Problem 43 - Algebra and Trigonometry

Question

\(21-48=\) 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, expressed as \((x, 5 - \frac{5}{6}x)\).

Step 1: Clear the Fractions

To make the equations easier to handle, we first eliminate the fractions by finding a common denominator. For the first equation, multiply every term by 10 (since 10 is a common multiple of 2 and 5), resulting in:\[ 5x + 6y = 30 \]For the second equation, multiply every term by 3 (since 3 is the denominator of the fraction in the equation):\[ 5x + 6y = 30 \]

Step 2: Simplify and Compare the Equations

After simplifying, both equations become:\[ 5x + 6y = 30 \]Since both equations are identical, this indicates that the lines represented by these equations are the same, meaning they are coincident.

Step 3: Determine the Nature of the Solution

Since both equations represent the same line, any point \((x, y)\) on this line will satisfy the equations, resulting in infinitely many solutions.

Step 4: Express the Solution as Ordered Pairs

To express the solutions in ordered-pair form, solve \(y\) in terms of \(x\) from the simplified equation:\[ 6y = 30 - 5x \]Divide through by 6:\[ y = 5 - \frac{5}{6}x \]Thus, the solutions can be expressed as ordered pairs:\[(x, 5 - \frac{5}{6}x)\] for any real number \(x\).

Key Concepts

Linear Equations

A linear equation is a mathematical expression that describes a straight line on a graph. These equations have no exponents higher than one and typically take the form \(ax + by = c\), where \(a\), \(b\), and \(c\) stand for constants. In our exercise, the system comprises two linear equations:

• \(\frac{1}{2}x + \frac{3}{5}y = 3\)
• \(\frac{5}{3}x + 2y = 10\)

The main goal when solving systems of linear equations is to find values of \(x\) and \(y\) that satisfy all given equations simultaneously.
Usually, students deal with systems graphically, algebraically, or numerically, aiming to find one single solution, no solution, or many solutions. Simplifying these equations often helps in visualizing their relationships.

Infinite Solutions

When a system of equations is said to have infinitely many solutions, it means that there isn't just one set of \((x, y)\) that works. Instead, there are countless pairs that fit. In our problem, both equations simplified to\(5x + 6y = 30\), showing they represent the same line. This means any point along this line will be a solution.
Such systems are referred to as having dependent equations, which lead to coincident lines when graphed.

• One way to recognize infinite solutions is when one equation can be manipulated to become the same as another in the system.
• In terms of logistic, convert one equation into the format resembling the other until both equations are indistinguishable.

This outcome highlights how infinite solutions can occur in practical problem-solving scenarios, owing to overlapping scenarios.

Clearing Fractions

Fractions in linear equations can be tricky to work with. Clearing fractions helps simplify the equations, making them easier to solve. In our exercise:

• The first equation was: \(\frac{1}{2}x + \frac{3}{5}y = 3\). We cleared the fractions by multiplying each term by 10, the least common multiple of 2 and 5. This resulted in \(5x + 6y = 30\).
• The second equation was: \(\frac{5}{3}x + 2y = 10\). Multiplying each term by 3 (the denominator) gave the same equation, \(5x + 6y = 30\).

Finding the least common multiple of denominators is a vital step to eliminate fractions, thus turning the equation into a more familiar and manageable form without changing its solution.

Ordered Pairs

Ordered pairs \((x, y)\) are used to express solutions to systems of equations, representing points on a graph. In our system with infinite solutions, the ordered pairs follow the same line described by the equation.

• After simplifying to \(y = 5 - \frac{5}{6}x\), this expression represents the dependency of \(y\) on \(x\).
• An example of an ordered pair solution may look like \(\left(x, 5 - \frac{5}{6}x\right)\).

These pairs convey that for each value chosen for \(x\), there corresponds a specific \(y\).
This way of expressing solutions is particularly useful in infinite or multiple solution scenarios, as it neatly captures the relationship between variables through basic substitution in algebraic form.