Problem 40 - Intermediate Algebra

Question

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} r-s+t=4 \\ r+2 s-t=-1 \\ r+s-3 t=-2 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( r = 2 \), \( s = -1 \), and \( t = 1 \).

Step 1: Arrange the System of Equations

We are given the system of linear equations:\[\begin{array}{c}r-s+t=4 \r+2s-t=-1 \r+s-3t=-2 \ \end{array}\]Our goal is to solve for variables \(r\), \(s\), and \(t\).

Step 2: Eliminate a Variable from Equations (1) and (2)

Start by subtracting Equation (1) from Equation (2) to eliminate \(r\):\[(r + 2s - t) - (r - s + t) = -1 - 4\]This simplifies to:\[3s - 2t = -5\]

Step 3: Eliminate a Variable from Equations (1) and (3)

Next, subtract Equation (1) from Equation (3) to eliminate \(r\):\[(r + s - 3t) - (r - s + t) = -2 - 4\]This simplifies to:\[2s - 4t = -6\]

Step 4: Simplify and Use the New Equations

Now we have the new system with two equations involving \(s\) and \(t\):\[\begin{array}{c}3s - 2t = -5 \2s - 4t = -6 \ \end{array}\]Divide the second equation by 2 to simplify it:\[s - 2t = -3\]

Step 5: Solve the Simplified System for One Variable

Using the simplified equations:1. \(3s - 2t = -5\)2. \(s - 2t = -3\)Subtract equation 2 from equation 1:\[(3s - 2t) - (s - 2t) = -5 - (-3)\]This gives us:\[2s = -2\]Divide both sides by 2:\[s = -1\]

Step 6: Solve for Another Variable

Substitute \(s = -1\) back into one of the simplified equations, for example, \(s - 2t = -3\):\[-1 - 2t = -3\]Add 1 to both sides:\[-2t = -2\]Divide by -2:\[t = 1\]

Step 7: Solve for the Final Variable

Now substitute \(s = -1\) and \(t = 1\) back into one of the original equations, such as \(r - s + t = 4\):\[r - (-1) + 1 = 4\]This simplifies to:\[r + 2 = 4\]Subtract 2 from both sides:\[r = 2\]

Step 8: Verify the Solution

Substitute \(r = 2\), \(s = -1\), and \(t = 1\) into all original equations to ensure they are satisfied:1. \(r - s + t = 4\) gives \(2 - (-1) + 1 = 4\) (holds true)2. \(r + 2s - t = -1\) gives \(2 + 2(-1) - 1 = -1\) (holds true)3. \(r + s - 3t = -2\) gives \(2 - 1 - 3 = -2\) (holds true)

Key Concepts

Linear Equations

Linear equations are the building blocks of systems of equations. They are equations of the first degree, which means they contain variables raised only to the power of one. Each term is either a constant or the product of a constant and a single variable.

When solving systems of linear equations, you are essentially finding the values of the variables that satisfy all equations in the system simultaneously. For example, in our original exercise, we have three linear equations with three unknowns, which are usually represented in a standard form like this:

• \(r - s + t = 4\)
• \(r + 2s - t = -1\)
• \(r + s - 3t = -2\)

These equations are solved by methods such as substitution, elimination, or matrix operations, to find the values of the variables \(r\), \(s\), and \(t\). In typical solutions, the goal is to simplify the system to the point where each variable can be easily solved one by one.

Inconsistent Systems

Not all systems of linear equations have a solution. When a system has no solution, it is termed "inconsistent." This occurs when the equations in the system contradict each other.

For instance, if we rearrange the equations and find conflicting conditions such as trying to solve something like:

• \(0 = 5\)

It implies a contradiction because such a statement is obviously false for any real number assignment to the variables. This misunderstanding indicates that no common set of values for the variables will satisfy all equations at once.

In our provided exercise, however, the system was consistent, showing a real set of solutions. If it were inconsistent, at some point in the solution process, contradictory results would arise. Spotting such inconsistencies is essential, especially in real-world problems, to avoid erroneous conclusions.

Dependent Equations

Dependent equations are another concept you may encounter in systems of linear equations. Sometimes, two or more equations in a system may not be independent of each other, meaning one can be derived from a linear combination of the other(s).

In terms of solutions, a system of linear equations with dependent equations may end up having infinite solutions rather than a unique solution. The equations essentially "overlap," and solving them will reveal that they are representing the same constraint.

If solving our given system had led to a valid equation like:

• \(0 = 0\)

across multiple simplifications without determining unique values for the variables, it would indicate dependent equations resulting in infinitely many solutions.

Knowing the difference between dependent and independent equations is critical to understanding why a system cannot be solved normally and helps in accurately interpreting the meaning behind the mathematical relationships in applied scenarios.