Solving a System of Linear Equations with Three Variables

Math Problem Statement

Find all solutions to the following system of equations: 3x + 4y + z = 1, 2x + 3y = 0, 4x + 3y - z = -2.

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The system of equations shown in the image is:

1. \quad 3x + 4y + z &= 1 \\ 2. \quad 2x + 3y &= 0 \\ 3. \quad 4x + 3y - z &= -2 \end{aligned}$$ ### Step-by-Step Solution: #### Step 1: Solve Equation (2) for $$x$$ The second equation is $$2x + 3y = 0$$, which can be solved for $$x$$ as follows: $$2x = -3y \quad \Rightarrow \quad x = \frac{-3y}{2}$$ #### Step 2: Substitute $$x = \frac{-3y}{2}$$ into Equations (1) and (3) Now substitute this expression for $$x$$ into equations (1) and (3). **Substitute into Equation (1):** $$3\left(\frac{-3y}{2}\right) + 4y + z = 1$$ $$-\frac{9y}{2} + 4y + z = 1$$ Multiply everything by 2 to eliminate the fractions: $$-9y + 8y + 2z = 2$$ Simplifying: $$-y + 2z = 2 \quad \Rightarrow \quad y = 2z - 2 \quad \text{(Equation 4)}$$ **Substitute

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Linear Algebra
Systems of Equations

Substitution Method
Elimination Method

Linear Independence
Uniqueness of Solutions

Grades 10-12

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