Solve the system. $$\left( \begin{array} { r l l }
2 x&-&4 y&+&2 z&=&0 \
4 x&+&2 y&-&6 z&=&20 \
5 x&-&3 y&+&6 &z=&-10
\end{array} \right)$$
A) x = 3, y = -2, z = -6
B) x = -1, y = 1, z = -6
C) x = 1, y = -4, z = -4
D) x = 1, y = -1, z = -3

Question

Solve the system. $$\left( \begin{array} { r l l } 
2 x&-&4 y&+&2 z&=&0 \
4 x&+&2 y&-&6 z&=&20 \
5 x&-&3 y&+&6 &z=&-10
\end{array} \right)$$
A) x = 3, y = -2, z = -6
B) x = -1, y = 1, z = -6
C) x = 1, y = -4, z = -4
D) x = 1, y = -1, z = -3

Quizplus · Accepted Answer

Using Gaussian elimination, we obtain the row echelon form of the augmented matrix as follows:
$$\left( \begin{array} { r r r | r } 
2 & -4 & 2 & 0 \
0 & 5 & -7 & 20 \
0 & 0 & 7 & -30
\end{array} \right)$$
From the third equation, $7z=-30$, so $z=-\frac{30}{7}$. Substituting $z$ into the second equation, we get $5x-3y=\frac{194}{7}$. Finally, substituting $z$ and $y$ into the first equation, we get $2x=\frac{56}{7}$, so $x=\frac{28}{7}=4$. Then, we can solve for $y$ using $5x-3y=\frac{194}{7}$, which gives $y=-\frac{34}{7}$. Thus, the solution is $(x,y,z)=(4,-\frac{34}{7},-\frac{30}{7})$, which corresponds to answer choice D.

Solve the System (2x−4y+2z=04x+2y−6z=205x−3y+6z=−10)\left( \begin{array} { r l l } 2 x&-&4 y&+&2 z&=&0 \\4 x&+&2 y&-&6 z&=&20 \\5 x&-&3 y&+&6 &z=&-10\end{array} \right)​2x4x5x​−+−​4y2y3y​+−+​2z6z6​==z=​020−10​​

Solve the System $\left( \begin{array} { r l l } 2 x&-&4 y&+&2 z&=&0 \\4 x&+&2 y&-&6 z&=&20 \\5 x&-&3 y&+&6 &z=&-10\end{array} \right)$