Solve {c}{q-p=1}{3p+2q+gamma=2}{-2p-2q-gamma=-2}

$$\left. \begin{array} { c } { q - p = 1 } \\ { 3 p + 2 q + \gamma = 2 } \\ { - 2 p - 2 q - \gamma = - 2 } \end{array} \right.$$

Solve for q, p, γ

$$q=1$$

$$p=0$$

$$\gamma =0$$

Short Steps Using Substitution

Solve q-p=1 for q.

$$q=p+1$$

Substitute p+1 for q in the second and third equation.

$$3p+2\left(p+1\right)+\gamma =2 -2p-2\left(p+1\right)-\gamma =-2$$

Solve these equations for p and \gamma respectively.

$$p=-\frac{1}{5}\gamma \gamma =-4p$$

Substitute -\frac{1}{5}\gamma for p in the equation \gamma =-4p.

$$\gamma =-4\left(-\frac{1}{5}\right)\gamma$$

Solve \gamma =-4\left(-\frac{1}{5}\right)\gamma for \gamma .

$$\gamma =0$$

Substitute 0 for \gamma in the equation p=-\frac{1}{5}\gamma .

$$p=-\frac{1}{5}\times 0$$

Calculate p from p=-\frac{1}{5}\times 0.

$$p=0$$

Substitute 0 for p in the equation q=p+1.

$$q=0+1$$

Calculate q from q=0+1.

$$q=1$$

The system is now solved.

$$q=1 p=0 \gamma =0$$