Q22. - Algebra 2

Question

Solve each system of equations 

$\begin{array}{c}r+s+t=5\\ 2r-7s-3t=13\\ \frac{1}{2}r-\frac{1}{3}s+\frac{2}{3}t=-1\end{array}$

Step-by-Step Solution

Answer

The solution set of the given system of equations is .$(8,3,-6)$

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

Multiply the equation $r+s+t=5$ by $3$and add the new resultant equation to the equation $2r-7s-3t=13$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}r+s+t=5\\ 2r-7s-3t=13\end{array}}\begin{array}{c}\text{multiply by 3}\to \\ \end{array}\underset{\_}{\begin{array}{l}3r+3s+3t=15\\ 2r-7s-3t=13\end{array}}\\ 5r-4s+0=28\end{array}$

So, the resultant equation is $5r-4s=28$.

 Next, multiply the equation $r+s+t=5$ by $4$ and multiply the equation $\frac{1}{2}r-\frac{1}{3}s+\frac{2}{3}t=-1$ by 6.

Then subtract the two new resultant equations

$\begin{array}{l}\underset{\_}{\begin{array}{l}r+s+t=5\\ \frac{1}{2}r-\frac{1}{3}s+\frac{2}{3}t=-1\end{array}}\begin{array}{c}\text{multiply by }4\to \\ \text{multiply by }6\to \end{array}\underset{\_}{\begin{array}{l}4r+4s+4t=20\\ (-)3r-2s+4t=-6\end{array}}\\ r+6s+0=26\end{array}$

So, the resultant equation is $r+6s=26$

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

Multiply $5r-4s=28$ by 3 and multiply the equation $r+6s=26$ by 2.

Then add the two new resultant equations.

$\begin{array}{l}\underset{\_}{\begin{array}{l}5r-4s=28\\ r+6s=26\end{array}}\begin{array}{c}\text{multiply by 3}\\ \text{multiply by 2}\end{array}\underset{\_}{\begin{array}{l}15r-12s=84\\ 2r+12s=52\end{array}}\\ 17r+0=136\end{array}$

Solve $17r=136$for $r$:

$\begin{array}{c}17r=136\\ \frac{17r}{17}=\frac{136}{17}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Divide both sides by }17\\ r=8\end{array}$

Step 3 – Find the values of s and t .

Substitute$\$r=8\$in \backslash [r+6s=26\backslash ]$ and find the value of $s$.

$\begin{array}{c}r+6s=26\\ 8+6s=26\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute }8\text{ for }r\\ 6s=18\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }8\text{ from both sides}\\ s=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Divide both sides by 6}\end{array}$

Substitute $r=8,s=3in r+s+t=5$ and find the value of $t$.

$\begin{array}{c}r+s+t=5\\ 8+3+t=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute }8\text{ for }r,3\text{ for }s\\ 11+t=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ t=-6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract }11\text{\hspace{0.17em}from both sides}\end{array}$

Hence, the solution of the given system of equations is$\left(r,s,t\right)=\left(8,3,-6\right)$.