Chapter 4: Q39. (page 158)

Solve each system of equations.
$\begin{matrix} 3 r - 15 s + 4 t = - 57 \\ 9 r + 45 s - t = 26 \\ - 6 r + 10 s + 3 t = - 19 \end{matrix}$

Short Answer

Expert verified

The solution of given system of equation is $(- \frac{4}{3}, \frac{3}{5}, - 11)$ .

Step by step solution

– Description of step.

Add $9 r + 45 s - t = 26$ and $- 6 r + 10 s + 3 t = - 19$ .

\begin{matrix} 9 r + 45 s - t - 6 r + 10 s + 3 t = 26 - 19 \\ 3 r + 55 s + 2 t = 7 \end{matrix}

– Description of step.

Subtract $3 r - 15 s + 4 t = - 57$ from the obtained equation in step 1 then divide each side of the obtained equation by 2 and solve for $s$ .

\begin{matrix} 3 r + 55 s + 2 t - (3 r - 15 s + 4 t) = 7 - (- 57) \\ 3 r + 55 s + 2 t - 3 r + 15 s - 4 t = 64 \\ \frac{70}{2} s - \frac{2}{2} t = \frac{64}{2} \\ 35 s - t = 32 \\ s = \frac{t}{35} + \frac{32}{35} \end{matrix}

– Description of step.

Multiply $3 r - 15 s + 4 t = - 57$ by 3 and add obtained equation to $9 r + 45 s - t = 26$ then solve for $r$ .

\begin{matrix} 9 r - 45 s + 12 t + 9 r + 45 s - t = - 171 + 26 \\ 18 r + 11 t = - 145 \\ 18 r = - 145 - 11 t \\ r = - \frac{145}{18} - \frac{11}{18} t \end{matrix}

– Description of step.

Substitute $\frac{t}{35} + \frac{32}{35}$ for $s$ and $- \frac{145}{18} - \frac{11}{18} t$ for $r$ into $- 6 r + 10 s + 3 t = - 19$ find the value of $t$ .

\begin{matrix} - 6 (- \frac{145}{18} - \frac{11}{18} t) + 10 (\frac{t}{35} + \frac{32}{35}) + 3 t = - 19 \\ (\frac{11}{3} + \frac{2}{7} + 3) t = - 19 - \frac{145}{3} - \frac{64}{7} \\ \frac{146}{21} t = - \frac{1606}{21} \\ t = - \frac{1606}{146} \\ = - 11 \end{matrix}

– Description of step.

Substitute $- 11$ for $t$ into $s = \frac{t}{35} + \frac{32}{35}$ and simplify.

\begin{matrix} s = \frac{- 11}{35} + \frac{32}{35} \\ = \frac{- 11 + 32}{35} \\ = \frac{3}{5} \end{matrix}

– Description of step.

Substitute $- 11$ for $t$ into $r = - \frac{145}{18} - \frac{11}{18} t$ and simplify.

\begin{matrix} r = - \frac{145}{18} - \frac{11}{18} \times (- 11) \\ = \frac{- 145 + 121}{18} \\ = - \frac{4}{3} \end{matrix}

Therefore, the solution of given system of equation is $(- \frac{4}{3}, \frac{3}{5}, - 11)$ .

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Solve each system of equations.
$\begin{matrix} 3 r - 15 s + 4 t = - 57 \\ 9 r + 45 s - t = 26 \\ - 6 r + 10 s + 3 t = - 19 \end{matrix}$

Short Answer

Step by step solution

– Description of step.

– Description of step.

– Description of step.

– Description of step.

– Description of step.

– Description of step.

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Solve each system of equations.3r-15s+4t=-579r+45s-t=26-6r+10s+3t=-19

Short Answer

Step by step solution

– Description of step.

– Description of step.

– Description of step.

– Description of step.

– Description of step.

– Description of step.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Applied Mathematics

Pure Maths

Decision Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Solve each system of equations.
$\begin{matrix} 3 r - 15 s + 4 t = - 57 \\ 9 r + 45 s - t = 26 \\ - 6 r + 10 s + 3 t = - 19 \end{matrix}$