Problem 29 - Precalculus: Mathematical for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 3. $$\left\\{\begin{array}{l} \frac{1}{2} x+\frac{3}{5} y=3 \\ \frac{5}{3} x+2 y=10 \end{array}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions, represented as \((x, y) = \left(x, \frac{30 - 5x}{6}\right)\).

Step 1: Clear Fractions in the First Equation

To make calculations easier, we first clear the fractions from the first equation \( \frac{1}{2}x + \frac{3}{5}y = 3 \). This can be done by finding the least common multiple (LCM) of the denominators 2 and 5, which is 10. Multiply the entire equation by 10:\[ 10 \left(\frac{1}{2}x + \frac{3}{5}y \right) = 10 \cdot 3 \]Simplifying, we get:\[ 5x + 6y = 30 \].

Step 2: Simplify the Second Equation

The second equation is \( \frac{5}{3}x + 2y = 10 \). To remove the fraction, multiply the entire equation by 3 (the denominator), giving:\[ 3 \left( \frac{5}{3}x + 2y \right) = 3 \cdot 10 \]Simplifying, we get:\[ 5x + 6y = 30 \].

Step 3: Analyze the Resulting System

Now we have the system:\[\begin{align*}5x + 6y &= 30 \5x + 6y &= 30 \end{align*}\]Both equations are identical, meaning the original system simplifies to a single equation. This indicates that every solution to \( 5x + 6y = 30 \) is a solution to the system.

Step 4: Express the Solutions as Ordered Pairs

Since the system reduces to one equation, it has infinitely many solutions. To express these solutions in ordered-pair form, solve for one variable in terms of the other. Solving \( 5x + 6y = 30 \) for \( y \), we get:\[ 6y = 30 - 5x \]\[ y = \frac{30 - 5x}{6} \]Thus, the solutions can be expressed as ordered pairs: \( (x, y) = \left(x, \frac{30 - 5x}{6}\right) \), where \( x \) is any real number.

Key Concepts

Understanding Fractions in Linear Equations

When working with linear equations that include fractions, it can make solving them a bit tricky. But it's manageable with some simple techniques. Here's how you deal with fractions effectively:

• First, identify the denominators in the equation, as these are the numbers beneath the fraction line. For example, in the equation \( \frac{1}{2}x + \frac{3}{5}y = 3 \), the denominators are 2 and 5.
• Next, find the least common multiple (LCM) of these denominators. The LCM is the smallest number that both denominators divide into without leaving a remainder. Here, the LCM of 2 and 5 is 10.
• Clear the fractions by multiplying both sides of the equation by this LCM. This action gets rid of the fractions, making the equation easier to solve. In our example, multiplying by 10 gives us \( 5x + 6y = 30 \).

By eliminating fractions, we simplify the arithmetic and bring clarity to the process of solving the equation.

Identifying Infinitely Many Solutions

A system of linear equations may sometimes have infinitely many solutions. This occurs when the equations represent the same line. Understanding this concept relies heavily on observing the process of simplifying the equations. In our example, after clearing the fractions from both equations, they both simplify to \( 5x + 6y = 30 \). Because both equations are identical, this does not define a unique intersection where they meet. Some key points to keep in mind include:

• When the two equations simplify into one, it indicates that they are essentially the same line represented twice.
• Every point on this line is a solution to the system, leading to infinitely many solutions.
• This situation is different from when equations represent parallel lines, which do not intersect and thus, have no solution.

Recognizing infinitely many solutions is crucial as it tells us there is a whole set of possible answers, rather than just one.

Expressing Solutions as Ordered Pairs

When a system of equations has infinitely many solutions, we often express these solutions as ordered pairs. An ordered pair \( (x, y) \) shows the values of \( x \) and \( y \) that solve the equation. To find these pairs, we express one variable in terms of the other. In our simplified equation \( 5x + 6y = 30 \), solving for \( y \) gives: \[ y = \frac{30 - 5x}{6} \] This expression allows you to generate numerous solutions by substituting any real number for \( x \). Let’s see how ordered pairs work:

• If \( x = 0 \), then \( y = 5 \), which gives the ordered pair \( (0, 5) \).
• If \( x = 6 \), then \( y = 0 \), creating the ordered pair \( (6, 0) \).
• Any chosen \( x \) value will result in a corresponding \( y \) value, producing a pair \( (x, y) = (x, \frac{30 - 5x}{6}) \).

Ordered pairs provide a structured way to describe all solutions to the system visually and clearly. By calculating multiple pairs, you can even sketch the graph of the equation, illustrating its infinite solutions.