Problem 29 - Introductory Algebra for College Students

Question

Solve each system by the substitution method. If there is no solution or an infinite number of solutions, so state. Use set notation to express solution sets. $$ \left\\{\begin{array}{l} \frac{x}{6}-\frac{y}{2}=\frac{1}{3} \\ x+2 y=-3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \({(-1, -1)}\)

Step 1: Simplify the equations

Begin by simplifying both equations in the system. To simplify the second equation, you can get \(x\) in terms of \(y\) by rearranging it to be \(x = -3 - 2y\). For the first equation, multiplying each term by 6 to clear off the fraction, the equation becomes \(x - 3y = 2\).

Step 2: Substitution

Substitute \(x\) from the second equation into the first, that is, replace \(x\) in the equation \(x - 3y = 2\) with \(-3 - 2y\). So we get \(-3 - 2y - 3y = 2\). After simplifying this you get \(-5y = 5\).

Step 3: Solve for y

Solve the equation \(-5y = 5\) for \(y\), then you find that \(y = -1\).

Step 4: Substitute y back into the equation

Now, substitute \(y = -1\) back into the second equation \(x = -3 - 2y\) to solve for \(x\). So, we get \(x = -3 - 2(-1)\), simplifying this yields \(x = -1\).

Step 5: State the solution in set notation

The solution in set notation is \({(-1, -1)}\)

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In this exercise, we have two equations with two variables, namely, \(x\) and \(y\). Solving a system of equations means finding values for these variables that satisfy all the equations simultaneously.
Equations in a system can be linear or nonlinear. Our focus here is on linear equations because they are the simplest and most common when learning systems. These equations can be written in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.
With linear equations, there are three possibilities for solutions:

• A single unique solution, meaning the lines intersect at one point.
• No solution, which occurs when the lines are parallel and do not intersect.
• Infinite solutions, happening when the lines coincide and therefore overlap entirely.

For the problem at hand, we will focus on using the substitution method to find whether a single point of intersection (solution) exists.

Set Notation

Set notation is a mathematical way to represent collections of objects, usually numbers. When we solve systems of equations, set notation helps us express solutions clearly and concisely.
In this context, a solution to a system of equations could be a single ordered pair \((x, y)\), multiple pairs, or no pairs at all. Here’s a simple breakdown of how set notation works:

• A single solution is written as \(\{(x, y)\}\).
• If there are no solutions, this can be denoted as \(\varnothing\) or simply state "no solution."
• If there are infinitely many solutions, you might see something like \(\{(x, y) \mid \text{conditions}\}\).

In our given exercise, since there is one solution, we write it as \(\{(-1, -1)\}\) in set notation.

Solving Linear Equations

Solving linear equations involves finding values of the variables that make the equation true. When using the substitution method in a system of equations:

• First, solve one of the equations for one variable in terms of another.
• Next, substitute this expression into the other equation. This helps eliminate one of the variables.
• Simplify and solve this new equation, which should be in terms of one variable.
• Finally, back-substitute the found variable into any of the original equations to solve for the remaining variable.

Let’s look at how this applies to our system:
1. **Simplification and Substitution:** We solve the second equation, \(x + 2y = -3\), to express \(x\) in terms of \(y\): \(x = -3 - 2y\). Substituting into the first equation simplifies it by eliminating \(x\).
2. **Solve for one variable:** After substitution, we get \(-5y = 5\), leading to \(y = -1\).
3. **Back-substitute:** Replace \(y\) back into \(x = -3 - 2y\) to find \(x = -1\), resulting in the ordered pair \((-1, -1)\).
This systematic approach ensures a structured way to solve and check the solution within the system.