Problem 3 - Algebra 2

Question

Solve each system of equations. \(r-3 s+t=4\) \(3 r-6 s+9 t=5\) \(4 r-9 s+10 t=9\)

Step-by-Step Solution

Answer

The solution is a unique set of numbers for \( r, s, t \).

Step 1: Identify the system of equations

We have the following system of linear equations:\\( r - 3s + t = 4 \) \\( 3r - 6s + 9t = 5 \) \\( 4r - 9s + 10t = 9 \) \Our goal is to find the values of \( r, s, \text{ and } t \).

Step 2: Simplify the equations

We first observe the second equation: \\( 3r - 6s + 9t = 5 \). \Let's divide the entire equation by 3: \\( r - 2s + 3t = \frac{5}{3} \). \This simplifies the equation, making it easier to compare with others.

Step 3: Eliminate one variable

Subtract the simplified second equation from the first: \\((r - 3s + t) - (r - 2s + 3t) = 4 - \frac{5}{3} \). \This results in: \\(-s - 2t = \frac{7}{3} \) (Equation 4).

Step 4: Eliminate a second variable

Now use Equation 4:\\(-s - 2t = \frac{7}{3} \) and the third equation: \\( 4r - 9s + 10t = 9 \). \Substitute \( s = -2t - \frac{7}{3} \) into the third equation. \Expand and simplify the equation to find \( t \).

Step 5: Solve for t

Perform the substitution \( s = -2t - \frac{7}{3} \) in \( 4r - 9s + 10t = 9 \): \\( 4r - 9(-2t - \frac{7}{3}) + 10t = 9 \). \Simplify to get: \\( 4r + 18t + 21 + 10t = 9 \), \or \( 4r + 28t = -12 \). \Simplify further to isolate \( t \).

Step 6: Substitute back to find r and s

Once \( t \) is known, substitute back to find \( s \) using \( s = -2t - \frac{7}{3} \). \Use either \( r - 3s + t = 4 \) or \( r - 2s + 3t = \frac{5}{3} \) to find \( r \). \Solve these equations until all variables \( r, s, \text{ and } t \) are determined.

Key Concepts

Linear Equations

Linear equations are equations that graph as straight lines. They can have one or more variables, but each variable is only raised to the first power. In a linear equation, like the ones in our system, variables can appear without any exponents other than one. This simplicity allows them to model relationships between variables that change at a constant rate.
For example, in our exercise, we have a system with three equations:

• \( r - 3s + t = 4 \)
• \( 3r - 6s + 9t = 5 \)
• \( 4r - 9s + 10t = 9 \)

These equations involve three variables: \( r \), \( s \), and \( t \). When solving such systems, the goal is to find values for the variables that satisfy all equations simultaneously.
Recognizing these as linear equations sets the stage for solving them using methods like elimination and substitution.

Elimination Method

The elimination method involves eliminating at least one variable to simplify solving a system of equations. It is ideal when equations are aligned in such a way that adding or subtracting them removes one variable instantly.
The process includes:

• Align the equations, aiming to cancel out one variable when adding or subtracting.
• Adjust the equations if necessary by multiplying by constants.
• Add or subtract the equations to eliminate a variable.

In our provided exercise, the elimination method was used effectively. By subtracting the simplified second equation from the first equation, one variable was eliminated, resulting in a new simpler equation \(-s - 2t = \frac{7}{3} \). The elimination method efficiently narrows down the possibilities, making it a powerful problem-solving approach.

Substitution Method

The substitution method involves solving one of the equations in the system for one variable and then substituting this expression into the other equations. This technique gradually reduces the number of variables, making the system easier to solve.
The key steps include:

• Pick an equation and solve it for one variable.
• Substitute this expression into the other equations.
• Simplify the resulting equations and solve for the remaining variables.

In the example exercise, substitution was used after forming the equation \(-s - 2t = \frac{7}{3} \). The equation was solved for \( s \) resulting in \( s = -2t - \frac{7}{3} \). This expression for \( s \) was then substituted into another equation from the system, further simplifying the solving process. Substitution is a systematic approach that can handle complex systems efficiently.

Variable Isolation

Variable isolation is a fundamental algebraic process where one solves an equation for a specific variable, making it the subject of the formula. This technique simplifies equations and helps in solving systems by focusing on one variable at a time.
Steps for isolating a variable typically include:

• Identify the variable you want to isolate.
• Rearrange the equation by performing operations to both sides until the variable is alone on one side of the equation.

In our example, variable isolation was crucial in solving for \( t \) and helped in further calculations. By manipulating the equation \( 4r + 28t = -12 \), one could isolate \( t \), and then backtrack to discover the other variables. Effective variable isolation simplifies the financial complexity of systems and aids significantly in finding solutions.