Problem 33 - 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. $$\left\\{\begin{array}{l} \frac{1}{2} x+\frac{1}{3} y=2 \\ \frac{1}{5} x-\frac{2}{3} y=8 \end{array}\right.$$

Step-by-Step Solution

Answer

The system has a unique solution: (10, -9).

Step 1: Clear Fractions

First, clear the fractions by finding a common multiple for each equation. Multiply the first equation by 6 and the second by 15 to eliminate the fractions:\[ 6 \left(\frac{1}{2}x + \frac{1}{3}y = 2\right) \Rightarrow 3x + 2y = 12 \]\[ 15 \left(\frac{1}{5}x - \frac{2}{3}y = 8\right) \Rightarrow 3x - 10y = 120 \]

Step 2: Subtract Equations

Subtract the first equation from the second to eliminate \(x\):\[ (3x - 10y) - (3x + 2y) = 120 - 12 \]This simplifies to:\[ -12y = 108 \]

Step 3: Solve for y

Solve \(-12y = 108\) for \(y\):\[ y = \frac{108}{-12} = -9 \].

Step 4: Substitute y back into one equation

Substitute \(y = -9\) into the first equation to solve for \(x\):\[ 3x + 2(-9) = 12 \]This simplifies to:\[ 3x - 18 = 12 \].

Step 5: Solve for x

Add 18 to both sides to solve for \(x\):\[ 3x = 30 \]\[ x = \frac{30}{3} = 10 \].

Step 6: Express the Solution

The solution to the system of equations is the ordered pair \((x, y) = (10, -9)\).

Key Concepts

Algebraic Methods

Algebraic methods provide a systematic way to solve equations and systems of equations. They typically include approaches such as substitution, elimination, and using matrices. Each method has its own advantages, depending on the characteristics of the equations involved.

In this specific problem, you used algebraic methods to find the solution to a system of two linear equations. This involves symbolic manipulation to isolate variables and find their values. Understanding how to manipulate equations algebraically can help in simplifying complex mathematical problems.

• Substitution involves solving one equation for a variable and substituting that expression into another equation.
• The elimination method focuses on eliminating one variable to solve for another, which was used here.

Algebraic methods require practice, but once mastered, they make solving systems of equations much more approachable.

Fractions in Equations

Working with fractions in equations can often be challenging, especially in systems of equations. They can make calculations cumbersome and complex. One effective strategy is to eliminate them early on by finding a common denominator and multiplying through to clear them.

In the provided system of equations, the first task was to eliminate the fractions. This makes the system easier to solve with standard algebraic techniques.

• A common multiple helps convert each term of the fraction into a whole number.
• In the exercise, multiplying the first equation by 6 and the second by 15 eliminated all fractional terms.

This simplifies subsequent steps and reduces the possibility of errors. Getting rid of fractions streamlines the solving process, making it more straightforward to apply methods like elimination or substitution.

Elimination Method

The elimination method is a technique used to solve systems of linear equations. This method involves adding or subtracting equations to cancel out one of the variables. The goal is to reduce the system to a single equation with one unknown, which is then solved with basic arithmetic.

In the given solution, after clearing fractions, the equations were structured such that the coefficient of one variable matched in magnitude but not in sign. By subtracting the first equation from the second, the variable \(x\) was eliminated, enabling the solution for \(y\).

• Use algebraic operations to create opposites for a variable.
• Once one variable is eliminated, solve for the remaining variable using simple arithmetic.
• Substitute back to find the other variable.

The elimination method is especially useful when equations have variables with conveniently matching coefficients.

Ordered Pairs

In the context of systems of equations, solutions are often expressed as ordered pairs. An ordered pair is a simple notation used to represent the solution of an equation or a system of equations, denoted as \((x, y)\) where \(x\) and \(y\) are the values calculated.

After solving the system, you found the solution \((x, y) = (10, -9)\). This ordered pair shows that when \(x = 10\), \(y\) must be \(-9\) to satisfy both equations in the system.

• Ordered pairs provide a concise way to present solutions.
• They make it easy to verify the correctness by substituting back into the original equations.
• Ordered pairs are the basic building blocks for graphing linear equations.

Expressing solutions in ordered pair form aids in interpretation and application, especially in more advanced topics such as graphing and analysis.