Question

Solve each system of equations $\begin{array}{c}3x+y+z=4\\ 2x+2y+3z=3\\ x+3y+2z=5\end{array}$.

Short answer

The solution set of the given system of equations is $\left(1,2,-1\right)$.

Step-by-step solution

Step 1. Use the elimination method to get the system of equations in two variables.

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

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

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

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

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

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

Step 2. Use the elimination method to solve the system of two equations.

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

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

Solve $12x=12$ for x:

$\begin{array}{c}12x=12\\ \frac{12x}{12}=\frac{12}{12}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Dividebothsidesby12}\\ x=1\end{array}$

Step 3. Find the values of y and z.

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

$\begin{array}{c}5x-y=3\\ 5\left(1\right)-y=3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Substitute1for}x\\ 5-y=3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Simplify}\\ -y=-2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Subtract5frombothsides}\\ y=2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Dividebothsidesby}-1\end{array}$

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

$\begin{array}{c}3x+y+z=4\\ 3\left(1\right)+2+z=4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}substitute1for}x,2\text{for}y\\ 5+z=4\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}simplify}\\ z=-1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}subtract5frombothsides}\end{array}$

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