Problem 27 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} 9 x^{2}+y^{2} &=9 \\ x^{2}+y^{2} &=5 \end{aligned}$$

Step-by-Step Solution

Answer

The system has four solutions: \[ \begin{aligned} (x_1,y_1) &= \left(\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right), \\ (x_2, y_2) &= \left(\frac{1}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right), \\ (x_3, y_3) &= \left(-\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right), \\ (x_4, y_4) &= \left(-\frac{1}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) \end{aligned} \]

Step 1: Subtract the second equation from the first equation

Subtract the second equation from the first one in order to eliminate y and get an equation in x: \[ (9x^2 + y^2) - (x^2 + y^2) = 9 - 5 \]

Step 2: Solve the resulting equation for x

Simplify the equation and solve for x: \\\[ 8x^2 = 4 \\\\ x^2 = \frac{4}{8} \\\\ x^2 = \frac{1}{2} \\\\ x = \pm \frac{1}{\sqrt{2}} \\\]

Step 3: Substitute x back into one of the original equations

Substitute the values of x obtained in Step 2 into the second equation to find the corresponding values for y: \\\[ y^2 = 5 - ( \pm\frac{1}{\sqrt{2}})^2 \\\]

Step 4: Solve for y

Simplify the equation for y and solve it: \\\[ y^2 = 5 - \frac{1}{2} \\\\ y^2 = \frac{9}{2} \\\\ y = \pm \frac{3}{\sqrt{2}} \\\]

Step 5: Write the solutions

The system has four solutions for x and y, given by: \[ \begin{aligned} (x_1,y_1) &= \left(\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right), \\ (x_2, y_2) &= \left(\frac{1}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right), \\ (x_3, y_3) &= \left(-\frac{1}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right), \\ (x_4, y_4) &= \left(-\frac{1}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right) \end{aligned} \]

Key Concepts

Quadratic Equations

A quadratic equation is a fundamental concept in algebra that involves variables raised to the power of two. The general form of a quadratic equation is given by \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not equal to zero.

Quadratic equations can have two solutions, one solution, or no real solution, depending on the value of the discriminant \(b^2 - 4ac\).

• If the discriminant is positive, the equation has two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If the discriminant is negative, there are no real solutions, but two complex solutions.

Solving quadratic equations can be done using various methods like factoring, completing the square, or the quadratic formula. In our exercise, we deal with the system of quadratic equations where we try to find the intersection points of two curves.

Algebraic Solution

Solving a system of equations algebraically involves finding the values of the variables that satisfy all equations simultaneously. It requires manipulating the equations to eliminate variables and simplify the problem.

In the given exercise, we started by subtracting the second equation from the first. This reduced the system to a single equation with one variable, which could then be easily solved for \(x\). This process illustrates the power of algebra in reducing complex problems into simpler, manageable ones.

• After isolation, the algebraic manipulation focused on solving for \(x\).
• Once \(x\) was known, substitution allowed us to find \(y\).

Using algebraic solutions, especially in quadratic systems, often involves clear, structured steps. This method ensures that every solution is logically sound and mathematically correct.

Substitution Method

The substitution method is a technique used to solve systems of equations, especially when one equation can be easily solved for one of the variables. It involves three major steps:

First, solve one of the equations for one variable in terms of the other variables. By doing this, you create an expression that can be substituted into another equation.

• In the exercise, after solving for \(x\), we substituted \(x\) into the second equation to find the corresponding values of \(y\).

Substituting helps in reducing the complexity of solving multiple equations at once.

Second, replace the variable from your newly created expression into another equation.

• This step is crucial as it transforms the original system into a more straightforward problem, solving one equation at a time.

Finally, back-substitute the value of the solved variable into one of the original equations to find other variable values. This method streamlines the solving process, making it efficient for quadratic systems, as evident in our exercise.