Respuesta :
Answer
• x = –1
• y = –2
• z = 1
Explanation
Given the system:
[tex]\begin{gathered} -5x+3z=8\text{, equation 1} \\ 5y-4z=-14,\text{ equation 2} \\ x-4y=7,\text{ equation 3} \end{gathered}[/tex]We have to divide equation 1 over 5 and add it to equation 2:
[tex]\begin{gathered} -(5x)+0y+3z=8 \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}[/tex][tex]\begin{gathered} \frac{-(5x)+0y+3z=8}{5} \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}[/tex][tex]\begin{gathered} -x+0y+\frac{3}{5}z=\frac{8}{5} \\ x-4y+0z=7 \\ 0x+5y-4z=-14 \end{gathered}[/tex]Now, we have to add 1/5(equation 1) to equation 2:
[tex]\begin{gathered} -x+0y+\frac{3}{5}z=\frac{8}{5} \\ x-4y+0z=7 \\ --------- \\ 0x-4y+\frac{3}{5}z=\frac{43}{5} \end{gathered}[/tex]Next, we multiply the equation 2 obtained previously times 5:
[tex]\begin{gathered} (0x-4y+\frac{3}{5}z=\frac{43}{5})\cdot5 \\ -20y+3z=43 \end{gathered}[/tex]Then, we divide equation 2 over 4:
[tex]\begin{gathered} \frac{-20y+3z=43}{4} \\ -5y+\frac{3}{4}z=\frac{43}{4} \end{gathered}[/tex]We add it to equation 3:
[tex]\begin{gathered} -5y+\frac{3}{4}z=\frac{43}{4} \\ 5y-4z=-14 \\ ------- \\ 0y-\frac{13}{4}z=-\frac{13}{4} \end{gathered}[/tex]Then, we are left with:
[tex]\begin{equation*} -\frac{13}{4}z=-\frac{13}{4} \end{equation*}[/tex]Simplifying:
[tex]-13z=-13[/tex][tex]z=\frac{-13}{-13}=1[/tex][tex]z=1[/tex]Now that we have the value of z (z = 1), we can replace it in the modified equation 2 and solve for y:
[tex]-20y+3z=43[/tex][tex]-20y+3(1)=43[/tex][tex]-20y+3=43[/tex][tex]-20y=43-3[/tex][tex]-20y=40[/tex][tex]y=\frac{40}{-20}[/tex][tex]y=-2[/tex]Finally, calculating the value of x with any of the equations (as we already have the other two values):
[tex]x-4(-2)=7[/tex][tex]x+8=7[/tex][tex]x=7-8[/tex][tex]x=-1[/tex]