Problem 13 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} x^{2}+2 y^{2} &=11 \\ x^{2}-y^{2} &=8 \end{aligned}$$

Step-by-Step Solution

Answer

The short answer is: The solutions to the system of equations are \((3, 1), (-3, 1), (3, -1), \) and \((-3, -1)\).

Step 1: Solve one equation for a variable

In this case, it is easiest to solve the second equation for x² since it has a simple form: \(x^2 - y^2 = 8\) Adding y² to both sides: \(x^2 = y^2 + 8\) Now we have an expression for x² in terms of y².

Step 2: Substitute in the other equation

Now we will substitute the expression for x² from Step 1 into the first equation of the system: \(x^2 + 2y^2 = 11\) Substituting the expression for x² from Step 1, we get: \((y^2 + 8) + 2y^2 = 11\)

Step 3: Solve for the remaining variable

We will now solve the equation from Step 2 for y²: \((y^2 + 8) + 2y^2 = 11\) Combining like terms, we have: \(3y^2 = 3\) Dividing both sides by 3: \(y^2=1\) Now, take the square root of both sides to find y: \(y = \pm \sqrt{1}\) Hence, \(y = 1\) or \(y = -1\)

Step 4: Solve for the other variable

Now we can substitute the values of y back into the expression for x² from Step 1 to find the corresponding values of x. Case 1: \(y = 1\) \(x^2 = (1)^2 + 8 = 9 \) \(x = \pm \sqrt{9}\) So, \(x = 3\) or \(x = -3\) Case 2: \(y = -1\) \(x^2 = (-1)^2 + 8 = 9\) \(x = \pm \sqrt{9}\) So, \(x = 3\) or \(x = -3\)

Step 5: Write the solution as ordered pairs

Finally, we will write the solution as ordered pairs (x, y). Since we have two values of x and two values of y, we will have a total of four ordered pairs: \((3, 1), (-3, 1), (3, -1), (-3, -1)\)

Key Concepts

Solving Algebraic Equations

When faced with a system of algebraic equations, the goal is generally to find the values of the variables that satisfy all equations simultaneously. Think of it as finding a point or points where all equations intersect.
To solve algebraic equations:

• Identify the equations involved, understanding their form and structure.
• Choose a suitable method for solving, based on the characteristics of the equations (in our case, using substitution).
• Perform operations such as addition, subtraction, multiplication, or division systematically to simplify and solve the equations.

In our exercise, both equations involve squares of variables, indirectly hinting at quadratic properties which might be tackled more effectively with substitution.

Substitution Method

The substitution method is a powerful tool when dealing with systems of equations. It involves isolating one variable in one of the equations and substituting this into the other equation. This creates a single equation with one unknown, making it much simpler to solve.
In our exercise, we start by taking the equation \(x^2 - y^2 = 8\) and solve for \(x^2\) in terms of \(y^2\). This gives us \(x^2 = y^2 + 8\). Using substitution, we place this expression for \(x^2\) into the other equation \(x^2 + 2y^2 = 11\).

• This substitution eliminates \(x\) momentarily and allows us to focus on solving for \(y^2\).
• Once \(y\)'s value is found, we substitute back to find \(x\)'s value.

This method is particularly useful when one equation is easier to solve for one variable than directly solving the entire system.

Quadratic Equations

Quadratic equations appear frequently in algebra and typically take the form \(ax^2 + bx + c = 0\). They are characterized by their highest degree being a square, meaning \(x^2\) or \(y^2\) is the term of interest.
Important points to remember with quadratic equations:

• They generally have two solutions, which could be real or complex numbers.
• The solutions are often found using factorization or the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
• You might also deal with them by finding the square root of both sides, as shown in this exercise.

In our system, after substitution, we end up solving \(3y^2 = 3\). From here, solving for \(y\) becomes straightforward, leading us to the values \(y = 1\) or \(y = -1\). Each value paves the way to figure out the corresponding \(x\), completing the solutions through the quadratic perspective.