Problem 27 - Beginning and Intermediate Algebra

Question

Solve each system using the elimination method. $$\begin{aligned}\frac{x}{4}+\frac{y}{2} &=-1 \\\\\frac{3}{8} x+\frac{5}{3} y &=-\frac{7}{12}\end{aligned}$$

Step-by-Step Solution

Answer

The short version of the solution is: The solution to the system of equations is \(x = -\dfrac{32}{9}\) and \(y = -\dfrac{7}{6}\).

Step 1: Rewrite equations with the same denominators

We start by rewriting the equations with the same denominators: \[ \begin{aligned} \frac{x}{4}+\frac{y}{2} =-1 \Rightarrow \frac{x}{8}+\frac{y}{4} =-\frac{1}{2} \\\\ \frac{3}{8} x+\frac{5}{3} y =-\frac{7}{12} \Rightarrow \frac{3}{8} x+\frac{5}{6} y =-\frac{7}{12} \end{aligned} \]

Step 2: Multiply equations so that one variable can be eliminated

Here, we will multiply the first equation by 3 and the second equation by 2 to make the coefficients of y the same: \[ \begin{aligned} 3(\frac{x}{8}+\frac{y}{4}) =3(-\frac{1}{2}) \Rightarrow \frac{3x}{8}+\frac{3y}{4} =-\frac{3}{2} \\\\ 2(\frac{3}{8} x+\frac{5}{6} y) =2(-\frac{7}{12}) \Rightarrow \frac{6}{8} x+\frac{5}{3} y =-\frac{7}{6} \end{aligned} \]

Step 3: Use elimination method to find one variable

To eliminate y, we can subtract the first equation from the second: \[ \begin{aligned} (\frac{6}{8} x+\frac{5}{3} y) - (\frac{3x}{8}+\frac{3y}{4}) = (-\frac{7}{6}) -(-\frac{3}{2}) \\\\ \frac{6 - 3}{8}x + \frac{5 - 9}{6}y = \frac{-7 + 9}{6} \\\\ \frac{3}{8}x - \frac{4}{6}y = \frac{2}{6} \end{aligned} \] Now, we can solve for x: \[ \begin{aligned} \frac{3}{8}x = \frac{4}{6}y + \frac{2}{6} \\\\ x = \frac{8}{3}(\frac{4}{6}y + \frac{2}{6}) \\\\ x = \frac{8}{3}(\frac{4}{6}y) + \frac{8}{3}(\frac{2}{6}) \end{aligned} \]

Step 4: Find the value of x

Now that we have x in terms of y, we can plug this expression for x back into either of the original equations. We'll plug it into the first rewritten equation: \[ \frac{1}{8}\left(\frac{8}{3}(\frac{4}{6}y) + \frac{8}{3}(\frac{2}{6})) + \frac{y}{4} = -\frac{1}{2} \] Solving for y, we get: \[ \begin{aligned} \frac{4}{3}y + \frac{4}{6} + \frac{y}{4} = -\frac{1}{2} \\\\ \frac{16}{12}y + \frac{8}{12} + \frac{3}{12}y = -\frac{6}{12} \\\\ \frac{19}{12}y = -\frac{14}{12} \\\\ y = -\frac{7}{6} \end{aligned} \]

Step 5: Find the value of x

With the value of y found, we insert it into the equation for x: \[ x = \frac{8}{3}(\frac{4}{6}(-\frac{7}{6})) + \frac{8}{3}(\frac{2}{6}) \] Solving for x, we get: \[ \begin{aligned} x = \frac{8}{3}(-\frac{28}{36}) + \frac{16}{9} \\\\ x = -\frac{8}{3} \cdot \frac{7}{9} + \frac{16}{9} \\\\ x = -\frac{16}{3} + \frac{16}{9} \\\\ x = -\frac{32}{9} \end{aligned} \] The solution to the system of equations is \(x = -\dfrac{32}{9}\) and \(y = -\dfrac{7}{6}\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In linear algebra, we often deal with systems of linear equations. These are equations that graph as straight lines and can intersect, be parallel, or coincide with each other.
Solving a system enables us to find the values of the unknown variables that satisfy all given equations simultaneously.

• When the equations intersect at a point, there is one unique solution, meaning the system is consistent and independent.
• If the equations are parallel, there is no solution, making the system inconsistent.
• When the equations coincide, they have infinitely many solutions, indicating the system is consistent and dependent.

In our exercise, we've solved a system using the elimination method, which helps in finding the common solution for the variables involved.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations and their representations through matrices and vector spaces. It deals with the properties and behavior of vector spaces and linear mappings between such spaces.
Linear algebra is fundamental in understanding systems of linear equations as it provides various methods and techniques for solving them, such as the elimination method used in our example.

• The key components include vectors, matrices, and linear transformations, which are instrumental in calculations involving linear systems.
• By translating the equations into matrix form, we can harness powerful algebraic tools to find solutions more efficiently than solving each equation individually.

Being a cornerstone of mathematical computations, linear algebra is utilized extensively in science and engineering fields for modeling and problem-solving.

Fractions in Equations

Fractions often appear in linear equations and can initially seem tricky. They require us to find a common denominator to add or subtract them, which can simplify solving systems of equations.
In our problem, we encountered fractions in the equations: \[ \frac{x}{4} + \frac{y}{2} = -1 \quad \text{and} \quad \frac{3}{8}x + \frac{5}{3}y = -\frac{7}{12} \]
To address these, we can rewrite the equations to have the same denominators, simplifying the arithmetic when using methods like elimination.

• Common denominators solve the problem of directly comparing or combining fractions.
• Applying these steps leads, step by step, to a solution without errors, even when handling intricate fractional equations.

Simplifying fractions early in the process can make subsequent calculations clearer and easier, ensuring accurate results when solving the system of equations.