Problem 20 - Algebra 2

Question

Solve each system of equations by using elimination. \(r+4 s=-8\) \(3 r+2 s=6\)

Step-by-Step Solution

Answer

The solution to the system is \( r = 4 \) and \( s = -3 \).

Step 1: Prepare the Equations

The given system of equations is: 1. \( r + 4s = -8 \) 2. \( 3r + 2s = 6 \). We aim to eliminate one of the variables by making the coefficients of one of the variables the same in both equations.

Step 2: Equalize Coefficients for Elimination

To eliminate variable \(s\), multiply the first equation by 2. This gives us: \( 2(r + 4s) = 2(-8) \) which simplifies to: \( 2r + 8s = -16 \). Now, the system of equations is: 1. \( 2r + 8s = -16 \) 2. \( 3r + 2s = 6 \).

Step 3: Subtract the Equations

Subtract the second equation from the modified first equation: \( (2r + 8s) - (3r + 2s) = -16 - 6 \) This simplifies to: \( -r + 6s = -22 \).This equation can be rearranged to find \( r \).

Step 4: Solve for Variable r

Rearrange the equation from the subtraction to solve for \( r \): \( -r = -22 - 6s \) So, \( r = 22 + 6s \).

Step 5: Substitute to Find s

Substitute \( r = 22 + 6s \) back into one of the original equations. Using the first equation: \( (22 + 6s) + 4s = -8 \) Simplifying gives: \( 22 + 10s = -8 \) Now, solve for \( s \): \( 10s = -8 - 22 \) \( 10s = -30 \) \( s = -3 \).

Step 6: Substitute to Find r

Now that we have \( s = -3 \), substitute back into \( r = 22 + 6s \): \( r = 22 + 6(-3) \) \( r = 22 - 18 \) \( r = 4 \).

Key Concepts

Understanding Systems of Equations

In mathematics, a system of equations is a set of two or more equations with the same set of unknowns. The task is to find values for these unknowns that make all the equations true simultaneously.
Systems can be solved using different methods, such as graphical methods, substitution, and elimination. Sometimes, they may have one solution, no solution, or many solutions.
For example, the system of equations in our original problem is \(r + 4s = -8\) and \(3r + 2s = 6\). The goal is to find values of \(r\) and \(s\) that satisfy both equations at the same time. Understanding the relationships between the equations is key to finding the solution.

Solving Equations Made Simple

Solving an equation involves finding the values of variables that make the equation true. This requires isolating the variable on one side of the equation.
There are a few steps to follow when solving equations:

• Simplify both sides of the equation if necessary.
• Use operations to eliminate numbers or terms from one side.
• Ensure you keep the equation balanced by performing the same operation on both sides.

In our original exercise, we rearrange the equations to isolate one variable, using algebraic methods like substitution and elimination. The ultimate goal is to find that single set of values that 'satisfies' or solves the entire system.

The Role of Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations to isolate and solve for variables. It's like solving a puzzle where you adjust pieces until they fit perfectly.
In our system of equations, we used multiplication to adjust coefficients, making elimination possible. By multiplying the entire first equation by 2, we created a new equation \(2r + 8s = -16\) that aligned with the coefficients of the second equation, \(3r + 2s = 6\).
Then, by subtracting the two modified equations, we effectively eliminated one variable, narrowing it down to a single linear equation with one variable. This opened the door for us to find the values for that variable and subsequently solve the system.

Mastering the Substitution Method

The substitution method is a key strategy for solving systems of equations, especially when one equation can easily be solved for one variable. Once this variable is found, you substitute its value back into another equation.
Consider the exercise at hand: after manipulating the equations, we expressed \(r\) in terms of \(s\) as \(r = 22 + 6s\). By substituting \(s = -3\) into this expression, we were able to find the value of \(r\).
Substitution involves these main steps:

• Solve one of the equations for one variable.
• Substitute that expression into the other equation.
• Solve the resulting equation for the remaining variable.

This method can simplify complex systems and often leads directly to the solution without requiring further manipulation.