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Magazine Chapter 3: Problem 39
$$ \begin{aligned} y+z &=1 \\ x+y+z &=1 \\ x+2 y+2 z &=2 \end{aligned} $$
Short Answer
Expert verified
x = 0, y = 1 - z, z is any real number.
Step by step solution
01
Understand the system of equations
There are three equations given: 1. \( y + z = 1 \) 2. \( x + y + z = 1 \) 3. \( x + 2y + 2z = 2 \) We need to find values for \( x, y, \) and \( z \) that satisfy all three equations.
02
Solve for one variable using equation 1
From the first equation, solve for \( y \) in terms of \( z \): \( y + z = 1 \). Thus, \( y = 1 - z \).
03
Substitute \( y = 1 - z \) into equation 2
Substitute \( y = 1 - z \) into the second equation: \( x + y + z = 1 \). This gives us \( x + (1 - z) + z = 1 \), which simplifies to \( x + 1 - z + z = 1 \) and further to \( x + 1 = 1 \). Solving for \( x \), we get \( x = 0 \).
04
Substitute \( y = 1 - z \) and \( x = 0 \) into equation 3
Substitute \( y = 1 - z \) and \( x = 0 \) into the third equation: \( x + 2y + 2z = 2 \). This becomes \( 0 + 2(1 - z) + 2z = 2 \). Simplify this: \( 0 + 2 - 2z + 2z = 2 \), which reduces to \( 2 = 2 \). This equation is automatically satisfied and does not provide additional information.
05
Verify the solution
We have found that \( x = 0 \), and \( y = 1 - z \). Thus, \( z \) can be any real number, and \( y \) will adjust accordingly to satisfy the first equation. Verify by substituting \( x = 0 \) and any \( z = k \) into all three equations and see if they hold true.Substitute into the second equation: \( x + y + z = 1 \) becomes \( 0 + (1 - k) + k = 1 \), which simplifies to \( 1 = 1 \).Substitute into the third equation: \( x + 2y + 2z = 2 \) becomes \( 0 + 2(1 - k) + 2k = 2 \) which simplifies to \( 2 = 2 \).Thus, the solution holds.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
substitution method
The substitution method is a technique used to solve systems of equations by isolating one variable and then substituting it into the other equations. This helps to reduce the system into a simpler one with fewer variables. Here's how it works step by step for our exercise:
- Start with the first equation: \( y + z = 1 \). Solve for one variable in terms of the other. In this case, we solve for \( y \): \( y = 1 - z \).
- Substitute this expression into the other equations. Substitute \( y = 1 - z \) into the second equation: \( x + y + z = 1 \). This simplifies to \( x + (1 - z) + z = 1 \), then to \( x + 1 = 1 \). Solving for \( x \), we get \( x = 0 \).
- Next, substitute \( y = 1 - z \) and \( x = 0 \) into the third equation: \( x + 2y + 2z = 2 \). This becomes \( 0 + 2(1 - z) + 2z = 2 \). Simplify this to \( 2 = 2 \), which is true and thus does not change the set of solutions.
consistent and dependent systems
A system of equations is consistent if there is at least one set of values for the variables that satisfies all equations simultaneously. Our exercise is an example of a consistent system because we found a solution. The system is also dependent, meaning there are infinitely many solutions. This happens because we have a situation where one equation is a combination of the others.Here鈥檚 why:
- First, note that we have three variables but effectively only two independent equations. This limits us to finding two of the variables in terms of the third.
- In the solutions, we found that \( x = 0 \), \( y = 1 - z \), and \( z \) can be any real number. So, there isn't a single set of values, but rather an infinite number satisfying all the equations.
variables
Variables are symbols used to represent unknown values. In a system of linear equations, each equation helps to define the possible values for these variables. Let's break down the role of variables in our exercise:
- We have three variables: \( x, y, \) and \( z \). Each one represents an unknown quantity we aim to find.
- Through the substitution method, we express one variable in terms of another, simplifying the system. For instance, we expressed \( y \) in terms of \( z \) as \( y = 1 - z \).
- In the last step, our generalized solution showed that for any real number \( z \), the values of \( x \) and \( y \) adjust accordingly to maintain the validity of the equations.
