Solve {l}{3x+y-z=27}{x-y+5z=100}{4x+5y-6z=-95}

$$\left. \begin{array} { l } { 3 x + y - z = 27 } \\ { x - y + 5 z = 100 } \\ { 4 x + 5 y - 6 z = - 95 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{793}{40} = 19\frac{33}{40} = 19.825$$

$$y = -\frac{411}{20} = -20\frac{11}{20} = -20.55$$

$$z = \frac{477}{40} = 11\frac{37}{40} = 11.925$$

Short Steps Using Substitution

Solve 3x+y-z=27 for y.

$$y=-3x+z+27$$

Substitute -3x+z+27 for y in the second and third equation.

$$x-\left(-3x+z+27\right)+5z=100 4x+5\left(-3x+z+27\right)-6z=-95$$

Solve these equations for x and z respectively.

$$x=-z+\frac{127}{4} z=-11x+230$$

Substitute -z+\frac{127}{4} for x in the equation z=-11x+230.

$$z=-11\left(-z+\frac{127}{4}\right)+230$$

Solve z=-11\left(-z+\frac{127}{4}\right)+230 for z.

$$z=\frac{477}{40}$$

Substitute \frac{477}{40} for z in the equation x=-z+\frac{127}{4}.

$$x=-\frac{477}{40}+\frac{127}{4}$$

Calculate x from x=-\frac{477}{40}+\frac{127}{4}.

$$x=\frac{793}{40}$$

Substitute \frac{793}{40} for x and \frac{477}{40} for z in the equation y=-3x+z+27.

$$y=-3\times \frac{793}{40}+\frac{477}{40}+27$$

Calculate y from y=-3\times \frac{793}{40}+\frac{477}{40}+27.

$$y=-\frac{411}{20}$$

The system is now solved.

$$x=\frac{793}{40} y=-\frac{411}{20} z=\frac{477}{40}$$