Solve {c}{x+y=42}{(x+z)+(y+z)=90}{x+z=y}

$$\left. \begin{array} { c } { x + y = 42 } \\ { ( x + z ) + ( y + z ) = 90 } \\ { x + z = y } \end{array} \right.$$

Solve for x, y, z

$$x=9$$

$$y=33$$

$$z=24$$

Short Steps Using Substitution

Reorder the equations.

$$x+z=y x+z+y+z=90 x+y=42$$

Solve x+z=y for y.

$$y=x+z$$

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

$$x+z+x+z+z=90 x+x+z=42$$

Solve these equations for x and z respectively.

$$x=45-\frac{3}{2}z z=42-2x$$

Substitute 45-\frac{3}{2}z for x in the equation z=42-2x.

$$z=42-2\left(45-\frac{3}{2}z\right)$$

Solve z=42-2\left(45-\frac{3}{2}z\right) for z.

$$z=24$$

Substitute 24 for z in the equation x=45-\frac{3}{2}z.

$$x=45-\frac{3}{2}\times 24$$

Calculate x from x=45-\frac{3}{2}\times 24.

$$x=9$$

Substitute 9 for x and 24 for z in the equation y=x+z.

$$y=9+24$$

Calculate y from y=9+24.

$$y=33$$

The system is now solved.

$$x=9 y=33 z=24$$