Solve {l}{3x+7z=7}{2x+3y+z=9}{5x-4y+1z=8}

$$\left\{ \begin{array} { l } { 3 x + 7 z = 7 } \\ { 2 x + 3 y + z = 9 } \\ { 5 x - 4 y + 1 z = 8 } \end{array} \right.$$

Solve for x, z, y

$$x = \frac{53}{20} = 2\frac{13}{20} = 2.65$$

$$y = \frac{179}{140} = 1\frac{39}{140} \approx 1.278571429$$

$$z=-\frac{19}{140}\approx -0.135714286$$

Short Steps Using Substitution

Reorder the equations.

$$2x+3y+z=9 3x+7z=7 5x-4y+1z=8$$

Solve 2x+3y+z=9 for z.

$$z=-2x-3y+9$$

Substitute -2x-3y+9 for z in the second and third equation.

$$3x+7\left(-2x-3y+9\right)=7 5x-4y+1\left(-2x-3y+9\right)=8$$

Solve these equations for x and y respectively.

$$x=\frac{56}{11}-\frac{21}{11}y y=\frac{3}{7}x+\frac{1}{7}$$

Substitute \frac{56}{11}-\frac{21}{11}y for x in the equation y=\frac{3}{7}x+\frac{1}{7}.

$$y=\frac{3}{7}\left(\frac{56}{11}-\frac{21}{11}y\right)+\frac{1}{7}$$

Solve y=\frac{3}{7}\left(\frac{56}{11}-\frac{21}{11}y\right)+\frac{1}{7} for y.

$$y=\frac{179}{140}$$

Substitute \frac{179}{140} for y in the equation x=\frac{56}{11}-\frac{21}{11}y.

$$x=\frac{56}{11}-\frac{21}{11}\times \frac{179}{140}$$

Calculate x from x=\frac{56}{11}-\frac{21}{11}\times \frac{179}{140}.

$$x=\frac{53}{20}$$

Substitute \frac{53}{20} for x and \frac{179}{140} for y in the equation z=-2x-3y+9.

$$z=-2\times \frac{53}{20}-3\times \frac{179}{140}+9$$

Calculate z from z=-2\times \frac{53}{20}-3\times \frac{179}{140}+9.

$$z=-\frac{19}{140}$$

The system is now solved.

$$x=\frac{53}{20} z=-\frac{19}{140} y=\frac{179}{140}$$