Problem 28 - College Algebra

Question

Solve each system by the addition method. \(\left\\{\begin{array}{l}2 x+3 y--16 \\ 5 x-10 y-30\end{array}\right.\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = \sqrt{\frac{40}{3}}, -\sqrt{\frac{40}{3}}\) and \(y = \frac{8}{3}\).

Step 1: Subtract the first equation from the second one

Using the addition method, which in this case involves subtraction because there are no like terms to add, subtract the first equation from the second equation: \(x^{2}+y^{2}-16 - (x^{2}-2y -8) = 0\). Simplify this to get: \(3y -8 = 0\).

Step 2: Solve the simplified equation for y

Isolate y in the simplified equation, we have: \(y = \frac{8}{3}\).

Step 3: Substitute y's value in the first equation

Substitute y's value in the first equation: \(x^{2} - 2* \frac{8}{3} -8 = 0\). This simplifies to: \(x^{2} - \frac{16}{3} - 8 = 0\).

Step 4: Solve the equation obtained in Step 3

The equation can be written as: \(x^{2} = \frac{16}{3} + 8 = \frac{40}{3}\). So the solutions for x are: \(x = \sqrt{\frac{40}{3}}\) and \(x = - \sqrt{\frac{40}{3}}\).

Key Concepts

Addition Method

The addition method, often known as the elimination method, is a technique for solving systems of equations. It involves adding or subtracting equations in order to eliminate one variable, making it easier to solve for the other. In this exercise, we use subtraction because the equations don't have like terms that can be directly added.

• First, take one equation and manipulate it to eliminate a variable when combined with the other equation.
• Here, the first equation, \(x^{2} - 2y - 8\), is subtracted from the second equation, \(x^{2} + y^{2} - 16\), to eliminate \(x^{2}\).
• This simplifies the system and allows us to solve for \(y\).

This method simplifies a complex system into something more manageable for further calculations.

Algebraic Substitution

After simplifying the system using the addition method, the next step typically involves substitution. Algebraic substitution means replacing a variable with its known value or expression in another equation.

• Once we found \(y = \frac{8}{3}\), we substituted this value back into one of the original equations.
• In our case, we substituted into the first equation \(x^{2} - 2 \cdot \frac{8}{3} - 8 = 0\).
• This substitution allowed us to obtain an equation with just one variable, \(x\), making it solvable.

Substitution helps simplify and transform the system, paving the way for solving the remaining variable.

Quadratic Equations

Quadratic equations appear frequently in algebra and are characterized by the form \(ax^2 + bx + c = 0\). In this exercise, we derived a quadratic equation after substitution.

• The equation \(x^{2} = \frac{40}{3}\) is a simplified quadratic form where \(b = 0\) and \(c = -\frac{40}{3}\).
• We then solved it using the square root method, finding \(x\) by computing both the positive and negative square roots.
• The solutions were \(x = \sqrt{\frac{40}{3}}\) and \(x = -\sqrt{\frac{40}{3}}\).

Understanding how to manipulate and solve quadratic equations is essential for uncovering all possible solutions to a problem.

Problem-Solving Steps

Solving a system of equations can be complex, but breaking it into clear steps simplifies the process. Here’s a structured approach to tackling such problems:

• **Identify:** Carefully understand what each equation represents.
• **Eliminate:** Use methods like addition or subtraction to eliminate a variable.
• **Substitute:** With one variable found, substitute back to solve for the other variable.
• **Simplify:** Perform calculations to simplify and solve the resultant equation.
• **Verify:** Double-check the solutions to ensure they satisfy both original equations.

Following these systematic steps helps efficiently reach a solution while ensuring accuracy and comprehension of the problem at hand.