Solve {l}{18x+4y+3z=3}{6x-15y-2z=49}{3x-9y-20z=-41}

$$\left. \begin{array} { l } { 18 x + 4 y + 3 z = 3 } \\ { 6 x - 15 y - 2 z = 49 } \\ { 3 x - 9 y - 20 z = - 41 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{642}{1835}\approx 0.34986376$$

$$y = -\frac{6651}{1835} = -3\frac{1146}{1835} \approx -3.624523161$$

$$z = \frac{6851}{1835} = 3\frac{1346}{1835} \approx 3.733514986$$

Short Steps Using Substitution

Solve 18x+4y+3z=3 for x.

$$x=-\frac{2}{9}y-\frac{1}{6}z+\frac{1}{6}$$

Substitute -\frac{2}{9}y-\frac{1}{6}z+\frac{1}{6} for x in the second and third equation.

$$6\left(-\frac{2}{9}y-\frac{1}{6}z+\frac{1}{6}\right)-15y-2z=49 3\left(-\frac{2}{9}y-\frac{1}{6}z+\frac{1}{6}\right)-9y-20z=-41$$

Solve these equations for y and z respectively.

$$y=-\frac{144}{49}-\frac{9}{49}z z=\frac{83}{41}-\frac{58}{123}y$$

Substitute -\frac{144}{49}-\frac{9}{49}z for y in the equation z=\frac{83}{41}-\frac{58}{123}y.

$$z=\frac{83}{41}-\frac{58}{123}\left(-\frac{144}{49}-\frac{9}{49}z\right)$$

Solve z=\frac{83}{41}-\frac{58}{123}\left(-\frac{144}{49}-\frac{9}{49}z\right) for z.

$$z=\frac{6851}{1835}$$

Substitute \frac{6851}{1835} for z in the equation y=-\frac{144}{49}-\frac{9}{49}z.

$$y=-\frac{144}{49}-\frac{9}{49}\times \frac{6851}{1835}$$

Calculate y from y=-\frac{144}{49}-\frac{9}{49}\times \frac{6851}{1835}.

$$y=-\frac{6651}{1835}$$

Substitute -\frac{6651}{1835} for y and \frac{6851}{1835} for z in the equation x=-\frac{2}{9}y-\frac{1}{6}z+\frac{1}{6}.

$$x=-\frac{2}{9}\left(-\frac{6651}{1835}\right)-\frac{1}{6}\times \frac{6851}{1835}+\frac{1}{6}$$

Calculate x from x=-\frac{2}{9}\left(-\frac{6651}{1835}\right)-\frac{1}{6}\times \frac{6851}{1835}+\frac{1}{6}.

$$x=\frac{642}{1835}$$

The system is now solved.

$$x=\frac{642}{1835} y=-\frac{6651}{1835} z=\frac{6851}{1835}$$