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In Exercises \(1-34\), find all real solutions of the system of equations. If no real solution exists, so state. $$\left\\{\begin{array}{l} x^{2}+y^{2}+4 y=1 \\ x^{2}-y^{2} \quad=3 \end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are \(x = ±\sqrt{5/3}\), \(y = 1/6\)

Step by step solution

01

Combine the Equations

Combining the two equations by adding them together to get rid of \(x^{2}\), which yields: \( x^{2} + y^{2} + 4y + x^{2} - y^{2} = 1 + 3, \) simplifying this gives: \(x^{2} + 4y = 4 \)
02

Solve for y

Next is to solve the equation derived from step 1 for y: \(2x^{2} = 4 - 4y \) divide both sides by 4: \(y = 1 - \frac{x^2}{2}\)
03

Substitute y into the second original equation

Substitute y from step 2 into the second equation: \(x^{2} - (1 - x^{2}/2)^2 = 3)\). Simplify and solve for \(x\), which gives \(x = ±\sqrt{5/3}\)
04

Substitute x back to find y

Substitute x from step 3 back into the equation from step 2 to find the values for y, which gives \(y = 1 - 5/6 = 1/6\) and \(y = 1 - 5/6 = 1/6\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Solutions
The process of finding algebraic solutions to systems of equations involves determining a set of values that satisfy all equations in the system simultaneously. This particular exercise deals with finding real solutions to a set of two quadratic equations. The first step in solving such systems is often to manipulate the equations to eliminate one variable, making it easier to solve for the other. By adding or subtracting the equations, as done in the given solution, one can simplify the system to a more manageable form.

In our textbook example, we combined the equations to eliminate the variable that did not change sign, in this case, the term with x2. This strategic move is essential in reducing the complexity of the equations at hand and easing the path to finding a solution. By understanding the principles of algebraic solutions, students can apply similar steps to solve various types of systems of equations.
Quadratic Equations
Quadratic equations are polynomials that are second-order, meaning they include a variable raised to the power of two, typically in the form ax2 + bx + c = 0, where a, b, and c are constants and x represents the variable. The textbook exercise presents a system with equations that are quadratic in nature due to the presence of x2 and y2.

To solve quadratic equations, one can factor the expression, complete the square, or use the quadratic formula. The example solution, however, takes advantage of the unique system setup to simplify and remove quadratic terms, demonstrating an alternate yet effective approach to finding solutions to quadratic systems.
Substitution Method
The substitution method is a fundamental technique in algebra for solving systems of equations. It involves isolating one variable in one equation and then substituting that expression into the other equation. This process converts the system into a single equation in one variable, which is typically easier to solve.

In our exercise, once the value for y was found in terms of x, it was substituted back into one of the original equations, leading to a solution for x. This substitution method is particularly useful for systems that aren't easily solved by other methods like elimination, and it is a powerful tool when the relationship between the variables can be cleanly expressed, facilitating the final step of finding the solution for the second variable.

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