Problem 12 - College Algebra

Question

Solve each system in Exercises \(5-18\). $$ \left\\{\begin{array}{c} x+\quad z=3 \\ x+2 y-z=1 \\ 2 x-y+z=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=1, y=1, z=2\).

Step 1: Express one variable in terms of other(s)

Resolve the first equation \(x+z=3\) into \(x=3-z\).

Step 2: Substitute in the other equations

Substitute the expression for x into the other two equations: \n For \(x+2y-z=1\), substituting \(x\) yields \(3-z+2y-z=1\) which simplifies to \(3+2y-2z=1\), then it becomes \(2y-2z=-2\) and finally, divide by 2 to get \(y-z=-1\). \n For \(2x-y+z=3\), substituting \(x\) yields \(2(3-z)-y+z=3\), simplifies to \(6-2z-y+z=3\), then it becomes \(-y-z=-3\) or \(y+z=3\).

Step 3: Solve the final system of equations

We now have a system of two equations with two variables \(y\) and \(z\): \(y-z=-1\) and \(y+z=3\). By adding these two equations, we get \(2y=2\), which upon simplification gives \(y=1\). Substituting \(y=1\) into the equation \(y+z=3\), we get \(z=2\). Finally, substitute \(z=2\) into the equation \(x=3-z\), we get \(x=1\).

Step 4: The solution

After solving, the solutions for \(x\), \(y\), \(z\) are \(1, 1, 2\), respectively.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves expressing one variable in terms of others and then substituting this expression into the other equations. This simplifies the problem to fewer equations with fewer unknown variables.
One great thing about this method is its simplicity. You start by solving one of the equations for one variable. In the example provided, we solved the first equation, \( x + z = 3 \), for \( x \) yielding \( x = 3 - z \).

• This is your pivot expression: it's critical because it provides a link between \( x \) and \( z \).
• It acts as a key to unlock the solution by enabling you to substitute and eliminate variables.

Once the expression is substituted into the remaining equations, the problem reduces to a simpler form. This lets you direct focus on fewer variables, making it easier to solve. Understanding substitution well can make solving equations far less daunting.

Linear Equations

Linear equations are equations where each term is either a constant or the product of a constant and a single variable. They form the basis of algebra and are fundamental in solving systems of equations.
Why are they called "linear"? Because if you graph them, they produce a straight line. In the context of systems of equations, you are looking for a point where all lines intersect, which is your solution.

• In the exercise, each equation represents a plane in three-dimensional space, and their intersection points define the solution.
• Linear equations are straightforward since they typically contain operations of addition, subtraction, and multiplication.

Once you grasp how to interact with linear equations, you can manage many mathematical challenges. Often, they are described in the form \( ax + by + cz = d \), expressing them just like in the provided equations. With practice, you'll find these fundamental structures are your trusty algebra tools.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to find solutions efficiently. It involves using a variety of algebraic properties and techniques to reformulate equations into forms that are easier to analyze and solve.
The key to mastering algebraic manipulation is to become comfortable with operations such as addition, subtraction, multiplication, division, and the application of the distributive law.

• For example, in the solution, from \( 2y - 2z = -2 \), dividing all terms by 2 simplifies to \( y - z = -1 \).
• Such simplification is vital because it makes equations more straightforward and manageable.

Through practice, these manipulations become almost second nature, making the path to solutions clearer. It often involves keeping equations balanced, adhering to the concept that whatever you do to one side, you must also do to the other. This principle is the heart of algebra and makes complex systems less intimidating.