Solve {l}{8x+y+6z=28}{3x+5y+7z=34}{4x+9y+2z=28}

$$\left. \begin{array} { l } { 8 x + y + 6 z = 28 } \\ { 3 x + 5 y + 7 z = 34 } \\ { 4 x + 9 y + 2 z = 28 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=2$$

$$z=3$$

Short Steps Using Substitution

Solve 8x+y+6z=28 for y.

$$y=28-8x-6z$$

Substitute 28-8x-6z for y in the second and third equation.

$$3x+5\left(28-8x-6z\right)+7z=34 4x+9\left(28-8x-6z\right)+2z=28$$

Solve these equations for x and z respectively.

$$x=\frac{106}{37}-\frac{23}{37}z z=\frac{56}{13}-\frac{17}{13}x$$

Substitute \frac{106}{37}-\frac{23}{37}z for x in the equation z=\frac{56}{13}-\frac{17}{13}x.

$$z=\frac{56}{13}-\frac{17}{13}\left(\frac{106}{37}-\frac{23}{37}z\right)$$

Solve z=\frac{56}{13}-\frac{17}{13}\left(\frac{106}{37}-\frac{23}{37}z\right) for z.

$$z=3$$

Substitute 3 for z in the equation x=\frac{106}{37}-\frac{23}{37}z.

$$x=\frac{106}{37}-\frac{23}{37}\times 3$$

Calculate x from x=\frac{106}{37}-\frac{23}{37}\times 3.

$$x=1$$

Substitute 1 for x and 3 for z in the equation y=28-8x-6z.

$$y=28-8-6\times 3$$

Calculate y from y=28-8-6\times 3.

$$y=2$$

The system is now solved.

$$x=1 y=2 z=3$$