Problem 5 - Algebra and Trigonometry

Question

$$ \left\\{\begin{aligned} x-2 y+4 z &=3 \\ y+2 z &=7 \\ z &=2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 3 \), \( z = 2 \).

Step 1: Solving for z

We start with the simplest equation given: \( z = 2 \). Since this equation is already solved for \( z \), we know that \( z = 2 \).

Step 2: Substitute z into the second equation

Substitute the value of \( z = 2 \) into the second equation \( y + 2z = 7 \):\[ y + 2(2) = 7 \]This simplifies to:\[ y + 4 = 7 \]Subtract 4 from both sides to solve for \( y \):\[ y = 3 \]

Step 3: Substitute y and z into the first equation

Now that we know \( y = 3 \) and \( z = 2 \), substitute these values into the first equation \( x - 2y + 4z = 3 \):\[ x - 2(3) + 4(2) = 3 \]Simplify this equation:\[ x - 6 + 8 = 3 \]\[ x + 2 = 3 \]Subtract 2 from both sides to solve for \( x \):\[ x = 1 \]

Key Concepts

Substitution Method

The substitution method is a widely used technique for solving systems of linear equations. It involves solving one of the equations for one variable, and then substituting this expression into the other equations. This method is effective because it reduces the original system to one with fewer equations and variables.

• Step 1: Choose an equation and solve for one of the variables. In our exercise, the simplest equation is already provided as \( z = 2 \). This makes it convenient to substitute.
• Step 2: Substitute the expression into the other equations. By inserting the value of \( z \) into other equations, we simplify the system significantly. Here, substituting \( z = 2 \) into the second equation makes solving for \( y \) straightforward.
• Step 3: Repeat until all variables are solved. After finding the value of \( y \), we substitute both \( y \) and \( z \) into the remaining equation to solve for \( x \).

The substitution method is particularly useful when one equation is already isolated, simplifying the process of finding solutions.

Solving Linear Equations

Solving linear equations involves finding the values of variables that make the equations true. Linear equations in a system have a degree of one. To find solutions, we apply algebraic operations until we isolate the variable. In this problem, the process involved a series of substitutions and simplifications.

• Use basic operations like addition, subtraction, multiplication, and division to rearrange the equation.
• The goal is to have the variable on one side of the equation and the constants on the other.

For example, in the exercise, after the substitution, we have:

• For \( y \): The equation \( y + 4 = 7 \) is simplified to solve for \( y \) by subtracting 4 from both sides.
• For \( x \): The equation \( x + 2 = 3 \) solves easily by subtracting 2 from each side.

Each step moves us closer to the solution by isolating the variables, replicating a pattern of logical deduction to reach correct outcomes.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations (like addition and subtraction). Understanding how to manipulate these expressions is key to solving equations. In systems of linear equations, the expressions can be simplified through substitution or by performing algebraic operations.
In the original exercise, we see how substituting \( z = 2 \) into \( y + 2z = 7 \) helps us break down and solve the expression:

• By replacing \( z \) with 2, the expression simplifies to \( y + 4 = 7 \).
• This further reduces to \( y = 3 \), an easily interpretable value.

Subsequently, substituting both \( y = 3 \) and \( z = 2 \) into the first equation helps isolate \( x \). These manipulations showcase the power of understanding and using algebraic expressions to dissect and piece together solutions in a structured way.