Problem 38 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} \quad \quad y +z &=1 ,\\\ x+y+z &=1 ,\\\ x+2 y+2 z &=2 \end{aligned}$$

Step-by-Step Solution

Answer

The system is dependent; solutions are (0, y, 1-y).

Step 1 - Write the system of equations

Start with the given system of equations: y + z = 1 x + y + z = 1 x + 2y + 2z = 2

Step 2 - Subtract the first equation from the second

Subtract the first equation from the second to eliminate z: (x + y + z) - (y + z) = 1 - 1 This simplifies to x = 0

Step 3 - Substitute x = 0 into the second equation

Substitute the value of x into the second equation: 0 + y + z = 1 This simplifies to y + z = 1

Step 4 - Substitute x = 0 into the third equation

Substitute the value of x into the third equation: 0 + 2y + 2z = 2 This simplifies to 2y + 2z = 2 Then divide the entire equation by 2: y + z = 1

Step 5 - Analyze the resulting equations

Now we have two equations: y + z = 1 y + z = 1 These two equations are identical, indicating that the system is dependent. The solution is any pair y and z where y + z = 1, and x = 0.

Key Concepts

dependent system

A system of equations is considered dependent if all equations represent the same line or plane. In other words, one equation can be derived from the others by simple algebraic manipulation. In such cases, there are infinitely many solutions, as every point on the line (or plane) will satisfy all equations in the system.
For example, in the given system:

• y + z = 1
• x + y + z = 1
• x + 2y + 2z = 2

We find that after simplification, two of the equations (y + z = 1) are the same. This indicates that we have a dependent system where there are infinitely many solutions. Any value of y and z satisfying y + z = 1 and x = 0 is a solution.

elimination method

The elimination method involves adding or subtracting equations to eliminate one of the variables, making it easier to solve the system. The idea is to focus on one variable at a time and simplify the system step by step.
For instance, let's eliminate z from the given system:

• Original equations: y + z = 1, x + y + z = 1, and x + 2y + 2z = 2.
• Step 1: Subtract the first equation from the second: (x + y + z) - (y + z) = 1 - 1 yields x = 0.
• Step 2: Substitute x = 0 back into the second and third equations to simplify.

This method helps to simplify and systematically solve the system, making the next steps easier to handle.

substitution method

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equations. This reduces the number of equations and makes it easier to solve for the remaining variables.
In the given system, we can use substitution after finding x = 0:

• Step 1: Substitute x = 0 into the second equation: 0 + y + z = 1 simplifies to y + z = 1.
• Step 2: Substitute x = 0 into the third equation: 0 + 2y + 2z = 2 simplifies to y + z = 1 after dividing by 2.

By substituting, we can see that both remaining equations result in y + z = 1, indicating a dependent system with infinitely many solutions where y + z = 1 and x = 0.

algebraic equations

Algebraic equations are mathematical statements that express the equality between two expressions involving variables and constants using operations like addition, subtraction, multiplication, and division.
In solving algebraic systems, whether linear or nonlinear, the primary goal is to find the values of the variables that satisfy all given equations simultaneously.
In the provided problem, our system of algebraic equations is:

• y + z = 1
• x + y + z = 1
• x + 2y + 2z = 2

To solve, we simplify and manipulate these equations through methods like substitution and elimination. Solving such systems provides valuable insight into possible variables’ values, revealing the nature and consistency of the system, whether independent, dependent, or inconsistent.