Problem 38 - College Algebra

Question

Solve each system by any method. $$ \begin{array}{l} \frac{1}{2} x+\frac{1}{3} y=\frac{1}{3} \\ \frac{3}{2} x+\frac{1}{4} y=-\frac{1}{8} \end{array} $$

Step-by-Step Solution

Answer

\( x = -\frac{1}{3} \), \( y = \frac{3}{2} \).

Step 1: Clear Fractions in the First Equation

To eliminate the fractions in the first equation, multiply the entire equation by 6 (the least common multiple of 2 and 3). The equation changes from \( \frac{1}{2}x + \frac{1}{3}y = \frac{1}{3} \) to \( 3x + 2y = 2 \).

Step 2: Clear Fractions in the Second Equation

To eliminate the fractions in the second equation, multiply the entire equation by 8 (the least common multiple of 2 and 4). The equation changes from \( \frac{3}{2}x + \frac{1}{4}y = -\frac{1}{8} \) to \( 12x + 2y = -1 \).

Step 3: Subtract Equations to Eliminate y

Subtract the first equation from the second equation. We have \( 12x + 2y = -1 \) and \( 3x + 2y = 2 \). Subtract: \( (12x + 2y) - (3x + 2y) = -1 - 2 \) to get \( 9x = -3 \).

Step 4: Solve for x

Divide both sides of the equation from Step 3 by 9 to solve for \( x \). So, we have \( x = \frac{-3}{9} = -\frac{1}{3} \).

Step 5: Substitute x into the First Equation

Substitute \( x = -\frac{1}{3} \) into the first equation \( 3x + 2y = 2 \). This gives \( 3(-\frac{1}{3}) + 2y = 2 \), which simplifies to \( -1 + 2y = 2 \).

Step 6: Solve for y

Add 1 to both sides of the equation from Step 5 to get \( 2y = 3 \). Then divide both sides by 2 to find \( y = \frac{3}{2} \).

Key Concepts

Solving Systems of Equations

In mathematics, a system of equations is a set of two or more equations with the same variables. The challenge is to find values for these variables that satisfy all the equations in the system simultaneously. This process is known as solving systems of equations. There are various methods to solve these systems, such as substitution, elimination, and using matrices. In this problem, we start with two linear equations with fractions. Solving such systems requires a clear understanding of algebraic manipulation to eliminate fractions and make the system easier to handle. By solving these systems successfully, we gain pivotal insights into understanding intersections in linear algebra.

Eliminating Fractions

Dealing with fractions in equations can be cumbersome, but there's a straightforward technique to handle them. Eliminating fractions involves multiplying each term in the equation by the least common multiple (LCM) of the denominators. This converts the equations into a more manageable integer format. For instance, in our exercise, the LCM of the denominators in the first equation is 6. By multiplying each term by 6, the equation becomes free of fractions, transforming into a simpler expression. Similarly, for the second equation, the LCM is 8. Eliminating fractions simplifies the equations considerably, making subsequent steps much more efficient. It's a crucial step in solving systems of equations smoothly.

Substitution Method

The substitution method is one of the core techniques for solving systems of equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation. In our solution, after eliminating fractions and simplifying the equations, we strategically choose to subtract one equation from the other to eliminate the variable 'y.'
After determining the value of 'x,' the substitution method kicks in, where we substitute the value of 'x' back into one of the simplified equations to find the value of 'y.'

• This method is particularly useful when one equation can be easily solved for one variable.
• It involves algebraic manipulation and is often used in conjunction with other methods to streamline the solving process.

Linear Algebra

Linear algebra is a branch of mathematics that focuses on vector spaces, linear mappings, and systems of linear equations. Solving systems of linear equations, as seen in our exercise, is fundamental to linear algebra. The techniques used, such as elimination and substitution, are practical applications of linear algebra concepts. In this exercise, linear equations represent lines, and solving the system determines the point of intersection of these lines, which is finding common solutions for the variables. Understanding the principles of linear algebra helps to tackle more complex systems and paves the way for studying matrices and transformations, which have vast applications in computer science, physics, and engineering. It serves as the foundation for many mathematical concepts and real-world problem-solving.