Problem 56 - Introductory Algebra for College Students

Question

In Exercises \(45-56,\) solve each system by the method of your choice. If there is no solution or an infinite number of solutions, so state. Use set notation to express solution sets. Explain why you selected one method over the other two. $$\left\\{\begin{aligned} 4 y &=-5 x \\ 5 x+8 y &=20 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is {(\( x, y \)) | \( x = 4 \), \( y = -5 \)}

Step 1: Express \( y \) in terms of \( x \) from the first equation

From \( 4y=-5x \), we divide by 4 on both sides to yield \( y = -\frac{5}{4}x \).

Step 2: Substitute \( y \) into the second equation

Plugging \( y = -\frac{5}{4}x \) into \( 5x+8y=20 \), we get \( 5x + 8(-\frac{5}{4}x) = 20 \), which simplifies to \( x = 4 \).

Step 3: Substitute \( x \) into the first equation

Substitute \( x = 4 \) into the first equation \( 4y = -5x \) to get \( 4y = -20 \), which means \( y = -5 \).

Step 4: Express solution set in set notation

The solution set is {(\( x, y \)) | \( x = 4 \), \( y = -5 \)}

Key Concepts

Set Notation

Set notation is a systematic way of expressing the solution set of a problem, particularly when dealing with systems of equations. It uses curly braces to denote the collection of solutions or elements that satisfy a given condition. Set notation is especially useful in mathematics to clearly indicate all possible solutions or specific points.
In our problem, after solving the system of equations, we reached the solution where both equations are satisfied. This results in a single point, which we express using set notation. The solution is written as \( \{(x, y) | x = 4, y = -5\} \). This tells us that in the context of the problem, the only values of \( x \) and \( y \) that work are 4 and -5 respectively.

• The curly braces \( \{ \} \) denote the beginning and end of the solution set.
• The expression \( (x, y) \) represents the ordered pair of solutions.
• Inside the braces, the pipe symbol \( | \) can be interpreted as "such that".

Understanding set notation makes it easier for you to express solutions concisely and verify if a given solution is valid.

Substitution Method

The substitution method is one of the most efficient techniques for solving a system of linear equations. This method is particularly effective when one of the equations can easily be rearranged to solve for one variable. In our case, the system was:
\[\begin{aligned}4y &=-5x \5x+8y &=20\end{aligned}\]
We used the first equation to express \( y \) in terms of \( x \), giving us \( y = -\frac{5}{4}x \). This expression for \( y \) is then substituted into the second equation to find the exact value of \( x \). This simplification process allows us to deal with one variable at a time and often results in a straightforward path to finding the solution.

• Step 1: Solve one equation for one variable.
• Step 2: Substitute the expression from Step 1 into the other equation.
• Step 3: Solve the resultant single-variable equation.
• Step 4: Use the found value to substitute back and solve for the other variable.

The substitution method is clear and minimizes errors, especially when working with simple linear equations.

Linear Equations

Linear equations are equations of the first order, meaning they involve terms up to the first power and create straight lines when graphed. These equations are fundamental in algebra and are expressed in the form \( ax + by = c \) where \( a \), \( b \), and \( c \) are constants.
In our exercise, the system \( \begin{aligned} 4y &=-5x \ 5x+8y &=20 \end{aligned} \) consists of two linear equations. Working with linear equations allows us to apply straightforward methods like substitution, as executed.

• They have one or more variables, often \( x \) and \( y \).
• The graph of a linear equation is always a straight line.
• Linear equations can form systems that result in no solution, one unique solution, or infinitely many solutions.

Understanding linear equations enables you to tackle complex problems by transforming them into simple, solvable components. They are the building blocks for more advanced topics in algebra.