Solve each system of equations

$\begin{array}{c}2a-b+3c=-7\\ 4a+5b+c=29\\ a-\frac{2}{3}b+\frac{1}{4}c=-10\end{array}$

Multiply the equation $2a-b+3c=-7$by $5$and add the new resultant equation to the equation $4a+5b+c=29$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}2a-b+3c=-7\\ 4a+5b+c=29\end{array}}\begin{array}{c}\text{multiplyby5}\to \\ \end{array}\underset{\_}{\begin{array}{l}10a-5b+15c=-35\\ 4a+5b+c=29\end{array}}\\ 14a+0+16c=-6\end{array}$

So, the resultant equation is $14a+16c=-6$.

Multiply the equation $2a-b+3c=-7$by $\frac{2}{3}$and subtract the new resultant equation from the equation $a-\frac{2}{3}b+\frac{1}{4}c=-10$

$\begin{array}{l}\underset{\_}{\begin{array}{l}2a-b+3c=-7\\ a-\frac{2}{3}b+\frac{1}{4}c=-10\end{array}}\begin{array}{c}\text{multiplyby}\frac{2}{3}\to \\ \end{array}\underset{\_}{\begin{array}{l}\frac{4}{3}a-\frac{2}{3}b+2c=-\frac{14}{3}\\ (-)a-\frac{2}{3}b+\frac{1}{4}c=-10\end{array}}\\ \frac{1}{3}a+0+\frac{7}{4}c=\frac{16}{3}\end{array}$

Multiply $\frac{1}{3}a+\frac{7}{4}c=\frac{16}{3}$by 42 and subtract the new resultant equation to $14a+16c=-6$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\frac{1}{3}a+\frac{7}{4}c=\frac{16}{3}\\ 14a+16c=-6\end{array}}\begin{array}{c}\text{multiplyby}42\\ \end{array}\underset{\_}{\begin{array}{l}14a+\frac{147}{2}c=224\\ (-)14a+16c=-6\end{array}}\\ 0+\frac{115}{2}c=230\end{array}$

Solve$\frac{115}{2}c=230$for c :

$\begin{array}{c}\frac{115}{2}c=230\\ \frac{2}{115}\cdot \frac{115c}{2}=\frac{2}{115}\cdot 230\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Multiplybothsidesby}\frac{2}{115}\\ c=4\end{array}$

Substitute $c=4in14a+16c=-6$and find the value of $a$

$\begin{array}{c}14a+16c=-6\\ 14a+16\left(4\right)=-6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Substitute}4\text{for}c\\ 14a+64=-6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Simplify}\\ 14a=-70\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract}64\text{frombothsides}\\ a=-5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividebothsidesby14}\end{array}$

Substitute$a=-5,c=4in2a-b+3c=-7$and find the value of $b$

$\begin{array}{c}2a-b+3c=-7\\ 2(-5)-b+3\left(4\right)=-7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substitute}-5\text{for}a,4\text{for}c\\ -10-b+12=-7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ 2-b=-7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}simplify}\\ -b=-9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Subtract}2\text{\hspace{0.17em}frombothsides}\\ b=9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividebothsidesby}-1\end{array}$

Hence, the solution of the given system of equations is$\left(a,b,c\right)=\left(-5,9,4\right)$.