Answer:
[tex](x,y,z)=\left(-\dfrac{1}{11},\dfrac{21}{11},\dfrac{21}{11}\right)[/tex]
Step-by-step explanation:
Given:
[tex]-2x+3y-z=-23x+y=4-2y+2z=4[/tex]
Therefore:
[tex]\begin{cases}-2x+3y-z=4\\-23x+y=4\\4-2y+2z=4\end{cases}[/tex]
Rewrite Equation 2 to make x the subject:
[tex]\implies -23x+y=4[/tex]
[tex]\implies -23x+y-y=4-y[/tex]
[tex]\implies -23x=4-y[/tex]
[tex]\implies \dfrac{-23x}{-23}=\dfrac{4}{-23}-\dfrac{y}{-23}[/tex]
[tex]\implies x=-\dfrac{4}{23}+\dfrac{y}{23}[/tex]
Rewrite Equation 3 to make z the subject:
[tex]\implies 4-2y+2z=4[/tex]
[tex]\implies 4-2y+2z-4=4-4[/tex]
[tex]\implies -2y+2z=0[/tex]
[tex]\implies -2y+2z+2y=0+2y[/tex]
[tex]\implies 2z=2y[/tex]
[tex]\implies \dfrac{2z}{2}=\dfrac{2y}{2}[/tex]
[tex]\implies z=y[/tex]
Substitute the found expressions for y and z into Equation 1 and solve for y:
[tex]\implies -2x+3y-z=4[/tex]
[tex]\implies -2\left(-\dfrac{4}{23}+\dfrac{y}{23}\right)+3y-y=4[/tex]
[tex]\implies \dfrac{8}{23}-\dfrac{2}{23}y+2y=4[/tex]
[tex]\implies \dfrac{8}{23}+\dfrac{44}{23}y=4[/tex]
[tex]\implies \dfrac{8}{23}+\dfrac{44}{23}y- \dfrac{8}{23}=4- \dfrac{8}{23}[/tex]
[tex]\implies \dfrac{44}{23}y=\dfrac{84}{23}[/tex]
[tex]\implies \dfrac{44}{23}y \cdot \dfrac{23}{44}=\dfrac{84}{23}\cdot \dfrac{23}{44}[/tex]
[tex]\implies y=\dfrac{84}{44}[/tex]
[tex]\implies y=\dfrac{21}{11}[/tex]
Therefore, as y = z:
[tex]\implies z=\dfrac{21}{11}[/tex]
Substitute the found value of y into the found expression for x and solve for x:
[tex]\implies x=-\dfrac{4}{23}+\dfrac{y}{23}[/tex]
[tex]\implies x=-\dfrac{4}{23}+\dfrac{\frac{21}{11}}{23}[/tex]
[tex]\implies x=-\dfrac{4}{23}+\dfrac{21}{253}[/tex]
[tex]\implies x=-\dfrac{1}{11}[/tex]
Therefore, the final solution is:
[tex](x,y,z)=\left(-\dfrac{1}{11},\dfrac{21}{11},\dfrac{21}{11}\right)[/tex]
