Question

Solve each system of equations.

$\begin{array}{c}3r-15s+4t=-57\\ 9r+45s-t=26\\ -6r+10s+3t=-19\end{array}$

Short answer

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

Step-by-step solution

– Description of step.

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

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

– Description of step.

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

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

– Description of step.

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

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

– Description of step.

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

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

– Description of step.

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

$\begin{array}{c}s=\frac{-11}{35}+\frac{32}{35}\\ =\frac{-11+32}{35}\\ =\frac{3}{5}\end{array}$

– Description of step.

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

$\begin{array}{c}r=-\frac{145}{18}-\frac{11}{18}\times \left(-11\right)\\ =\frac{-145+121}{18}\\ =-\frac{4}{3}\end{array}$

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