Problem 39 - College Algebra with Modeling and Visualization

Question

Solve the system, if possible. $$ \begin{array}{r} x-y+z=1 \\ x+2 y-z=2 \\ y-z=0 \end{array} $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (1, 1, 1)\).

Step 1: Write the system of equations

We have the following system of linear equations: 1. \(x - y + z = 1\) 2. \(x + 2y - z = 2\) 3. \(y - z = 0\)

Step 2: Solve for one variable

Start with equation 3: \(y - z = 0\). We can solve for one variable in terms of the other, let's solve for \(z\): \(z = y\).

Step 3: Substitute into other equations

Substitute \(z = y\) into equations 1 and 2. For equation 1: \(x - y + y = 1\), which simplifies to \(x = 1\). For equation 2: \(x + 2y - y = 2\), which simplifies to \(x + y = 2\).

Step 4: Solve the simplified equations

From Step 3, we have \(x = 1\) and \(x + y = 2\). Substitute \(x = 1\) into \(x + y = 2\): \(1 + y = 2\). Solving for \(y\), we get \(y = 1\).

Step 5: Find the final value of z

Since we have \(z = y = 1\), \(z\) is also 1.

Key Concepts

Linear Equations

A linear equation is any equation that can be written in the form of \[a_1x_1 + a_2x_2 + ... + a_nx_n = b\] where \(a_1, a_2, ..., a_n\) and \(b\) are constants. The solutions to linear equations form straight lines when graphed on a coordinate plane. In our specific case, we dealt with three linear equations:

• \(x - y + z = 1\)
• \(x + 2y - z = 2\)
• \(y - z = 0\)

These equations are considered linear because each term is either a constant or the product of a constant and a single variable. Additionally, each variable is raised to the power of one, which is a defining property of linear equations. These equations were used to form a system to find the values of \(x\), \(y\), and \(z\). Understanding the nature of linear equations helps in recognizing them and solving them using various methods.

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves expressing one variable in terms of another and substituting this expression into the other equations. This method simplifies the original system to make it easier to solve.In our problem, we started with the equation:\[y - z = 0\]From here, we solved for \(z\):\[z = y\]This simple relationship between \(z\) and \(y\) allowed us to substitute \(z\) with \(y\) in the remaining equations. By doing so, we were able to reduce the number of variables and simplify the equations:

• In \(x - y + z = 1\), substituting gives \(x = 1\)
• In \(x + 2y - z = 2\), substituting results in \(x + y = 2\)

This effectively reduced the system to simpler equations with only one or two variables, which were much more manageable. Substitution is a powerful method, especially when one equation is simple enough to express a variable easily.

Solving Algebraic Equations

Solving algebraic equations involves finding the values of variables that satisfy the equation. In our system, after applying the substitution method, we ended up with simpler equations:

• \(x = 1\)
• \(x + y = 2\)

To find the value of \(y\), we substituted \(x = 1\) into \(x + y = 2\):\[1 + y = 2\]Subtracting 1 from both sides gives us:\[y = 1\]Finally, since \(z = y\), we conclude: \[z = 1\]Algebraic equations might initially seem intimidating, but reducing complex systems into simple forms by using algebraic techniques like substitution makes them much easier to handle. Practicing these steps helps to reinforce understanding and improve problem-solving skills, making it more intuitive over time.