Solve {c}{x+y+z=3243}{y=x+167}{10x+20y+30z=62710}

$$\left. \begin{array} { c } { x + y + z = 3243 } \\ { y = x + 167 } \\ { 10 x + 20 y + 30 z = 62710 } \end{array} \right.$$

Solve for x, y, z

$$x=1097$$

$$y=1264$$

$$z=882$$

Short Steps Using Substitution

Reorder the equations.

$$y=x+167 x+y+z=3243 10x+20y+30z=62710$$

Substitute x+167 for y in the second and third equation.

$$x+x+167+z=3243 10x+20\left(x+167\right)+30z=62710$$

Solve these equations for x and z respectively.

$$x=1538-\frac{1}{2}z z=1979-x$$

Substitute 1538-\frac{1}{2}z for x in the equation z=1979-x.

$$z=1979-\left(1538-\frac{1}{2}z\right)$$

Solve z=1979-\left(1538-\frac{1}{2}z\right) for z.

$$z=882$$

Substitute 882 for z in the equation x=1538-\frac{1}{2}z.

$$x=1538-\frac{1}{2}\times 882$$

Calculate x from x=1538-\frac{1}{2}\times 882.

$$x=1097$$

Substitute 1097 for x in the equation y=x+167.

$$y=1097+167$$

Calculate y from y=1097+167.

$$y=1264$$

The system is now solved.

$$x=1097 y=1264 z=882$$