Problem 31 - Intermediate Algebra

Question

Solve each system using substitution. $$ \left\\{\begin{array}{l} r+s-3 t=21 \\ r+4 s=9 \\ 5 s+t=-4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( r = 17 \), \( s = -2 \), \( t = 6 \).

Step 1: Choose an Equation for Substitution

First, it's helpful to manipulate one of the equations to express one variable in terms of the others. Let's take the second equation: \( r + 4s = 9 \), and solve it for \( r \).

Step 2: Solve for the Chosen Variable

From the equation \( r + 4s = 9 \), solve for \( r \) by isolating it on one side: \( r = 9 - 4s \).

Step 3: Substitute into Another Equation

Substitute \( r = 9 - 4s \) found from Step 2 into the first equation \( r + s - 3t = 21 \). Doing so gives: \( (9 - 4s) + s - 3t = 21 \).

Step 4: Simplify and Solve for One Variable

After substitution, simplify the equation: \( 9 - 3s - 3t = 21 \). Solving for \( t \) gives: \( t = \frac{12 - 3s}{3} \) or \( t = 4 - s \).

Step 5: Substitute Back into a Different Equation

Now use the expression for \( t \) in terms of \( s \), \( t = 4 - s \), and substitute it into the third equation \( 5s + t = -4 \): \( 5s + (4 - s) = -4 \).

Step 6: Solve for the Second Variable

Solve \( 5s + 4 - s = -4 \), this simplifies to \( 4s = -8 \). Solving for \( s \) gives \( s = -2 \).

Step 7: Solve for the Remaining Variables

Substitute \( s = -2 \) back into \( t = 4 - s \): \( t = 4 + 2 = 6 \). Now substitute \( s = -2 \) into \( r = 9 - 4s \): \( r = 9 - (-8) = 17 \).

Step 8: Verify the Solution

Check the solution \( r = 17 \), \( s = -2 \), \( t = 6 \) against the original equations to ensure they hold true. 1. Equation 1: \( 17 + (-2) - 3(6) = 17 - 2 - 18 = 21 \), which is correct. 2. Equation 2: \( 17 + 4(-2) = 17 - 8 = 9 \), which is correct. 3. Equation 3: \( 5(-2) + 6 = -10 + 6 = -4 \), which is correct.

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations that share common variables. In the exercise, we are given a system with three equations:

• \( r + s - 3t = 21 \)
• \( r + 4s = 9 \)
• \( 5s + t = -4 \)

The goal is to find values for the variables \( r \), \( s \), and \( t \) that satisfy all given equations simultaneously. Solving such systems often involves methods like substitution or elimination. Here, we've used substitution to gradually express each variable in terms of others until we find a common set of solutions. By handling each equation as a piece of a bigger puzzle, we aim to reach values for \( r \), \( s \), and \( t \) that keep the mathematical balance intact, maintaining the truth of each original equation.

Algebraic Manipulation

Algebraic manipulation is key to solving systems of equations. It involves rearranging equations, combining like terms, and simplifying expressions to make solving easier. In this exercise, we started with the equation \( r + 4s = 9 \).By isolating \( r \), we rewrote it as \( r = 9 - 4s \), turning this equation into an expression that can substitute "r" in the others.There are several algebraic techniques used in this exercise:

• Isolating Variables: This is done by moving terms to one side using addition or subtraction. We isolated \( r \) and later solved models for \( t \) and \( s \) in similar ways.
• Substitution: Carefully replacing one variable with its equivalent expression found in another equation can simplify a system significantly.
• Simplification: Reducing complex expressions by combining like terms. For instance, simplifying \( 9 - 3s - 3t = 21 \) allowed us to solve for \( t \).

Algebraic manipulation is an iterative process. It frequently involves moving back and forth between equations, gradually simplifying them to make each step clearer and more manageable. This skill is essential in mathematics and aids in handling more complex systems.

Solving Equations Step by Step

Solving equations involves patiently working through each equation one step at a time. A systematic, step-by-step method helps in isolating and finding the value of each variable. Here's how it was done in our exercise:

• Select a Starting Point: We chose to start with \( r + 4s = 9 \), as isolating \( r \) seemed straightforward.
• Substitute Wisely: We substituted the expression for \( r \) into \( r + s - 3t = 21 \), and later did similarly with \( t \) into the other equations. This strategy keeps equations simpler, allowing gradual easing into a solution.
• Solve Iteratively: After isolating terms, we worked the calculations one at a time: solving for \( t \) led directly to solving for \( s \), and finally for \( r \).
• Verify the Solution: It's vital to substitute the found values back into the original equations to check if they truly solve all equations. Verification confirmed our solutions \( r = 17 \), \( s = -2 \), and \( t = 6 \).

Each step is like climbing a stair. Checking each box ensures stability before moving to the next. Always remember, a methodical approach in solving equations builds a strong foundation for tackling more challenging algebraic tasks.