Solve {l}{x+2y+3z=1}{x+4y+3z=0}{2x+7y+7z=3}

$$\left. \begin{array} { l } { x + 2 y + 3 z = 1 } \\ { x + 4 y + 3 z = 0 } \\ { 2 x + 7 y + 7 z = 3 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{11}{2} = -5\frac{1}{2} = -5.5$$

$$y=-\frac{1}{2}=-0.5$$

$$z = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Short Steps Using Substitution

Solve x+2y+3z=1 for x.

$$x=-2y-3z+1$$

Substitute -2y-3z+1 for x in the second and third equation.

$$-2y-3z+1+4y+3z=0 2\left(-2y-3z+1\right)+7y+7z=3$$

Solve these equations for y and z respectively.

$$y=-\frac{1}{2} z=-3y+1$$

Substitute -\frac{1}{2} for y in the equation z=-3y+1.

$$z=-3\left(-\frac{1}{2}\right)+1$$

Calculate z from z=-3\left(-\frac{1}{2}\right)+1.

$$z=\frac{5}{2}$$

Substitute -\frac{1}{2} for y and \frac{5}{2} for z in the equation x=-2y-3z+1.

$$x=-2\left(-\frac{1}{2}\right)-3\times \frac{5}{2}+1$$

Calculate x from x=-2\left(-\frac{1}{2}\right)-3\times \frac{5}{2}+1.

$$x=-\frac{11}{2}$$

The system is now solved.

$$x=-\frac{11}{2} y=-\frac{1}{2} z=\frac{5}{2}$$