91Ó°ÊÓ

Multiply the equation $3x+y+z=4$by 3 and subtract the new resultant equation from $2x+2y+3z=3$.

$\begin{array}{l}\underset{¯}{\begin{array}{l}3x+\text{ â¶Ä„}y+\text{ â¶Ä„ }z=4\\ 2x+2y+3z=3\end{array}}\text{ â¶Ä„ â¶Ä„}\begin{array}{c}\text{multiplyby3}\\ \end{array}\text{ â¶Ä„}\underset{¯}{\begin{array}{l}\text{ â¶Ä„ â¶Ä„}9x+3y+3z=12\\ \left(−\right)2x+2y+3z=\text{ â¶Ä„}3\end{array}}\\ \text{ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„}7x+\text{ â¶Ä„}y+\text{ â¶Ä„}0\text{ }=\text{ â¶Ä„ }9\end{array}$

So, the resultant equation is $7x+y=9$.

Multiply the equation $3x+y+z=4$by 2 and subtract the new resultant equation from $x+3y+2z=5$.

$\begin{array}{l}\underset{¯}{\begin{array}{l}3x+\text{ â¶Ä„}y+\text{ â¶Ä„ }z=4\\ \text{ â¶Ä„}x+3y+2z=5\end{array}}\text{ â¶Ä„ â¶Ä„}\begin{array}{c}\text{multiplyby2}\\ \end{array}\text{ â¶Ä„}\underset{¯}{\begin{array}{l}\text{ â¶Ä„ â¶Ä„}6x+2y+2z=8\\ \left(−\right)\text{ â¶Ä„}x+3y+2z=\text{ }5\end{array}}\\ \text{ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„ â¶Ä„}5x−\text{ â¶Ä„}y+\text{ â¶Ä„}0\text{ }=3\end{array}$

So, the resultant equation is $5x-y=3$.

Add $7x+y=9$and $5x-y=3$.

$\begin{array}{l}\text{ }\underset{¯}{\begin{array}{l}7x+y=9\\ 5x−y=3\end{array}}\\ 12x+0=12\end{array}$

Solve $12x=12$ for x:

$\begin{array}{c}12x=12\\ \frac{12x}{12}=\frac{12}{12}\text{ â¶Ä„ â¶Ä„ â¶Ä„Dividebothsidesby12}\\ x=1\end{array}$

Substitute $x=1$in $5x-y=3$ and find the value of y.

Substitute $x=1,y=2$in $3x+y+z=4$and find the value ofz.

Hence, the solution of the given system of equations is $\left(x,y,z\right)=\left(1,2,-1\right)$.