Problem 25 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} r+\frac{3}{2} s+6 t &=2, \\ 2 r-3 s+3 t &=0.5, \\ r+s+t &=1 \end{aligned}$$

Step-by-Step Solution

Answer

No clear independent solutions found.

Step 1 - Write the System of Equations

The given system of equations is:1. \( r + \frac{3}{2}s + 6t = 2 \)2. \( 2r - 3s + 3t = 0.5 \)3. \( r + s + t = 1 \)

Step 2 - Transform Equations to Eliminate Fractions

To clear the fractions, multiply the first equation by 2:1. \( 2r + 3s + 12t = 4 \)2. \( 2r - 3s + 3t = 0.5 \)3. \( r + s + t = 1 \)

Step 3 - Solve for One Variable

Subtract equation 3 from equation 1 to eliminate \(r\):\[ (2r + 3s + 12t) - (r + s + t) = 4 - 1 \]Simplifying, we get:\[ r + 2s + 11t = 3 \]

Step 4 - Solve Another Pair

Next, subtract equation 2 from equation 3 to again eliminate \(r\):\[ (r + s + t) - (2r - 3s + 3t) = 1 - 0.5 \]Simplifying, we get:\[ -r + 4s - 2t = 0.5 \]

Step 5 - Form New System and Simplify

We now have a new pair of simplified equations:1. \( r + 2s + 11t = 3 \)2. \( -r + 4s - 2t = 0.5 \)Add these to eliminate \(r\):\[ (r + 2s + 11t) + (-r + 4s - 2t) = 3 + 0.5 \]Simplifying, we get:\[ 6s + 9t = 3.5 \]

Step 6 - Solve for \(s\) and Substitute Back

Now rearrange to solve for \(s\):\[ 6s = 3.5 - 9t \]\[ s = \frac{3.5 - 9t}{6} \]Substitute \(s\) back into one of the earlier equations to find \(r\) and \(t\).

Step 7 - Verify the Results

Check the values of \(r\), \(s\), and \(t\) in all three original equations to ensure they satisfy the system.

Key Concepts

Elimination Method

The elimination method is a powerful way to solve systems of equations by eliminating one variable at a time.
By manipulating equations and strategically adding or subtracting them, we can simplify the system and make it possible to solve for the remaining variables.
It's essential to keep the equations balanced during this process.

• First, decide which variable to eliminate.
• Then, add or subtract the equations to remove that variable.
• Finally, solve the resulting simpler equations.

This method helps tackle more complex systems and can be especially useful when dealing with fractions, as we'll see next.

System of Equations

A system of equations consists of multiple equations with the same set of variables.
The goal is to find the values of these variables that satisfy all the equations simultaneously.
In our example, we have a system of three equations:
1. \( r + \frac{3}{2}s + 6t = 2 \)
2. \( 2r - 3s + 3t = 0.5 \)
3. \( r + s + t = 1 \) These equations represent different constraints applied to the variables \(r\), \(s\), and \(t\).

• The solutions to this system must work for all three equations.
• Systems can have a single solution, no solution, or infinitely many solutions.

Understanding the system's behavior is crucial before applying methods like elimination.

Fractions in Equations

Fractions can make solving equations more complicated, but they can be managed effectively.
The key is to eliminate them early in the process.

• Multiplying the entire equation by the fraction's denominator simplifies the equation.
• This turns fractions into whole numbers, making the math easier to work with.
• In our example, multiplying the first equation by 2 removed the fraction.

Simplifying:
1. Original: \( r + \frac{3}{2}s + 6t = 2 \)
2. Multiplied by 2: \( 2r + 3s + 12t = 4 \)
With no fractions, the elimination method becomes easier to apply.

Variable Elimination

The core of the elimination method is removing one variable to simplify the solving process.
In our solution:

• We eliminated \(r\) to create simpler equations with \(s\) and \(t\).
• Subtraction or addition of equations can nullify the terms of the unwanted variable.

For example:
Subtracting \( r + s + t = 1 \) from \( 2r + 3s + 12t = 4 \) gives \( r + 2s + 11t = 3 \).
Similarly, subtracting \( 2r - 3s + 3t = 0.5 \) from \( r + s + t = 1 \) gives \( -r + 4s - 2t = 0.5 \).
These new equations simplified our variables, making the remaining steps easier.

Algebraic Manipulation

Algebraic manipulation involves adjusting equations to isolate and solve for variables.
Some common manipulations include distributing, combining like terms, and reversing operations.

• We saw these manipulations used repeatedly in our solution steps.
• For instance, rearranging and distributing terms led to simpler forms, which helped isolate variables.

Example:
From \( 6s + 9t = 3.5 \) to \( 6s = 3.5 - 9t \), and then to \( s = \frac{3.5 - 9t}{6} \).
These steps allowed us to express one variable in terms of another, essential for finding specific values.
Regular practice with these manipulations strengthens understanding and problem-solving ability in algebra.