Question

Solve each system of equations

$\begin{array}{c}2r+s+t=14\\ -r-3s+2t=-2\\ 4r-6s+3t=-5\end{array}$.

Short answer

The solution set of the given system of equations is $\left(1,5,7\right)$.

Step-by-step solution

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

Multiply the equation $-r-3s+2t=-2$by 2 and add the new resultant equation to $2r+s+t=14$.

$\begin{array}{l}\underset{¯}{\begin{array}{l}-r-3s+2t=-2\\ 2r+s+t=14\end{array}}\begin{array}{c}\text{multiplyby2}\to \\ \end{array}\underset{¯}{\begin{array}{l}-2r-6s+4t=-4\\ 2r+s+t=14\end{array}}\\ 0-5s+5t=10\end{array}$

Simplify $-5s+5t=10$:

$\begin{array}{c}-5s+5t=10\\ -s+t=2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}dividebothsidesby5}\end{array}$

So, the resultant equation is $-s+t=2$.

Next, multiply the equation $-r-3s+2t=-2$by 4and add the new resultant equation to $4r-6s+3t=-5$.

$\begin{array}{l}\underset{¯}{\begin{array}{l}-r-3s+2t=-2\\ 4r-6s+3t=-5\end{array}}\begin{array}{c}\text{multiplyby4}\to \\ \end{array}\underset{¯}{\begin{array}{l}-4r-12s+8t=-8\\ 4r-6s+3t=-5\end{array}}\\ 0-18s+11t=-13\end{array}$

So, the resultant equation is $-18s+11t=-13$.

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

Multiply $-s+t=2$by 18 and subtract the new resultant equation from $-18s+11t=-13$.

$\begin{array}{l}\underset{¯}{\begin{array}{l}-s+t=2\\ -18s+11t=-13\end{array}}\begin{array}{c}\text{multiplyby18}\\ \end{array}\underset{¯}{\begin{array}{l}-18s+18t=36\\ \left(-\right)-18s+11t=-13\end{array}}\\ 0+7t=49\end{array}$

Solve $7t=49$for $t$:

$\begin{array}{c}7t=49\\ \frac{7t}{7}=\frac{49}{7}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Dividebothsidesby7}\\ t=7\end{array}$

Step 3. Find the values of r and s.

Substitute $t=7$in $-s+t=2$and find the value of $s$.

$\begin{array}{c}-s+t=2\\ -s+7=2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Substitute7for}t\\ -s=-5\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Subtract7frombothsides}\\ s=5\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Dividebothsidesby-1}\end{array}$

Substitute $s=5,t=7$in $-r-3s+2t=-2$and find the value of r.

$\begin{array}{c}-r-3s+2t=-2\\ -r-3\left(5\right)+2\left(7\right)=-2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}substitute5for}s,7\text{for}t\\ -r-15+14=-2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}simplify}\\ -r-1=-2\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}simplify}\\ -r=-1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}add1onbothsides}\\ r=1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}dividebothsidesby}-\text{1}\end{array}$

Hence, the solution of the given system of equations is $\left(r,s,t\right)=\left(1,5,7\right)$.