Problem 24 - College Algebra Essentials

Question

In Exercises \(19-28,\) solve each system by the addition method. $$ \left\\{\begin{aligned} 16 x^{2}-4 y^{2}-72 &=0 \\ x^{2}-y^{2}-3 &=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

\(x = \sqrt{5}, -\sqrt{5}\) and \(y = \sqrt{2}, -\sqrt{2}\)

Step 1: Multiply the second equation

To cancel out one of the variables, the equations should be in the same form. Therefore, first, multiply the second equation by -4: \[-4*(x^{2}-y^{2}-3)=0\] results in \[-4x^{2}+4y^{2}+12=0\]

Step 2: Add the two equations

Now add the two equations together: \(16x^2 - 4x^2 = 12x^2\) and \(-4y^2 + 4y^2 = 0y^2\) and \(-72 + 12 = -60\). This leads to a new equation: \(12x^2 - 60 = 0\)

Step 3: Simplify the new equation

Simplify the equation by dividing all terms by 12: \(x^2 - 5 = 0\). This gives the equation in a simpler form.

Step 4: Solve for x

Now that the equation is in its simplest form, solve it to find the value of x: \(x^2 = 5\), so \(x = \sqrt{5}, -\sqrt{5}\)

Step 5: Substitute x into the second equation

Substitute the value of x into the second equation to solve for y. Plugging \(x = \sqrt{5}\) and \(x = -\sqrt{5}\) into the second equation \(x^{2}-y^{2}-3 =0\) gives \(y = \sqrt{2}, -\sqrt{2}\)

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations that have common variables. In the context of this exercise, the given system consists of two equations involving the variables \(x\) and \(y\). The aim is to find the values of these variables that satisfy both equations simultaneously. This set of equations is quadratic, meaning it involves terms with the variables raised to a power of two. Solving systems like these requires specific techniques because guessing the solution isn't practical.

• Each equation represents a condition (constraint) on the variables.
• Solutions are values that make all equations true.
• Systems can have one solution, no solution, or infinitely many solutions, depending on their nature.

Understanding systems of equations is vital in algebra as it forms the basis for much-complex mathematical modeling and real-world problem-solving.

Algebraic Manipulation

Algebraic manipulation involves transforming equations to isolate variables or simplify them. In this exercise, the \'addition method\' is used, which is also known as elimination. This technique is particularly useful when you want to cancel out one of the variables.
Here’s a look at how this works:

• The goal is to align equations such that adding or subtracting them eliminates one variable.
• In this exercise, the second equation is multiplied by \(-4\) so that the terms can effectively cancel out when added to the first equation.
• This manipulation makes it easier to solve for one variable.

Algebraic manipulation is essential because it provides a methodical way to transform equations into a solvable form without changing their inherent properties. Mastering these techniques enhances problem-solving abilities.

Problem-Solving Strategies

Approaching algebraic problems efficiently requires strategic thinking. This is where problem-solving strategies come into play. In this exercise, the combination of manipulation and substitution provided a clear path to the solution.

• Identify what method best suits the problem, like addition, which cancels out a variable.
• Break down the problem into smaller steps, such as transforming equations and solving incrementally.
• Substitution, like in step 5, is often used after one variable value is found, allowing you to uncover the other variable values.

By following such strategies, you accelerate the solving process and minimize errors. Strategies in algebra are akin to a toolkit; knowing which tool to use and when streamlines solving even complex equations.