Problem 32 - Intermediate Algebra

Question

Solve each system using substitution. $$ \left\\{\begin{array}{l} r-s+6 t=12 \\ r+6 s=-28 \\ 7 s+t=-26 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( r = -4 \), \( s = -4 \), \( t = 2 \).

Step 1: Express one variable in terms of the others from the second equation

Start with the second equation, which is \( r + 6s = -28 \). Solve for \( r \) in terms of \( s \):\[ r = -28 - 6s \]

Step 2: Substitute the expression for r into the first equation

Substitute \( r = -28 - 6s \) into the first equation \( r - s + 6t = 12 \):\[ (-28 - 6s) - s + 6t = 12 \]Simplify this to:\[ -28 - 7s + 6t = 12 \]Add 28 to both sides:\[ -7s + 6t = 40 \]

Step 3: Solve for t in terms of s from the third equation

Use the third equation \( 7s + t = -26 \) to solve for \( t \) in terms of \( s \):\[ t = -26 - 7s \]

Step 4: Substitute t from Step 3 into the modified first equation

Substitute \( t = -26 - 7s \) into \( -7s + 6t = 40 \):\[ -7s + 6(-26 - 7s) = 40 \]Simplify and solve for \( s \):\[ -7s - 156 - 42s = 40 \]Combine terms:\[ -49s - 156 = 40 \]Add 156 to both sides:\[ -49s = 196 \]Divide by \(-49\):\[ s = -4 \]

Step 5: Substitute s back into expression for t

Substitute \( s = -4 \) into \( t = -26 - 7s \):\[ t = -26 - 7(-4) \]Calculate \( t \):\[ t = -26 + 28 = 2 \]

Step 6: Substitute s back into expression for r

Substitute \( s = -4 \) into \( r = -28 - 6s \):\[ r = -28 - 6(-4) \]Calculate \( r \):\[ r = -28 + 24 = -4 \]

Key Concepts

Systems of Equations

Systems of equations involve multiple equations that are considered simultaneously. The goal is often to find the values of the variables that make all given equations true at the same time.
In our exercise, we are working with a system of three linear equations with three variables: \(r\), \(s\), and \(t\). These equations are:

• \(r - s + 6t = 12\)
• \(r + 6s = -28\)
• \(7s + t = -26\)

Analyzing the system of equations involves using methods like substitution or elimination to work towards the solution. The ultimate aim is to isolate and determine the value for each variable. This ensures the given system is satisfied by these values. Each equation gives us a different piece of the puzzle, and we have to cleverly use each to find the solution.

Algebraic Manipulation

Algebraic manipulation is a crucial skill when dealing with systems of equations. It involves rearranging and simplifying equations to isolate and solve for the variables.
In this exercise, algebraic manipulation was used to express one variable in terms of the others.
For example:

• From the equation \(r + 6s = -28\), we solved for \(r\) as \(r = -28 - 6s\).
• In the equation \(-7s + 6t = 40\), which arose from substitution, we used basic operations to get this simplified form: \(-28 - 7s + 6t = 12 \rightarrow -7s + 6t = 40\).

Being comfortable with these processes allows for seamless navigation through more complex equations or systems. Mastery of algebraic manipulation simplifies the steps to the solution by making it easier to substitute and solve for remaining variables.

Linear Equations

Linear equations are equations of the first degree, which means they include variables raised only to the power of one.
Each equation in our system is a linear equation involving the variables \(r\), \(s\), and \(t\). The general form of a linear equation is \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(x\), \(y\), and \(z\) are the variables.

• In \(r - s + 6t = 12\), the coefficients are \(1, -1,\) and \(6\) for \(r, s,\) and \(t\), respectively, and the equation is equal to 12.
• Each linear equation describes a plane in a three-dimensional space, and the solution for the system is the point where all three planes intersect.

Understanding the structure and form of linear equations facilitates easier implementation of systematic techniques like substitution to solve the equations simultaneously.

Problem Solving Steps

Solving a system of equations using the substitution method involves a series of organized steps. Following a structured approach ensures clarity and efficiency in finding the solution.
Here’s a breakdown of the steps applied in the exercise:

• **Step 1**: Begin by isolating one variable. From \(r + 6s = -28\), solve for \(r\), giving \(r = -28 - 6s\).
• **Step 2**: Substitute this expression into another equation. This was done with the first equation \( r - s + 6t = 12\) by replacing \(r\).
• **Step 3**: Simplify and then solve for another variable. From the resulting equation and the third equation, simplify to find a variable, in this case, \(t\) in terms of \(s\).
• **Step 4**: Substitute back into equations to find definitive values. This involves substituting found expressions back into equations to solve for all variable values.

Following these steps logically allows for solving systems of equations systematically, helping to prevent errors and ensuring that all variables are accounted for in the solution.