Solve {l}{4x+5y+3z=2}{2x+4y-5z=1}{4x+3y+5z=5}

$$\left. \begin{array} { l } { 4 x + 5 y + 3 z = 2 } \\ { 2 x + 4 y - 5 z = 1 } \\ { 4 x + 3 y + 5 z = 5 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{131}{40} = 3\frac{11}{40} = 3.275$$

$$y = -\frac{39}{20} = -1\frac{19}{20} = -1.95$$

$$z=-\frac{9}{20}=-0.45$$

Short Steps Using Substitution

Solve 4x+5y+3z=2 for x.

$$x=-\frac{5}{4}y-\frac{3}{4}z+\frac{1}{2}$$

Substitute -\frac{5}{4}y-\frac{3}{4}z+\frac{1}{2} for x in the second and third equation.

$$2\left(-\frac{5}{4}y-\frac{3}{4}z+\frac{1}{2}\right)+4y-5z=1 4\left(-\frac{5}{4}y-\frac{3}{4}z+\frac{1}{2}\right)+3y+5z=5$$

Solve these equations for y and z respectively.

$$y=\frac{13}{3}z z=\frac{3}{2}+y$$

Substitute \frac{13}{3}z for y in the equation z=\frac{3}{2}+y.

$$z=\frac{3}{2}+\frac{13}{3}z$$

Solve z=\frac{3}{2}+\frac{13}{3}z for z.

$$z=-\frac{9}{20}$$

Substitute -\frac{9}{20} for z in the equation y=\frac{13}{3}z.

$$y=\frac{13}{3}\left(-\frac{9}{20}\right)$$

Calculate y from y=\frac{13}{3}\left(-\frac{9}{20}\right).

$$y=-\frac{39}{20}$$

Substitute -\frac{39}{20} for y and -\frac{9}{20} for z in the equation x=-\frac{5}{4}y-\frac{3}{4}z+\frac{1}{2}.

$$x=-\frac{5}{4}\left(-\frac{39}{20}\right)-\frac{3}{4}\left(-\frac{9}{20}\right)+\frac{1}{2}$$

Calculate x from x=-\frac{5}{4}\left(-\frac{39}{20}\right)-\frac{3}{4}\left(-\frac{9}{20}\right)+\frac{1}{2}.

$$x=\frac{131}{40}$$

The system is now solved.

$$x=\frac{131}{40} y=-\frac{39}{20} z=-\frac{9}{20}$$