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Sagot :
Given
Solve the following system of linear equations.
x + 3y + z = - 4
2x – 4y – 3z = 7
3x – 3y + 4z = 13
Solution
[tex]\begin{bmatrix}x+3y+z=-4 \\ 2x-4y-3z=7 \\ 3x-3y+4z=13\end{bmatrix}[/tex]Substitute x= -4-3y-z
[tex]\begin{bmatrix}2\mleft(-4-3y-z\mright)-4y-3z=7 \\ 3\mleft(-4-3y-z\mright)-3y+4z=13\end{bmatrix}[/tex]Simplify
[tex]\begin{bmatrix}-10y-5z-8=7 \\ -12y+z-12=13\end{bmatrix}[/tex]Make y the subject
[tex]\begin{gathered} -10y-5z-8=7 \\ -10y\text{ -5z=7+8} \\ -10y-5z=15 \\ \text{divide all through by 5} \\ -2y-z=3 \\ y=-\frac{z+3}{2} \end{gathered}[/tex]Now substitute
[tex]\begin{bmatrix}-12\mleft(-\frac{z+3}{2}\mright)+z-12=13\end{bmatrix}[/tex]Simplify
[tex]\begin{gathered} \\ \begin{bmatrix}7z+6=13\end{bmatrix} \\ \text{Make z the subject} \\ 7z=13-6 \\ 7z=7 \\ \text{divide both sides by 7} \\ \frac{7z}{7}=\frac{7}{7} \\ z=1 \end{gathered}[/tex]Now substitute z=1
[tex]\begin{gathered} y=-\frac{z+3}{2} \\ y=-\frac{1+3}{2}=-\frac{4}{2}=-2 \end{gathered}[/tex]Finally, to find x
when z =1 and y =-2
[tex]\begin{gathered} x+3y+z=-4 \\ x+3(-2)+1=-4 \\ x-6+1=-4 \\ \text{collect the like terms} \\ x-5=-4 \\ x=-4+5 \\ x=1 \end{gathered}[/tex]The final answer
[tex]\begin{gathered} x=1 \\ y=-2 \\ z=1 \end{gathered}[/tex]
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