Solve each system of equations.7.2r+3s−4t=204r−s+5t=133r+2s+4t=15

Q7. - Algebra 2 | StudyQuestionHub

Solve each system of equations.

$\begin{array}{l} 2 r + 3 s - 4 t = 20 \\ 4 r - s + 5 t = 13 \\ 3 r + 2 s + 4 t = 15 \end{array}$

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Answer

The solution is $r = 5, s = - 2, t = - 1 .$

1Step-1 –Using elimination to obtain two equations with two variables

Given system of equations are

$\begin{array}{l} 2 r + 3 s - 4 t = 20 \\ 4 r - s + 5 t = 13 \\ 3 r + 2 s + 4 t = 15 \end{array}$

Multiplying second equation by $3$ and adding to the first equation, we get

$\begin{array}{l} 2 r + 12 r - 4 t + 15 t = 20 + 39 \\ 14 r + 11 t = 59 \end{array}$

Again multiplying second equation by $2$ and adding to the third equation, we get

$\begin{array}{l} 8 r + 3 r + 10 t + 4 t = 26 + 15 \\ 11 r + 14 t = 41 \end{array}$

2Step-2 –Solving the system of two equations with two variables

System of two equations with two variables are

$\begin{array}{l} 114 r + 11 t = 59 \\ 11 r + 14 t = 41 \end{array}$

Multiplying the first equation by $11$ and second by $14$ and subtracting them we get

$\begin{matrix} 121 t - 196 t = 649 - 574 \\ - 75 t = 75 \\ t = - 1 \end{matrix}$

Putting the value of $t$ in equation $14 r + 11 t = 59$ we get

$\begin{array}{l} 14 r - 11 = 59 \\ 14 r = 59 + 11 \\ 14 r = 70 \\ r = 5 \end{array}$

3Step-3 –Finding the value of s using the value of r and t

Putting the value of $r$ and $t$ in the equation $\ [4 r - s + 5 t \] \ [4 r - s + 5 t \]$ we get,

$\begin{array}{l} 4 (5) - s + 5 (- 1) = 13 \\ 20 - s - 5 = 13 \\ - s = 13 - 15 \\ - s = - 2 \\ s - 2 \end{array}$