Problem 9 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution is \(x = 1\) and \(y = -1\).

Step 1: Set Up the System of Equations

We have the following system of linear equations: \( 3x - 2y = 5 \) and \( 2x + 5y = -3 \). Our goal is to use the elimination-by-addition method to find values for \(x\) and \(y\).

Step 2: Equalize Coefficients for Elimination

To eliminate one of the variables, we aim to equalize the coefficients of either \(x\) or \(y\) by manipulating the equations. Let's multiply the first equation by 5 and the second equation by 2 to eliminate \(y\):\[ 5(3x - 2y) = 5(5) \] becomes \( 15x - 10y = 25 \).\[ 2(2x + 5y) = 2(-3) \] becomes \( 4x + 10y = -6 \).

Step 3: Eliminate One Variable

Now, add the two new equations together to eliminate \(y\):\[ (15x - 10y) + (4x + 10y) = 25 + (-6) \] simplifies to \( 19x = 19 \).

Step 4: Solve for the Remaining Variable

Solve for \(x\):\[ 19x = 19 \] \[ x = \frac{19}{19} = 1 \]. Now we know \( x = 1 \).

Step 5: Substitute to Find the Other Variable

Substitute \(x = 1\) back into one of the original equations to find \(y\). We'll use the first equation:\[ 3(1) - 2y = 5 \] becomes \( 3 - 2y = 5 \).Solve for \(y\):\[ -2y = 5 - 3 \] or \( -2y = 2 \).\[ y = \frac{2}{-2} = -1 \].

Step 6: Confirm the Solution

Substitute \(x = 1\) and \(y = -1\) into the second original equation to confirm:\[ 2(1) + 5(-1) = 2 - 5 = -3 \], which holds true.

Key Concepts

System of Linear Equations

A system of linear equations is a set of two or more linear equations with the same variables. In our exercise, we have the system:

• \(3x - 2y = 5\)
• \(2x + 5y = -3\)

A solution to this system is a pair of values for \(x\) and \(y\) that satisfy both equations simultaneously. In graphical terms, it is the intersection point of the lines represented by these equations. Systems of equations can have one solution, no solution, or infinitely many solutions.
Understanding the nature of these solutions can help in choosing the appropriate method to solve them, such as the substitution method, graphing method, or elimination method, which we focus on here.

Solving Equations

Solving a linear equation means finding the value of the variable that makes the equation true. For our system, once we decide to eliminate one variable, we target organizing both equations in a way that makes solving straightforward.
This involves:

• Manipulating equations to create equal coefficients for one variable.
• Adding or subtracting the equations to eliminate one variable completely.
• Solving the simpler equation for the remaining variable.

This step-by-step approach allows us to systematically break down the system into manageable parts, first solving for one variable and then the other. This is because, by reducing one variable, we simplify the complex system into a single linear equation, which is much easier to solve.

Elimination Method

The elimination method is a popular technique used for solving systems of linear equations. It involves adjusting the equations to eliminate one of the variables, making it easier to solve for the other. Here's a step-by-step breakdown of how it works:

• Adjust coefficients: Multiply the equations by certain numbers to align coefficients of one of the variables. In our example, we multiplied the first equation by 5 and the second by 2.
• Add or subtract equations: The goal is to cancel out one variable by adding or subtracting the equations. In our case, adding eliminated \(y\) completely.
• Solve the reduced equation: With one variable gone, solve for the remaining one. Here, we found \(x = 1\).
• Substitute back:\ Use the found value to substitute back into one of the original equations, solving for the second variable, yielding \(y = -1\).

This method is systematic and efficient, especially for larger systems, and ensures precision by minimizing potential arithmetic errors often encountered in more intricate algebraic methods.