Problem 43 - Precalculus Mathematics 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 6. $$\left\\{\begin{array}{l}\frac{1}{2} x+\frac{3}{5} y=3 \\\\\frac{5}{3} x+2 y=10\end{array}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Clear Fractions

Multiply the first equation by 10 to eliminate the fractions:\[ 10 \left( \frac{1}{2} x + \frac{3}{5} y \right) = 10 \cdot 3 \]Resulting in:\[ 5x + 6y = 30 \]

Step 2: Multiply to Align Coefficients

Next, we will transform the second equation by multiplying it by 3 to align the coefficients of \(x\).\[ 3 \left( \frac{5}{3} x + 2y \right) = 3 \cdot 10 \]Which results in:\[ 5x + 6y = 30 \]

Step 3: Compare the Equations

Now we compare the two equations:\[ 5x + 6y = 30 \] and\[ 5x + 6y = 30 \]Both equations are identical.

Step 4: Determine the Nature of Solutions

Since both equations are identical, the system has infinitely many solutions. To express these solutions, assume one variable as arbitrary. Let \(y = t\), where \( t \) is any real number.

Step 5: Express Solutions in Ordered Pair

Substitute \(y = t\) into one of the equations, say \(5x + 6y = 30\):\[ 5x + 6t = 30 \]Solve for \(x\):\[ 5x = 30 - 6t \]\[ x = \frac{30 - 6t}{5} \]Thus, the solutions can be expressed as:\[ (x, y) = \left( \frac{30 - 6t}{5}, t \right) \]

Step 6: Conclusion

The system has infinitely many solutions. Each solution can be represented by choosing any real number \(t\) and calculating \(x\). The ordered pair is \(\left( \frac{30 - 6t}{5}, t \right)\).

Key Concepts

Infinitely Many Solutions

When solving systems of linear equations, one possible outcome is that there are infinitely many solutions. This happens when the equations represent the same line, meaning they have all points in common.
In the given system of equations, both equations were simplified to the same equation:

• 5x + 6y = 30

This result means that every solution to one equation is also a solution to the other. There is no unique intersecting point; instead, there are infinitely many points that satisfy both equations, resulting in infinitely many solutions.
To express these solutions in a simple form, one of the variables—often labeled as a "parameter" like \( t \)—is chosen to vary freely over all real numbers. This approach showcases the continuous nature of these solutions.

Ordered Pairs

Ordered pairs are a way to express solutions to equations. In a two-variable system like the one we are dealing with, solutions are expressed as ordered pairs \((x, y)\).
Each ordered pair represents a specific point on a graph that satisfies the equation.
For example, given any value of \( y = t \), you can find the corresponding \( x \) value using:

• \( x = \frac{30 - 6t}{5} \)
• \((x, y) = (\frac{30 - 6t}{5}, t) \)

This displays how ordered pairs can also show an infinite number of solutions by allowing one variable to be any real number while calculating the other accordingly. Each pair fits perfectly into the equation's framework, detailing a line on the graph.

Solving Linear Equations

Linear equations are math sentences that describe straight lines when graphed. Solving these equations involves finding values for the variables that make the sentence true.
Here is how you solve systems of linear equations:

• Start by simplifying the equation, such as eliminating fractions by multiplying all terms by the least common multiple.
• Align coefficients (the numbers multiplying the variables) to easily compare and solve equations.
• If the final equations are identical, the system has infinitely many solutions, as each point on the line is a solution.

In this problem, the step of clearing fractions played a crucial role in recognizing that both equations transformed into the same line. Once this happened, it became clear that there were infinitely many solutions. This systematic approach is designed to unravel even complex linear equations, providing a clear path to their solutions.