Solve {l}{3p_{1}-p_{2}+p_{3}=2}{-15p_{1}+6p_{2}-5p_{3}=5}{5p_{1}-2p_{2}+2p_{3}=3}

$$\left. \begin{array} { l } { 3 p _ { 1 } - p _ { 2 } + p _ { 3 } = 2 } \\ { - 15 p _ { 1 } + 6 p _ { 2 } - 5 p _ { 3 } = 5 } \\ { 5 p _ { 1 } - 2 p _ { 2 } + 2 p _ { 3 } = 3 } \end{array} \right.$$

Solve for p_1, p_2, p_3

$$p_{1}=1$$

$$p_{2}=15$$

$$p_{3}=14$$

Short Steps Using Substitution

Solve 3p_{1}-p_{2}+p_{3}=2 for p_{2}.

$$p_{2}=3p_{1}+p_{3}-2$$

Substitute 3p_{1}+p_{3}-2 for p_{2} in the second and third equation.

$$-15p_{1}+6\left(3p_{1}+p_{3}-2\right)-5p_{3}=5 5p_{1}-2\left(3p_{1}+p_{3}-2\right)+2p_{3}=3$$

Solve these equations for p_{3} and p_{1} respectively.

$$p_{3}=-3p_{1}+17 p_{1}=1$$

Substitute 1 for p_{1} in the equation p_{3}=-3p_{1}+17.

$$p_{3}=-3+17$$

Calculate p_{3} from p_{3}=-3+17.

$$p_{3}=14$$

Substitute 14 for p_{3} and 1 for p_{1} in the equation p_{2}=3p_{1}+p_{3}-2.

$$p_{2}=3\times 1+14-2$$

Calculate p_{2} from p_{2}=3\times 1+14-2.

$$p_{2}=15$$

The system is now solved.

$$p_{1}=1 p_{2}=15 p_{3}=14$$