Solve {rrr}{p}&{2}&{-1}{q}&{-2}&{2q}{l}{2}{1}{3}={r}{5}{-10}

$$\left( \begin{array} { r r r } { p } & { 2 } & { - 1 } \\ { q } & { - 2 } & { 2 q } \end{array} \right) \left( \begin{array} { l } { 2 } \\ { 1 } \\ { 3 } \end{array} \right) = \left( \begin{array} { r } { 5 } \\ { - 10 } \end{array} \right)$$

Solve for p, q

$$p=3$$

$$q=-1$$

Solution Steps

Consider the first equation. Add 1 to both sides.

$$2p=5+1$$

Add 5 and 1 to get 6.

$$2p=6$$

Divide both sides by 2.

$$p=\frac{6}{2}$$

Divide 6 by 2 to get 3.

$$p=3$$

Consider the second equation. Add 2 to both sides.

$$8q=-10+2$$

Add -10 and 2 to get -8.

$$8q=-8$$

Divide both sides by 8.

$$q=\frac{-8}{8}$$

Divide -8 by 8 to get -1.

$$q=-1$$

The system is now solved.

$$p=3 q=-1$$