Problem 13 - Intermediate Algebra

Question

For Problems \(1-18\), use the elimination-by-addition method to solve each system. (Objective 1 ) $$ \left(\begin{array}{l} 5 x+4 y=1 \\ 3 x-2 y=-1 \end{array}\right) $$

Step-by-Step Solution

Answer

\( x = -\frac{1}{11} \), \( y = \frac{4}{11} \)

Step 1: Identify the System of Equations

We have two equations that form a system: \( 5x + 4y = 1 \) and \( 3x - 2y = -1 \). Our goal is to solve for \( x \) and \( y \) using the elimination-by-addition method.

Step 2: Adjust Equations for Elimination

To eliminate one of the variables, we need to make the coefficients of either \( x \) or \( y \) opposites in the two equations. We will eliminate \( y \) by multiplying the entire first equation by 2, giving us: \( 10x + 8y = 2 \), and multiplying the entire second equation by 4, giving us: \( 12x - 8y = -4 \).

Step 3: Add the Equations

With the equations \( 10x + 8y = 2 \) and \( 12x - 8y = -4 \), we add them together to eliminate \( y \). The new equation becomes: \( 10x + 8y + 12x - 8y = 2 - 4 \), resulting in \( 22x = -2 \).

Step 4: Solve for x

Solve the equation \( 22x = -2 \) by dividing both sides by 22. We find \( x = \frac{-2}{22} = -\frac{1}{11} \).

Step 5: Substitute x into One of the Original Equations

Substitute \( x = -\frac{1}{11} \) into the first equation: \( 5(-\frac{1}{11}) + 4y = 1 \). This simplifies to \( -\frac{5}{11} + 4y = 1 \).

Step 6: Solve for y

Add \( \frac{5}{11} \) to both sides to isolate \( 4y \): \( 4y = 1 + \frac{5}{11} = \frac{11}{11} + \frac{5}{11} = \frac{16}{11} \). Divide both sides by 4 to solve for \( y \): \( y = \frac{16}{11} \div 4 = \frac{16}{11} \times \frac{1}{4} = \frac{4}{11} \).

Step 7: Verify the Solution

Substitute \( x = -\frac{1}{11} \) and \( y = \frac{4}{11} \) back into the original equations to ensure they hold true. Both equations should reflect: \( 5(-\frac{1}{11}) + 4(\frac{4}{11}) = 1 \) and \( 3(-\frac{1}{11}) - 2(\frac{4}{11}) = -1 \).

Key Concepts

System of Equations

A "system of equations" is essentially a collection of two or more equations with a common set of variables. In algebra, solving these systems means finding the values that satisfy all equations simultaneously. Consider the given system:

- Equation 1: \( 5x + 4y = 1 \)- Equation 2: \( 3x - 2y = -1 \)
The values for \(x\) and \(y\) should not only work for the first equation but also the second. This means the solutions of the system represent the point of intersection between the two lines when graphed. Solving systems like these helps identify relationships and changes among different variables, making it ideal for real-world applications like finance, engineering, and physics.

Substitution

Substitution is one method used to solve systems of equations, but it takes a different approach than elimination. In substitution, one of the equations is solved for one variable in terms of the other. Then, this expression is substituted into the other equation. This can be particularly useful when one equation is easily rearranged, allowing you to express one variable explicitly.

For example, if we had solved for \(x\) in terms of \(y\) from Equation 1, we could then substitute this expression into Equation 2 to solve for \(y\). Once \(y\) is known, we could go back to find \(x\). This method is useful when one equation is easier to manipulate than the other, such as when a variable already has a coefficient of 1. However, in our exercise, the elimination method was more straightforward due to the need for simple arithmetic adjustments to balance coefficients of variables.

Linear Algebra

Linear algebra provides the theoretical foundation for systems of equations. It explores the concepts surrounding vectors, matrices, and linear transformations. Our system from the exercise can be represented as a matrix equation:

\[\begin{pmatrix}5 & 4 \3 & -2\end{pmatrix}\begin{pmatrix}x \y\end{pmatrix}=\begin{pmatrix}1 \-1\end{pmatrix}\]In this form, each equation correlates to a row, and the solutions for \(x\) and \(y\) can be found by employing matrix operations like row reduction or matrix inversion, especially when dealing with larger systems. Understanding linear algebra enhances comprehension of solving systems of equations by providing insights into how equations interact and depend on one another.

Variables Elimination

The "variables elimination" technique is a preferred strategy when you want to remove a variable to solve a system. This method involves manipulating the system so that adding the equations together cancels one variable, simplifying the system to an equation with a single unknown.

Here's how it worked in the solved exercise:- By multiplying the first equation by 2 and the second by 4, the coefficients of \(y\) become opposites: \(8y\) and \(-8y\).- Adding these modified equations together eliminates \(y\), leaving \(22x = -2\).This strategy is useful in balancing an equation easily by straightforward arithmetic manipulation, which simplifies complex systems, making it an efficient method especially when equations are neither easy to substitute nor symmetrically structured.