Question

Solve each system of equations.

$\begin{array}{c}3a+2b=27\\ 5a-7b+c=5\\ -2a+10b+5c=-29\end{array}$

Short answer

The solution of given system of equation is $\left(\mathbf{7}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{9}\right)$.

Step-by-step solution

– Description of step.

Multiply $5a-7b+c=5$ by 5 and subtract $-2a+10b+5c=-29$ from the obtained equation.

$\begin{array}{c}25a-35b+5c-\left(-2a+10b+5c\right)=25-\left(-29\right)\\ 25a-35b+5c+2a-10b-5c=54\\ 25a-35b+2a-10b=54\\ 27a-45b=54\end{array}$

– Description of step.

Multiply $3a+2b=27$by 9 and subtract $27a-45b=54$from the obtained equation.

$\begin{array}{c}27a+18b-\left(27a-45b\right)=243-54\\ 27a+18b-27a+45b=189\\ \frac{63b}{63}=\frac{189}{63}\\ b=3\end{array}$

– Description of step.

Substitute 3 for $b$into $3a+2b=27$and simplify to obtain the value of$a$.

$\begin{array}{c}3a+2\left(3\right)=27\\ 3a+6=27\\ \frac{3a}{3}=\frac{21}{3}\\ a=7\end{array}$

– Description of step.

Substitute 3 for $b$ and 7 for $a$into $5a-7b+c=5$and simplify to obtain the value of $c$.

$\begin{array}{c}5\left(7\right)-7\left(3\right)+c=5\\ 35-21+c=5\\ 14+c=5\\ c=-9\end{array}$

Therefore, the solution of given system of equation is $\left(7,3,-9\right)$.