Problem 54 - 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. $$\left\\{\begin{array}{l} 2(x+y)=4 x+1 \\ 3(x-y)=x+y-3 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \((x, y) = (0.5, 1)\).

Step 1: Rearrangement of equations

Begin by rearranging both equations: From \(2(x+y)=4x+1\), we get \(2x + 2y = 4x + 1\) or \(2y = 2x + 1\) or \(y = x+ \frac{1}{2}\). From \(3(x-y) = x + y - 3\), we get \(3x - 3y = x + y - 3\) or \(2x = 4y - 3\), after simplifying we see that \(x = 2y - \frac{3}{2}\).

Step 2: Substitute x in the first equation

Substitute the expression of \(x\) from step 1 in the first equation: \(y = 2y - \frac{3}{2} + \frac{1}{2}\), after simplifying we will get \(y = 1\).

Step 3: Substitute y in the second equation

Substitute the value of \(y\) in the second equation to get:\(x = 2(1) - \frac{3}{2} = 0.5\) which gives the final value of \(x\).

Step 4: Write Solution Set

The solution set in set notation is \(\{(x, y)\} = \{(0.5, 1)\}\)

Key Concepts

Algebraic Methods

Solving a system of equations can be done using various algebraic methods. These methods provide a framework to find the values of the variables that satisfy all equations simultaneously.

There are several algebraic strategies for tackling these problems, like substitution, elimination, and graphical methods. The choice of strategy often depends on convenience or simplicity for the given set of equations.

• Substitution: Involves solving one equation for a variable and substituting this expression into another equation.
• Elimination: Aims to eliminate one variable by adding or subtracting equations, reducing the system to fewer variables.
• Graphical Method: Entails plotting each equation on a graph and identifying intersection points.

Sometimes, equations might suggest one method over the others for efficiency. For example, a substitution might be chosen if one equation is already solved for a single variable.

Substitution Method

The substitution method involves replacing one variable with an expression from another equation. It transforms a system of equations into a single-variable equation.

Here is how substitution was used in this exercise:
1. Begin with the equations:
* Given: \(2(x+y)=4 x+1\) and \(3(x-y)=x+y-3\).
2. Rearrange the first equation to express one variable in terms of another:
* From \(2y = 2x + 1\), we simplify to \(y = x + \frac{1}{2}\).
3. Use this expression in the second equation:
* Substitute \(x = 2y - \frac{3}{2}\) into \(y = x + \frac{1}{2}\). This substitution solves for \(y\).
4. Once \(y\) is determined, plug it back into either equation to solve for \(x\).

This approach is often preferred when one equation can be easily expressed in terms of one variable, streamlining the solution process.

Solution Set

The solution set of a system of equations is the collection of all ordered pairs (or triples, etc., in higher dimensions) that satisfy every equation in the system.

Solution sets can be expressed in set notation, which clearly and concisely demonstrates the final answer to a system of equations.

• Unique Solution: A single set of values, as in this exercise, e.g., \(\{(0.5, 1)\}\).
• Infinite Solutions: Often occurs when two equations represent the same line; they overlap completely.
• No Solution: Represents parallel lines that never intersect. There are no ordered pairs that satisfy all equations together.

In the provided exercise, solving the equations using substitution yielded a unique solution, \(\{(0.5, 1)\}\), indicating one point of intersection for these two lines in the graph of the equations.