Use Cramer’s Rule to solve each system of equations.

$x+y+z=6$

$2x+y-4z=-15$

$5x-3y+z=-10$

Apply the Cramer’s rule for the value of x.

$x=\frac{\left|\begin{array}{ccc}j& b& c\\ k& e& f\\ l& h& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$

Use the provided equations and substitute the value of a, b, c, d, e, f, g, h, i, j, k and l into role="math" localid="1647246488390" $x=\frac{\left|\begin{array}{ccc}j& b& c\\ k& e& f\\ l& h& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of x.

$\begin{array}{c}x=\frac{\left|\begin{array}{ccc}6& 1& 1\\ -15& 1& -4\\ -10& -3& 1\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -4\\ 5& -3& 1\end{array}\right|}\\ =\frac{44}{-44}\\ =-1\end{array}$

Apply the Cramer’s rule for the value of y.

$y=\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$

Use the provided equations and substitute the value of a, b, c, d, e, f, g, h, i, j, k and l into role="math" localid="1647246618728" $y=\frac{\left|\begin{array}{ccc}a& j& c\\ d& k& f\\ g& l& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$after it perform simplification of determinants to write the value of y.

$\begin{array}{c}y=\frac{\left|\begin{array}{ccc}1& 6& 1\\ 2& -15& -4\\ 5& -10& 1\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -4\\ 5& -3& 1\end{array}\right|}\\ =\frac{-132}{-44}\\ =3\end{array}$

Apply the Cramer’s rule for the value of z.

$z=\frac{\left|\begin{array}{ccc}a& b& j\\ d& e& k\\ g& h& l\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}$

Use the provided equations and substitute the value of a, b, c, d, e, f, g, h, i, j, k and l into role="math" localid="1647246777197" $m=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$after it perform simplification of determinants to write the value of z.

$\begin{array}{c}z=\frac{\left|\begin{array}{ccc}1& 1& 6\\ 2& 1& -15\\ 5& -3& -10\end{array}\right|}{\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -4\\ 5& -3& 1\end{array}\right|}\\ =\frac{-176}{-44}\\ =4\end{array}$

The obtained value of x is $-1$, y is 3 and z is 4 so the solution is $\left(-1,3,4\right)$.