Solve {cc}{x}&{y}{2}&{4}{l}{6}{x}={r}{-2}{4}

$$\left( \begin{array} { c c } { x } & { y } \\ { 2 } & { 4 } \end{array} \right) \left( \begin{array} { l } { 6 } \\ { x } \end{array} \right) = \left( \begin{array} { r } { - 2 } \\ { 4 } \end{array} \right)$$

Solve for x, y

$$x=-2$$

$$y=-5$$

Solution Steps

Consider the second equation. Subtract 12 from both sides.

$$4x=4-12$$

Subtract 12 from 4 to get -8.

$$4x=-8$$

Divide both sides by 4.

$$x=\frac{-8}{4}$$

Divide -8 by 4 to get -2.

$$x=-2$$

Consider the first equation. Insert the known values of variables into the equation.

$$6\left(-2\right)+y\left(-2\right)=-2$$

Multiply 6 and -2 to get -12.

$$-12+y\left(-2\right)=-2$$

Add 12 to both sides.

$$y\left(-2\right)=-2+12$$

Add -2 and 12 to get 10.

$$y\left(-2\right)=10$$

Divide both sides by -2.

$$y=\frac{10}{-2}$$

Divide 10 by -2 to get -5.

$$y=-5$$

The system is now solved.

$$x=-2 y=-5$$