Problem 6 - Algebra 2

Question

Solve each system of equations by using elimination. \(2 r-3 s=11\) \(2 r+2 s=6\)

Step-by-Step Solution

Answer

The solution is \(r = 4\) and \(s = -1\).

Step 1: Set Up for Elimination

To eliminate one of the variables, we can align the two equations given: 1. \(2r - 3s = 11\)2. \(2r + 2s = 6\).Notice that both equations have the term \(2r\), making it easier to use elimination.

Step 2: Subtract the Equations

Subtract the second equation from the first to eliminate \(2r\):\[(2r - 3s) - (2r + 2s) = 11 - 6\]Simplify the terms:\[-5s = 5\].

Step 3: Solve for s

Divide both sides of the equation \(-5s = 5\) by \(-5\) to solve for \(s\):\[s = -1\].

Step 4: Substitute s Back into One Equation

Use the value of \(s\) to substitute back into one of the original equations to find \(r\). Let's choose the second equation:\(2r + 2(-1) = 6\).Simplify and solve for \(r\):\[2r - 2 = 6\]\[2r = 8\]\[r = 4\].

Step 5: Solution of the System

The solution to the system of equations is \(r = 4\) and \(s = -1\).

Step 6: Verification

Verify the solution by substituting \(r = 4\) and \(s = -1\) back into the original equations:For the first equation: \(2(4) - 3(-1) = 8 + 3 = 11\) (Correct).For the second equation: \(2(4) + 2(-1) = 8 - 2 = 6\) (Correct).Both equations are satisfied, so the solution is correct.

Key Concepts

Understanding System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In simpler terms, these are equations that work together to find the values that satisfy all of them simultaneously. In our exercise, we have:

• Equation 1: \( 2r - 3s = 11 \)
• Equation 2: \( 2r + 2s = 6 \)

These equations are linear, meaning they graph as straight lines. The solution to a system of equations is the point where the graphs of these equations intersect. In an algebraic approach, these common solutions provide the values of the variables that satisfy all the equations in the system. This exercise uses the elimination method to solve for these values.

Exploring Substitution as a Method

Substitution is a useful technique in solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation.
This can simplify finding the values of the variables. However, the given exercise uses elimination, a complementary method to substitution. Although different, these methods aim to achieve the same goal: reducing the number of variables to solve more efficiently.
In practical scenarios, substitution is particularly handy when one of the variables is easily isolated. For example, if one equation is already in the form \( x = ... \) or \( y = ... \), substitution can directly use this into the other equation. Though our solution didn't use substitution, understanding it enriches comprehension of solving systems of equations.

Breaking Down Solving Equations

Solving equations involves finding the unknown values that make the equation true. Once we simplify the system using elimination, our goal is to solve for each variable.
In the provided exercise, the subtraction of equations results in:

• \(-5s = 5\)

This equation is simplified in one step, allowing us to solve for \( s \):

• Dividing both sides by \(-5\), we find \( s = -1 \).

After finding \( s \), we substitute it back into one of the original equations to solve for \( r \). This involves:

• Substituting into the second equation: \( 2r + 2(-1) = 6 \).
• Solving the equation, we simplify to find \( r = 4 \).

By consistently simplifying and executing arithmetic operations systematically, solving becomes straightforward. Each step requires attention to operations and simplifications.

Importance of Verification in Solutions

Verification is a crucial step in solving systems of equations. It ensures that our found solutions indeed satisfy all original equations.
For verification:

• Substitute \( r = 4 \) and \( s = -1 \) back into the original equations.
• Check that both equations hold true when these values are inserted.

For Equation 1: \( 2(4) - 3(-1) = 8 + 3 = 11 \), which is correct.
For Equation 2: \( 2(4) + 2(-1) = 8 - 2 = 6 \), which also holds.
Successful verification means the solution is accurate. Even if the solution seems correct, always double-check by plugging the values back into the original equations. Doing so confirms your accuracy and understanding of the problem.