Math Problem Statement
Consider the following system of equations. w + x + y + z = 6 w + y + z = 4 w + y = 2 (∗) (a) List the leading variables . (b) List the free variables . (c) The general solution of (∗) (expressed in terms of the free variables) is ( , , , ) . (d) Suppose that a fourth equation −2w + y = 5 is included in the system (∗). What is the solution of the resulting system? Answer: ( , , , ). (e) Suppose that instead of the equation in part (d), the equation −2w − 2y = −3 is included in the system (∗). Then what can you say about the solution(s) of the resulting system? Answer: .
Solution
We are given the following system of equations:
w + x + y + z = 6 \\ w + y + z = 4 \\ w + y = 2 \end{cases}$$ Let's solve each part step by step. --- ### (a) List the **leading variables**: Leading variables are variables that can be solved directly from the system of equations without any free parameters. By analyzing the system: 1. From $$w + y = 2$$, we can directly solve for $$w$$ or $$y$$. 2. From $$w + y + z = 4$$, once we have $$w$$ or $$y$$, we can solve for $$z$$. 3. From $$w + x + y + z = 6$$, we can solve for $$x$$ once $$w, y, z$$ are determined. Thus, **leading variables**: $$w$$, $$y$$, $$z$$, and $$x$$. --- ### (b) List the **free variables**: Free variables are variables that we cannot uniquely determine, and whose values can be arbitrary. In this case, there are no free variables, as all variables $$w, x, y, z$$ can be uniquely determined by solving the system. Thus, there are **no free variables**. --- ### (c) The general solution of $$(*)$$ (expressed in terms of the free variables): Let's solve the system to find the general solution. Start with the third equation: 1. $$w + y = 2$$ ⟶ solve for $$w = 2 - y$$. 2. Substitute $$w = 2 - y$$ into the second equation: $$(2 - y) + y + z = 4 \implies 2 + z = 4 \implies z = 2.$$ 3. Now substitute $$w = 2 - y$$ and $$z = 2$$ into the first equation: $$(2 - y) + x + y + 2 = 6 \implies 4 + x = 6 \implies x = 2.$$ So, the solution in terms of free variables is: $$(w, x, y, z) = (2 - y, 2, y, 2).$$ --- ### (d) Adding the fourth equation $$-2w + y = 5$$: Now, let's consider adding this fourth equation: $$-2w + y = 5.$$ Substitute $$w = 2 - y$$ into this equation: $$-2(2 - y) + y = 5 \implies -4 + 2y + y = 5 \implies 3y = 9 \implies y = 3.$$ Once $$y = 3$$, substitute into the previous expressions for $$w, x, z$$: - $$w = 2 - 3 = -1$$, - $$z = 2$$ (from before), - $$x = 2$$ (from before). Thus, the solution is: $$(w, x, y, z) = (-1, 2, 3, 2).$$ --- ### (e) Adding the equation $$-2w - 2y = -3$$: Now, consider the new fourth equation: $$-2w - 2y = -3.$$ Substitute $$w = 2 - y$$ into this equation: $$-2(2 - y) - 2y = -3 \implies -4 + 2y - 2y = -3 \implies -4 = -3,$$ which is a **contradiction**. Therefore, the system is **inconsistent**, and there is **no solution**. --- ### Summary of answers: - (a) **Leading variables**: $$w, x, y, z$$. - (b) **Free variables**: None. - (c) **General solution**: $$(w, x, y, z) = (2 - y, 2, y, 2)$$. - (d) **Solution of the system with $$-2w + y = 5$$**: $$(-1, 2, 3, 2)$$. - (e) **Solution of the system with $$-2w - 2y = -3$$**: The system is **inconsistent**, so there is **no solution**. --- Let me know if you would like more details or have any further questions! Here are 5 related questions: 1. How do you determine if a variable is leading or free in a system of equations? 2. Can a system with no free variables have infinitely many solutions? 3. How does a contradiction in a system of equations lead to no solution? 4. What conditions must a system of linear equations satisfy to be consistent? 5. How can you represent the general solution of a consistent system in terms of parameters? **Tip**: To identify if a system is inconsistent, look for equations that reduce to a false statement (e.g., $$0 = 1$$).Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Equations
Systems of Equations
Inconsistency in Systems
Leading and Free Variables
Formulas
w + x + y + z = 6
w + y + z = 4
w + y = 2
-2w + y = 5
-2w - 2y = -3
Theorems
Solving Systems of Equations
Consistency and Inconsistency of Linear Systems
Suitable Grade Level
Grades 10-12