Problem 95 - Algebra and Trigonometry

Question

Solve the system: $$ \left\\{\begin{array}{rr} {x^{2}-2 y^{2}=} & {-1} \\ {2 x^{2}-y^{2}=} & {1} \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \( (x = 1, y = -1), (x = 1, y = 1), (x = -1, y = -1), (x = -1, y = 1) \)

Step 1: Isolate one variable

Let's isolate \( y^{2} \) in the first equation. This gives us \( y^{2} = (x^{2} + 1)/2 \)

Step 2: Substitute into the second equation

Substitute \( (x^{2} + 1)/2 \) for \( y^{2} \) in the second equation, which gives: \( 2x^{2} - (x^{2} + 1)/2 = 1 \). Simplifying, we get \( 3x^{2}/2 - 1/2 = 1 \).

Step 3: Solve for x

The equation simplifies to \( 3x^{2} = 3 \), which leads to \( x^{2} = 1 \), thus \( x = 1 \) or \( x = -1 \)

Step 4: Solve for y

Substituting \( x = 1 \) into the equation \( y^{2} = (x^{2} + 1)/2 \), we get \( y^{2} = 1 \), resulting in \( y = 1 \) or \( y = -1 \). The same result is obtained for \( x = -1 \)

Step 5: Conclusion

The solution to the system is \( (x = 1, y = -1), (x = 1, y = 1), (x = -1, y = -1), (x = -1, y = 1) \)

Key Concepts

Quadratic Equations

When we talk about quadratic equations, we're dealing with polynomials where the highest power of one or both variables is squared – in this case, an equation of the form \( ax^2 + by^2 + c = 0 \). Quadratic equations can model various phenomena, from the shape of a trajectory of a thrown ball to the design of parabolic dishes used for satellite signals.
To solve these equations in a system, like in the exercise above, it's essential to manage them carefully because they can yield multiple solutions. This occurs because squaring a variable allows for both positive and negative roots. For example, if \( x^2 = 1 \), then \( x \) could be either 1 or -1. This characteristic is crucial when solving systems involving quadratic equations as you should expect more than one answer in most situations.

Substitution Method

The substitution method is a strategy to solve systems of equations wherein you solve one equation for one variable and then substitute that expression into the other equation(s). This method is particularly useful in systems with linear and quadratic equations, as it allows for isolating variables promptly.
In the exercise at hand, we first isolated \( y^2 \) from the first equation: \( y^2 = \frac{x^2 + 1}{2} \). Doing this enables us to express \( y^2 \) solely in terms of \( x \), making the subsequent equations easier to work with. By substituting this expression into the second equation, we can consolidate variables, reducing a system to just one variable.

• Isolate one variable in one equation.
• Substitute this expression into the other equations.
• Solve for the remaining variables.

This process systematically reduces the complexity of solving systems, especially those involving quadratics.

Solution Sets

Solution sets are the collection of all possible solutions to a given system of equations. In systems involving quadratic equations, it's common to have multiple solution sets due to the nature of quadratic roots.
Each solution set represents a unique pair or set of values that satisfy all equations simultaneously. In our exercise, after solving, we determined four sets: \((1, -1), (1, 1), (-1, -1), (-1, 1)\), meaning there are four distinct pairs of \( x \) and \( y \) values satisfying both equations in the system.
When you find solution sets:

• Check each pair against both equations in the system to verify accuracy.
• Remember that these sets may include complex numbers in other contexts.
• Ensure you've explored all possible permutations of solutions due to squared variables.

Practicing how to determine solution sets swiftly will aid not only in mathematics but in problem-solving and logical reasoning across disciplines.