c) {:[4x_(2)+x_(3)=0],[x_(1)+x_(2)+x_(3)=0],[2x_(1)+4x_(2)+3x_(3)=0]:}

Question

c)  {:[4x_(2)+x_(3)=0],[x_(1)+x_(2)+x_(3)=0],[2x_(1)+4x_(2)+3x_(3)=0]:}

Bytelearn · Accepted Answer

Given System of Equations: We are given a system of linear equations: $$ \begin{align*} 4x_2 + x_3 &= 0 \quad \text{(Equation 1)} \ x_1 + x_2 + x_3 &= 0 \quad \text{(Equation 2)} \ 2x_1 + 4x_2 + 3x_3 &= 0 \quad \text{(Equation 3)} \end{align*} $$ We will use the method of substitution or elimination to solve for $ x_1, x_2, $ and $ x_3 $.Solving Equation 1: First, we can solve Equation 1 for $ x_3 $ in terms of $ x_2 $: $ x_3 = -4x_2 $ We will use this expression for $ x_3 $ to substitute into the other equations.Substitute into Equation 2: Substitute $ x_3 = -4x_2 $ into Equation 2: $ x_1 + x_2 - 4x_2 = 0 $ Simplify the equation: $ x_1 - 3x_2 = 0 $ Now, we can solve for $ x_1 $ in terms of $ x_2 $: $ x_1 = 3x_2 $Solving for x1: Substitute $ x_3 = -4x_2 $ and $ x_1 = 3x_2 $ into Equation 3: $ 2(3x_2) + 4x_2 + 3(-4x_2) = 0 $ Simplify the equation: $ 6x_2 + 4x_2 - 12x_2 = 0 $ $ -2x_2 = 0 $ Now, solve for $ x_2 $: $ x_2 = 0 $Substitute into Equation 3: Since $ x_2 = 0 $, we can substitute back into the expressions for $ x_1 $ and $ x_3 $: $ x_1 = 3(0) = 0 $ $ x_3 = -4(0) = 0 $Solving for x2: We have found the values for $ x_1, x_2, $ and $ x_3 $: $ x_1 = 0 $ $ x_2 = 0 $ $ x_3 = 0 $ These values satisfy all three equations in the system.

c) $\begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array}$

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c) 4x2+x3=0x1+x2+x3=02x1+4x2+3x3=0 \begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array} 4x2​+x3​=0x1​+x2​+x3​=02x1​+4x2​+3x3​=0​

c) $\begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array}$