Solve each system of equations 2r+s+t=14−r−3s+2t=−24r−6s+3t=−5.

Q. 17PA - Algebra 2 | StudyQuestionHub

Q. 17PA

Question

Solve each system of equations

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

Step-by-Step Solution

Verified

Answer

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

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

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

$\begin{array}{l} \underline{\begin{array}{l} - r - 3 s + 2 t = - 2 \\ 2 r + s + t = 14 \end{array}} \begin{matrix} multiply by 2 \to \end{matrix} \underline{\begin{array}{l} - 2 r - 6 s + 4 t = - 4 \\ 2 r + s + t = 14 \end{array}} \\ 0 - 5 s + 5 t = 10 \end{array}$

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

$\begin{matrix} - 5 s + 5 t = 10 \\ - s + t = 2 divide both sides by 5 \end{matrix}$

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

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

$\begin{array}{l} \underline{\begin{array}{l} - r - 3 s + 2 t = - 2 \\ 4 r - 6 s + 3 t = - 5 \end{array}} \begin{matrix} multiply by 4 \to \end{matrix} \underline{\begin{array}{l} - 4 r - 12 s + 8 t = - 8 \\ 4 r - 6 s + 3 t = - 5 \end{array}} \\ 0 - 18 s + 11 t = - 13 \end{array}$

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

2Step 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 $- 18 s + 11 t = - 13$ .

$\begin{array}{l} \underline{\begin{array}{l} - s + t = 2 \\ - 18 s + 11 t = - 13 \end{array}} \begin{matrix} multiply by 18 \end{matrix} \underline{\begin{array}{l} - 18 s + 18 t = 36 \\ (-) - 18 s + 11 t = - 13 \end{array}} \\ 0 + 7 t = 49 \end{array}$