Problem 72 - Algebra 2

Question

Find the exact solution(s) of each system of equations. \(x^{2}+2 y^{2}=33\) \(x^{2}-y^{2}=9\)

Step-by-Step Solution

Answer

\((x, y) = (\pm\sqrt{17}, \pm 2\sqrt{2})\).

Step 1: Rewrite the system of equations

We have the system:Equation 1: \( x^2 + 2y^2 = 33 \)Equation 2: \( x^2 - y^2 = 9 \). We will use these equations to eliminate and solve for one variable.

Step 2: Eliminate one variable

Subtract Equation 2 from Equation 1 to eliminate \( x^2 \): \[ (x^2 + 2y^2) - (x^2 - y^2) = 33 - 9 \] This simplifies to \( 3y^2 = 24 \).

Step 3: Solve for y

From \( 3y^2 = 24 \), divide both sides by 3: \[ y^2 = 8 \] Take the square root of both sides to find \( y \): \( y = \pm \sqrt{8} = \pm 2\sqrt{2} \).

Step 4: Solve for x using y-values

Substitute \( y = 2\sqrt{2} \) and \( y = -2\sqrt{2} \) into Equation 2 to solve for \( x^2 \):For \( y = 2\sqrt{2} \),\[ x^2 - (2\sqrt{2})^2 = 9 \] \[ x^2 - 8 = 9 \] \[ x^2 = 17 \rightarrow x = \pm \sqrt{17} \] For \( y = -2\sqrt{2} \),\[ x^2 - (-2\sqrt{2})^2 = 9 \] \[ x^2 - 8 = 9 \] \[ x^2 = 17 \rightarrow x = \pm \sqrt{17} \].

Step 5: Write the solution pairs

The solutions are pairs of \( (x, y) \) values that satisfy both equations. From our computations:1. \( (x, y) = (\sqrt{17}, 2\sqrt{2}) \)2. \( (x, y) = (\sqrt{17}, -2\sqrt{2}) \)3. \( (x, y) = (-\sqrt{17}, 2\sqrt{2}) \)4. \( (x, y) = (-\sqrt{17}, -2\sqrt{2}) \).Each of these pairs is a valid solution.

Key Concepts

quadratic equations

Quadratic equations are fundamental in algebra and are expressed in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The key component, the variable squared, is what makes it quadratic. In this exercise, you see quadratics in a system with two equations. Each equation involves squared terms, such as \( x^2 \) and \( y^2 \). This format allows for unique problems where we solve not just for one variable but two, as is the case here. Solving quadratic equations often involves factoring, using the quadratic formula, or another method, like substitution or elimination, as we'll discuss next.

elimination method

The elimination method is a strategic approach used to solve systems of equations, especially when they involve quadratics. The goal is to remove one of the variables from the system by combining the equations in a specific way. In our problem, we subtract the second equation from the first:

• Equation 1: \( x^2 + 2y^2 = 33 \)
• Equation 2: \( x^2 - y^2 = 9 \)

By subtracting Equation 2 from Equation 1, you eliminate \( x^2 \) and obtain a simpler equation \( 3y^2 = 24 \). This is powerful because it reduces the complexity of solving for one variable at a time. With one variable eliminated, solving for the other becomes quite straightforward.

solution pairs

In systems of equations like these, the goal is to find solution pairs \((x, y)\) that satisfy both equations simultaneously. Once you've solved for \( y = \pm 2\sqrt{2} \), these values are substituted back into one of the original equations to find the corresponding \( x \) values. Each \( y \) generates two \( x \) values, \( \sqrt{17} \) and \( -\sqrt{17} \), leading to four sets of solutions in total:

• \((\sqrt{17}, 2\sqrt{2})\)
• \((\sqrt{17}, -2\sqrt{2})\)
• \((-\sqrt{17}, 2\sqrt{2})\)
• \((-\sqrt{17}, -2\sqrt{2})\)

These solution pairs depict all the values that make both equations true at the same time, showcasing the interconnected nature of these quadratic equations within a system.

square roots

Taking square roots is a common step in dealing with equations involving squared terms. After simplifying the equation \( y^2 = 8 \), finding \( y \) involves taking the square root of both sides, resulting in \( y = \pm 2\sqrt{2} \). The \( \pm \) symbol indicates that both the positive and negative roots are valid. Similarly, this concept applies when determining \( x \) from \( x^2 = 17 \), leading to solutions \( x = \pm \sqrt{17} \). Understanding square roots is essential for accurately solving such problems, as it involves recognizing that each squared term yields two possible values (one positive and one negative), which influence the final solution pairs.