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

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

Solve for x, y, z

$$x=-1,95$$

$$y=7,9$$

$$z=1,05$$

Short Steps Using Substitution

Solve 2x+y=4 for y.

$$y=4-2x$$

Substitute 4-2x for y in the equation x+y+z=7.

$$x+4-2x+z=7$$

Solve the second equation for x and the third equation for z.

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

Substitute -3z+1,2 for x in the equation z=x+3.

$$z=-3z+1,2+3$$

Solve z=-3z+1,2+3 for z.

$$z=\frac{21}{20}$$

Substitute \frac{21}{20} for z in the equation x=-3z+1,2.

$$x=-3\times \frac{21}{20}+1,2$$

Calculate x from x=-3\times \frac{21}{20}+1,2.

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

Substitute -\frac{39}{20} for x in the equation y=4-2x.

$$y=4-2\left(-\frac{39}{20}\right)$$

Calculate y from y=4-2\left(-\frac{39}{20}\right).

$$y=\frac{79}{10}$$

The system is now solved.

$$x=-\frac{39}{20} y=\frac{79}{10} z=\frac{21}{20}$$