Solve {l}{10x+20y+30z=41}{12x+20y+28z=42}{14x+25y+40z=54}

$$\left. \begin{array}{l}{ 10 x + 20 y + 30 z = 41 }\\{ 12 x + 20 y + 28 z = 42 }\\{ 14 x + 25 y + 40 z = 54 }\end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=\frac{4}{5}=0.8$$

$$z=\frac{1}{2}=0.5$$

Short Steps Using Substitution

Solve 10x+20y+30z=41 for x.

$$x=-2y-3z+\frac{41}{10}$$

Substitute -2y-3z+\frac{41}{10} for x in the second and third equation.

$$12\left(-2y-3z+\frac{41}{10}\right)+20y+28z=42 14\left(-2y-3z+\frac{41}{10}\right)+25y+40z=54$$

Solve these equations for y and z respectively.

$$y=\frac{9}{5}-2z z=\frac{17}{10}-\frac{3}{2}y$$

Substitute \frac{9}{5}-2z for y in the equation z=\frac{17}{10}-\frac{3}{2}y.

$$z=\frac{17}{10}-\frac{3}{2}\left(\frac{9}{5}-2z\right)$$

Solve z=\frac{17}{10}-\frac{3}{2}\left(\frac{9}{5}-2z\right) for z.

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

Substitute \frac{1}{2} for z in the equation y=\frac{9}{5}-2z.

$$y=\frac{9}{5}-2\times \frac{1}{2}$$

Calculate y from y=\frac{9}{5}-2\times \frac{1}{2}.

$$y=\frac{4}{5}$$

Substitute \frac{4}{5} for y and \frac{1}{2} for z in the equation x=-2y-3z+\frac{41}{10}.

$$x=-2\times \frac{4}{5}-3\times \frac{1}{2}+\frac{41}{10}$$

Calculate x from x=-2\times \frac{4}{5}-3\times \frac{1}{2}+\frac{41}{10}.

$$x=1$$

The system is now solved.

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