Solve {c}{x+y+z=280}{x+3y+4z=945}{x=y+10}

$$\left.\begin{array} { c } { x + y + z = 280 } \\ { x + 3 y + 4 z = 945 } \\ { x = y + 10 } \end{array} \right\}$$

Solve for x, y, z

$$x = \frac{185}{4} = 46\frac{1}{4} = 46.25$$

$$y = \frac{145}{4} = 36\frac{1}{4} = 36.25$$

$$z = \frac{395}{2} = 197\frac{1}{2} = 197.5$$

Short Steps Using Substitution

Reorder the equations.

$$x=y+10 x+3y+4z=945 x+y+z=280$$

Substitute y+10 for x in the second and third equation.

$$y+10+3y+4z=945 y+10+y+z=280$$

Solve these equations for y and z respectively.

$$y=\frac{935}{4}-z z=270-2y$$

Substitute \frac{935}{4}-z for y in the equation z=270-2y.

$$z=270-2\left(\frac{935}{4}-z\right)$$

Solve z=270-2\left(\frac{935}{4}-z\right) for z.

$$z=\frac{395}{2}$$

Substitute \frac{395}{2} for z in the equation y=\frac{935}{4}-z.

$$y=\frac{935}{4}-\frac{395}{2}$$

Calculate y from y=\frac{935}{4}-\frac{395}{2}.

$$y=\frac{145}{4}$$

Substitute \frac{145}{4} for y in the equation x=y+10.

$$x=\frac{145}{4}+10$$

Calculate x from x=\frac{145}{4}+10.

$$x=\frac{185}{4}$$

The system is now solved.

$$x=\frac{185}{4} y=\frac{145}{4} z=\frac{395}{2}$$