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${:[x_(1)+3x_(2)+5x_(3)+x_(4)=0],[4x_(1)+7x_(2)-3x_(3)-x_(4)=0],[3x_(1)+2x_(2)+7x_(3)+8x_(4)=0]:}$

$\begin{array}{l}x_{1}+3 x_{2}+5 x_{3}+x_{4}=0 \\ 4 x_{1}+7 x_{2}-3 x_{3}-x_{4}=0 \\ 3 x_{1}+2 x_{2}+7 x_{3}+8 x_{4}=0\end{array}$

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Math Problems

Grade 8

Solve a system of equations using any method

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\begin{array}{l}x_{1}+3 x_{2}+5 x_{3}+x_{4}=0 \\ 4 x_{1}+7 x_{2}-3 x_{3}-x_{4}=0 \\ 3 x_{1}+2 x_{2}+7 x_{3}+8 x_{4}=0\end{array}

Write Equations: First, let's write down the system of equations: \{x_{$1$}+\(3x_{ $2$ }+ $5$ x_{ $3$ }+x_{ $4$ }= $0$ , $4$ x_{ $1$ }+ $7$ x_{ $2$ } $-3$ x_{ $3$ }-x_{ $4$ }= $0$ , $3$ x_{ $1$ }+ $2$ x_{ $2$ }+ $7$ x_{ $3$ }+ $8$ x_{ $4$ }= $0$ \}
Check Independence: We need to check if these equations are independent or dependent. To do this, we can use the determinant or row reduction, but since we have $4$ variables and only $3$ equations, we can't directly calculate a determinant. Let's try to reduce it to row-echelon form and see if we have any free variables.
Reduce to Row-Echelon Form: Let's start with the first equation as our pivot. We'll keep it as is and modify the other two equations to eliminate $x_{1}$ from them.
Eliminate $x_{1}$ : Multiply the first equation by $-4$ and add it to the second equation to eliminate $x_{1}$ from the second equation: $\newline$ \(-4(x_{ $1$ }+ $3$ x_{ $2$ }+ $5$ x_{ $3$ }+x_{ $4$ }) + ( $4$ x_{ $1$ }+ $7$ x_{ $2$ } $-3$ x_{ $3$ }-x_{ $4$ }) = $-4$ ( $0$ ) + $0$ $\newline$ This simplifies to: $\newline$ $-4$ x_{ $1$ } $-12$ x_{ $2$ } $-20$ x_{ $3$ } $-4$ x_{ $4$ } + $4$ x_{ $1$ }+ $7$ x_{ $2$ } $-3$ x_{ $3$ }-x_{ $4$ } = $0$ $\newline$ Which simplifies to: $\newline$ $-5$ x_{ $2$ } $-23$ x_{ $3$ } $-3$ x_{ $4$ } = $0$
New System of Equations: Now, multiply the first equation by $-3$ and add it to the third equation to eliminate $x_{1}$ from the third equation: $\newline$ $-3(x_{1}+3x_{2}+5x_{3}+x_{4}) + (3x_{1}+2x_{2}+7x_{3}+8x_{4}) = -3(0) + 0$ $\newline$ This simplifies to: $\newline$ $-3x_{1}-9x_{2}-15x_{3}-3x_{4} + 3x_{1}+2x_{2}+7x_{3}+8x_{4} = 0$ $\newline$ Which simplifies to: $\newline$ $-7x_{2}-8x_{3}+5x_{4} = 0$
Check Consistency: Now we have a new system of equations: \{x_{$1$}+\(3x_{ $2$ }+ $5$ x_{ $3$ }+x_{ $4$ }= $0$ , $-5$ x_{ $2$ } $-23$ x_{ $3$ } $-3$ x_{ $4$ }= $0$ , $-7$ x_{ $2$ } $-8$ x_{ $3$ }+ $5$ x_{ $4$ }= $0$ \}
Infinite Solutions: We can see that we have $3$ equations and $4$ unknowns, which means we have more unknowns than equations. This usually means there are infinitely many solutions or no solution if the equations are inconsistent.
Infinite Solutions: We can see that we have $3$ equations and $4$ unknowns, which means we have more unknowns than equations. This usually means there are infinitely many solutions or no solution if the equations are inconsistent.To determine if the system has infinitely many solutions or no solution, we need to check if the equations are consistent. If they are consistent, then there are infinitely many solutions. If not, there is no solution.
Infinite Solutions: We can see that we have $3$ equations and $4$ unknowns, which means we have more unknowns than equations. This usually means there are infinitely many solutions or no solution if the equations are inconsistent.To determine if the system has infinitely many solutions or no solution, we need to check if the equations are consistent. If they are consistent, then there are infinitely many solutions. If not, there is no solution.Looking at the reduced system, we can see that none of the equations are multiples of each other, and there's no obvious contradiction. This suggests that the system is consistent and has infinitely many solutions.