the square root of both sides: \(x = \pm 4\)

Step 4: Find the corresponding values of y

Now that we have the values for x, we can find the corresponding values for y using the expression we found in Step 1: When \(x = 4\), \(y^2 = 2(4^2) - 7 = 2(16) - 7 = 32 - 7 = 25 \Rightarrow y = \pm 5\) When \(x = -4\), \(y^2 = 2(-4)^2 - 7 = 2(16) - 7 = 32 - 7 = 25 \Rightarrow y = \pm 5\)

Step 5: Write down the solutions

Now we can write down the solutions for the given system of equations: Solution 1: \(x = 4\), \(y = 5\) Solution 2: \(x = 4\), \(y = -5\) Solution 3: \(x = -4\), \(y = 5\) Solution 4: \(x = -4\), \(y = -5\)

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions where the highest exponent of the variable is 2. These equations occur frequently in systems of equations like the one we're solving here. A standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). In our exercise, 'quadratic equations' appear as expressions such as \( 2x^2 - y^2 = 7 \) and \( 2y^2 - 3x^2 = 2 \).
Quadratic equations can have real or complex roots, and they can be solved using various methods like factoring, completing the square, and the quadratic formula. In this exercise, we use algebraic substitution and then solve for variables systematically to find the solution.

Algebraic Substitution

Algebraic substitution is a method used to solve systems of equations, where one variable is replaced with an expression derived from another equation in the system. This method helps simplify complex equations, making them easier to solve.
In our example, we first solved the first equation for \( y^2 \), obtaining the expression \( y^2 = 2x^2 - 7 \). This expression was then substituted into the second equation: \( 2y^2 - 3x^2 = 2 \). By doing this, we reduce the number of variables in one of the equations, leaving us with a single variable to solve for, which, in this case, is \( x \).

• Obtain an expression for one variable from one equation.
• Substitute this expression into another equation in the system.
• Solve the resulting equation.

This technique is a powerful tool for dealing with multiple equations and is particularly useful in problems involving quadratic relationships.

Equation Solving

Solving equations involves finding the value(s) of the variable(s) that satisfy the equation. In this exercise, we make use of the expression obtained from algebraic substitution: \( y^2 = 2x^2 - 7 \).
Once substituted into the second equation, it simplifies to a single-variable quadratic equation: \( 4x^2 - 14 - 3x^2 = 2 \). By combining like terms and isolating \( x^2 \), we ended up with \( x^2 = 16 \). This step is crucial as it takes us closer to finding the values of \( x \).
The process of solving involves:

• Rewriting and simplifying the equation as necessary.
• Using algebraic operations to isolate the variable.
• Verifying the solutions satisfy the original equations.

This discipline helps ensure accurate solutions, especially when dealing with complex systems.

Square Roots

Square roots, one of the basic operations in algebra, help us solve equations involving squares of variables. When we have an equation like \( x^2 = 16 \), taking the square root on both sides helps determine the values for \( x \).
It's important to remember that every positive number has two square roots: a positive and a negative one. Therefore, when solving \( x^2 = 16 \), the solutions are \( x = 4 \) and \( x = -4 \).
This concept is pivotal when dealing with quadratic equations:

• Square roots allow us to find variable values from squared expressions.
• They emphasize symmetrical solutions (positive and negative).

Always ensure, after calculating the square roots, that your answers satisfy the conditions set forth in your original equations, confirming their validity.