Problem 57 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}+y^{2}=4 \\ 9 x^{2}+y^{2}=9 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \(\left(\pm \sqrt{\frac{5}{8}}, \pm \sqrt{\frac{27}{8}}\right)\).

Step 1: Analyze and Simplify the Equations

The system of equations contains two equations: \( x^2 + y^2 = 4 \) and \( 9x^2 + y^2 = 9 \). Notice that both equations have \( y^2 \), which suggests substitution can simplify the solution process.

Step 2: Subtract the First Equation From the Second

Subtract the first equation from the second equation:\[ 9x^2 + y^2 - (x^2 + y^2) = 9 - 4 \] This simplifies to: \[ 8x^2 = 5 \].

Step 3: Solve for \( x^2 \)

From \( 8x^2 = 5 \), solve for \( x^2 \) by dividing both sides by 8:\[ x^2 = \frac{5}{8} \].

Step 4: Substitute \( x^2 \) into the First Equation

Substitute \( x^2 = \frac{5}{8} \) into the first equation \( x^2 + y^2 = 4 \):\[ \frac{5}{8} + y^2 = 4 \].

Step 5: Solve for \( y^2 \)

Subtract \( \frac{5}{8} \) from 4:\[ y^2 = 4 - \frac{5}{8} \] Simplify this to:\[ y^2 = \frac{32}{8} - \frac{5}{8} = \frac{27}{8} \].

Step 6: Find Possible Values for \( x \) and \( y \)

Solve for \( x \) and \( y \) using the values of \( x^2 \) and \( y^2 \):Since \( x^2 = \frac{5}{8} \), \( x = \pm \sqrt{\frac{5}{8}} \).Since \( y^2 = \frac{27}{8} \), \( y = \pm \sqrt{\frac{27}{8}} \).

Step 7: Construct the Solution Set

Thus, the solutions for the system of equations are the combinations of potential values for \( x \) and \( y \), forming the points:\( \left( \sqrt{\frac{5}{8}}, \sqrt{\frac{27}{8}} \right) \), \( \left( \sqrt{\frac{5}{8}}, -\sqrt{\frac{27}{8}} \right) \), \( \left( -\sqrt{\frac{5}{8}}, \sqrt{\frac{27}{8}} \right) \), \( \left( -\sqrt{\frac{5}{8}}, -\sqrt{\frac{27}{8}} \right) \).

Key Concepts

Understanding Algebra in System of Equations

Algebra is a powerful tool we use to solve equations, such as systems of equations. A system of equations consists of multiple equations that you solve at the same time. The main goal is to find values of the unknowns that satisfy all the equations simultaneously. These systems can have infinite solutions, no solution, or a distinct number of solutions. In algebra, we often use techniques such as substitution and elimination to solve these systems. In the given problem, we used subtraction to eliminate one of the variables, which simplified the system significantly. By simplifying in this manner, we solve for one variable at a time, making the problem less complex. Remember, the key in algebra is to manipulate equations so that they become simple enough to reveal the unknown variables. Pay attention to equal signs and ensure that whatever operation you perform on one side, you do to the other as well.

Exploring Quadratic Equations

Quadratic equations form the backbone of the problem. A quadratic equation is any equation that can be rearranged into the form \( ax^2 + bx + c = 0 \). In our exercise, each equation is in terms of squares of unknowns, such as \( x^2 + y^2 \). This reveals that we are dealing with conic sections, specifically circles, based on their structure.A crucial aspect of quadratic equations involves finding the square roots to solve for individual variable values. Once you isolate a variable's square, like \( x^2 = \frac{5}{8} \), you solve by taking square roots, leading to both positive and negative solutions. Similarly, for \( y^2 = \frac{27}{8} \), you apply the same process. Remember, because of the equation's power (squared terms), solutions appear in pairs both positive and negative, revealing a symmetry that is charactertistic of quadratics.

Arriving at Solutions of Systems

Solving systems, especially with quadratic equations, involves arriving at sets of solutions that satisfy all the initial conditions. The final solutions in our problem elaborate this well, showing every possible combination of the variables. After solving for individual squares, combination of each positive and negative square root value gives you different solutions. Therefore, for \( x = \sqrt{\frac{5}{8}} \) and \( x = -\sqrt{\frac{5}{8}} \) combined with \( y = \sqrt{\frac{27}{8}} \) and \( y = -\sqrt{\frac{27}{8}} \), we have four solutions:

• \( \left(\sqrt{\frac{5}{8}}, \sqrt{\frac{27}{8}}\right) \)
• \( \left(\sqrt{\frac{5}{8}}, -\sqrt{\frac{27}{8}}\right) \)
• \( \left(-\sqrt{\frac{5}{8}}, \sqrt{\frac{27}{8}}\right) \)
• \( \left(-\sqrt{\frac{5}{8}}, -\sqrt{\frac{27}{8}}\right) \)

These solutions represent the points where both equations intersect. Always remember, when dealing with systems, check each solution to ensure they fit into the original equations. This process is an essential step to confirm the validity of answers in algebra.