Problem 5 - College Algebra

Question

Solve each system. $$\begin{aligned}&x+y+2 z=11\\\&x+y+3 z=14\\\&x+2 y-z=5\end{aligned}$$

Step-by-Step Solution

Answer

The solutions for the system are \(x = 1, y = 3, z = 3\)

Step 1: Identify the Relationships

When comparing the first two equations: \(x + y + 2z = 11\) and \(x + y + 3z = 14\), it becomes apparent that they can be subtracted from each other to eliminate the \(x\) and \(y\) variables, leaving an equation with only \(z\). This is a strategy for simplifying the system. Similarly, the second and third equations can be subtracted to find another singular variable equation.

Step 2: Simplify

The aforementioned strategy leads to two simple equations. By subtracting the first equation from the second, the result is \(z = 3\). By subtracting the second equation from the third, it leads to \(y - 4z = -9\). Plugging the earlier found \(z = 3\) into it gives \(y - 4*3 = -9\) which simplifies to \(y = 3\). The third equation can now be used for finding \(x\).

Step 3: Finalize the Solution

Substituting the found values of \(y\) and \(z\) into the third equation gives \(x + 2*3 - 3 = 5\) and simplifying result in \(x = 1\). Thus the solutions are \(x = 1\), \(y = 3\), and \(z = 3\).

Key Concepts

Elimination Method

The elimination method is a powerful technique for solving systems of linear equations. It involves manipulating the equations in a system to eliminate one variable at a time, simplifying the process of finding the solution. This method is particularly useful when the system consists of more than two equations and variables.

The key idea is to add or subtract equations in order to cancel out one of the variables. In the provided exercise, let's look at how this applies:

• Consider the equations: \(x + y + 2z = 11\) and \(x + y + 3z = 14\). By subtracting one from the other, we eliminate both \(x\) and \(y\), simplifying down to \(z = 3\).
• Similarly, by manipulating other combinations of equations, we simplify the system progressively until finding a solution for each variable.

Using the elimination method requires careful attention to ensure the operations are applied correctly, maintaining the equality across the equations.

Ultimately, this process leads us to simpler equations involving fewer variables, making it easier to solve the system step by step.

Solving Linear Equations

Linear equations are equations of the first order; they graph as straight lines. Solving them involves finding the values of the variables that make the equation true. In systems of linear equations, multiple linear equations are involved, and the solution is where these lines intersect.

For example, in the original exercise:

• We start with three equations and three unknowns: \(x + y + 2z = 11\), \(x + y + 3z = 14\), and \(x + 2y - z = 5\).
• The goal is to determine the set of values \((x, y, z)\) that satisfies all three equations simultaneously.

The steps typically involve:

• Isolating variables through strategies like substitution or elimination to reduce the system to simpler equations.
• Once reduced, solving for one variable at a time and substituting back to find the remaining variables.

In this problem, systematic elimination helped to achieve simplified forms, allowing the discovery of solutions \(x = 1\), \(y = 3\), and \(z = 3\). Efficient solving is about reducing complexity gradually until solutions are evident.

Substitution Method

The substitution method is another powerful approach to solve systems of equations, particularly when one equation in the system is easy to solve for one of the variables. It involves substituting this expression back into the other equations, reducing the number of variables step by step.

In our example, though primarily solved using elimination, we can see the potential application of substitution:

• After finding \(z = 3\), this value was substituted into the derived equation \(y - 4z = -9\). Solving for \(y\) became straightforward, leading to \(y = 3\).
• Then, these known values of \(y\) and \(z\) were substituted back into one of the original equations to solve for \(x\), finalizing it as \(x = 1\).

The substitution method uniquely breaks down the problem, simplifying each equation to involve one less variable with each substitution.

This method is particularly useful when one equation is already solved for one variable, making it intuitive to apply and fostering a clear pathway to solve for others.