Solve {c}{p+q+r=12}{2p-q+2r=12}{3p+2q+r=8}

$$\left. \begin{array} { c } { p + q + r = 12 } \\ { 2 p - q + 2 r = 12 } \\ { 3 p + 2 q + r = 8 } \end{array} \right.$$

Solve for p, q, r

$$r=12$$

$$p=-4$$

$$q=4$$

Short Steps Using Substitution

Solve p+q+r=12 for p.

$$p=-q-r+12$$

Substitute -q-r+12 for p in the second and third equation.

$$2\left(-q-r+12\right)-q+2r=12 3\left(-q-r+12\right)+2q+r=8$$

Solve these equations for q and r respectively.

$$q=4 r=-\frac{1}{2}q+14$$

Substitute 4 for q in the equation r=-\frac{1}{2}q+14.

$$r=-\frac{1}{2}\times 4+14$$

Calculate r from r=-\frac{1}{2}\times 4+14.

$$r=12$$

Substitute 4 for q and 12 for r in the equation p=-q-r+12.

$$p=-4-12+12$$

Calculate p from p=-4-12+12.

$$p=-4$$

The system is now solved.

$$p=-4 q=4 r=12$$