Problem 12 - College Algebra

Question

Solve each system. $$\begin{array}{r}x+\quad z=3 \\\x+2 y-z=1 \\\2 x-y+z=3\end{array}$$

Step-by-Step Solution

Answer

The solution for (x, y, z) is (2, 2, 1).

Step 1: Arrange the equations

Rearrange the equations so that the variables line up and the equals side are on the same side. In this case, the equations are already arranged nicely with the coefficients of the variables in order.

Step 2: Simplify the Equations

From the first equation \( x + z = 3 \), simplify it to find the value of x in terms of z. So, \( x = 3 - z \).

Step 3: Substitute x in second and third equations

Substitute the value of x from Step 2 into the second and third equations. This gives: 2(3 - z) + z - 1 = 2y (Substituting in the second equation) 2(3 - z) - y + z = 3 (Substituting in the third equation)

Step 4: Simplify the new equations

Solving the two new equations gives: y = 5 - 3z (From the second equation) y = 3 - z (From the third equation)

Step 5: Equate the two y values and solve for z

Now equate 5 - 3z = 3 - z and solve for z. This gives: z = 1.

Step 6: Substitute z into x and y

Substitute z = 1 into x = 3 - z and y = 5 - 3z. This gives: x = 3 - 1 = 2 y = 5 - 3(1) = 2.

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of linear equations. It involves solving one of the equations for one of the variables, and then substituting that expression into the other equations. This reduces the number of equations and variables, making it easier to find the solution.

In our exercise, we used the substitution method by first rearranging the equation \( x + z = 3 \) to solve for \( x \), resulting in \( x = 3 - z \). This step is crucial as it allows us to replace \( x \) in the remaining equations with \( 3 - z \).

By doing this substitution in the second \( x + 2y - z = 1 \) and third equations \( 2x - y + z = 3 \), the system of equations is simplified and the variables reduced, making the overall solving process more straightforward.

• Solve one variable in terms of the others.
• Substitute this expression into the other equations.
• Reduce the system to a more manageable form.

Simultaneous Equations

Simultaneous equations involve finding common solutions for two or more equations that have the same set of unknowns. These equations are solved together because their solutions must satisfy all the equations simultaneously.

In the given problem, we are dealing with three equations:

• \( x + z = 3 \)
• \( x + 2y - z = 1 \)
• \( 2x - y + z = 3 \)

These equations form a system where we must find values for \( x \), \( y \), and \( z \) that satisfy all three equations at the same time.

Each variable represents a dimension in the respective equations, and the goal is to find the point in the three-dimensional space where all these planes intersect. By using methods like substitution, we can work through the system to find these common solutions.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them easier to solve. This often includes operations such as adding, subtracting, multiplying, or dividing both sides of an equation by the same number, or substituting one variable's expression from one equation into another.

In the solution provided, algebraic manipulation plays a big role. We start by expressing \( x \) in terms of \( z \) from the first equation, which simplifies the system when substituted in the other equations. This manipulation allows us to handle the equations more effectively.

We also performed additional algebraic steps, simplifying equations involving \( y \) from:

• \( 2(3-z) + z - 1 = 2y \)
• \( 2(3-z) - y + z = 3 \)

and equating the results to further isolate \( z \). By simplifying and equating, we are actively using algebra to distill the initial complex system down to something solvable, demonstrating the power of algebraic techniques to break down complex systems into manageable parts.