Problem 40 - Algebra and Trigonometry

Question

Solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their 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: Analyze the system

Verify if the system does not have any solutions or if it has infinitely many solutions. This achieved by comparing the coefficients. This system is in the form of \(ax + by = c\) and \(dx + ey = f\), where \((a,b,c) \neq (d,e,f)\). Since the coefficients are not proportional, this system should have a unique solution.

Step 2: Set one variable in terms of the other

Rearrange one equation to express one variable in terms of the other. Choosing the second equation, we get \(y = -\frac{x}{2} - \frac{3}{2}\). This will be used to substitute into the first equation.

Step 3: Substitute and solve for the other variable

Substitute \(y\) into the first equation, we get \(\frac{x}{6} + \frac{(x/2 + 3/2)}{2} = 1/3\). Solving for \(x\), we get \(x = -1\).

Step 4: Substitute and solve for the first variable

Substitute \(x = -1\) into the equation from Step 2, and solve for \(y\). We get \(y = -\frac{-1}{2} - \frac{3}{2} = -1\).

Step 5: Solution in set notation

Expressing the solution in set notation, we get \((-1, -1)\)

Key Concepts

unique solution

A system of linear equations consists of two or more linear equations with the same set of variables. In this problem, we are dealing with two equations:\

• \(\frac{x}{6} - \frac{y}{2} = \frac{1}{3}\)
• \(x + 2y = -3\)

A system is said to have a "unique solution" if there is exactly one set of values for the variables that satisfies all equations simultaneously. For a system of two equations with two variables, a unique solution occurs when the lines represented by these equations intersect at one point on a plane.
To determine if a unique solution exists, check the coefficients of the system. If they are not proportional, as evidenced by different ratios when comparing each equation's coefficients, the intersection results in a single solution. This was confirmed in Step 1 of the original solution, as the coefficients of the two equations are not proportional, indicating that the system has a unique solution.

substitution method

The substitution method is an algebraic way to find solutions for a system of equations. It involves expressing one variable in terms of the other, then substituting this expression into the other equation. This reduces the system to one equation with one variable.
In Step 2 of the original solution, we rearranged the second equation \(x + 2y = -3\) to express \(y\) in terms of \(x\):\

• \(y = -\frac{x}{2} - \frac{3}{2}\)

This expression for \(y\) is then substituted into the first equation \(\frac{x}{6} - \frac{y}{2} = \frac{1}{3}\). By this substitution, the equation becomes\

• \(\frac{x}{6} + \frac{(x/2 + 3/2)}{2} = \frac{1}{3}\)

which upon solution gives \(x = -1\).
Substituting \(x = -1\) back into the expression for \(y\), we find the value of \(y\), giving us \(y = -1\).
This method is particularly useful when one of the equations is easily rearrangeable, or when the system has coefficients making obvious elimination cumbersome.

set notation

Set notation is a concise way of representing mathematical expressions, specifically highlighting solutions in a structured manner. When solving systems of equations, expressing solutions in set notation provides clarity and formality.
In the original problem, after using the substitution method and solving for both variables \(x\) and \(y\), we found the solution to be \((-1, -1)\).
In set notation, this solution is expressed as a set containing the ordered pair:\

• \{(-1, -1)\}

This representation clearly denotes the specific values for \(x\) and \(y\) that satisfy both equations in the system. It’s a compact way to communicate, especially useful in advanced mathematical discussions or when solutions need to be methodically categorized, such as having solution sets of systems with infinite solutions or no solutions at all.