Solve {r}{6x+3y=0}{3y-y=3}

$$\left. \begin{array} { r } { 6 x + 3 y = 0 } \\ { 3 y - y = 3 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{3}{4}=-0.75$$

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Consider the second equation. Combine 3y and -y to get 2y.

$$2y=3$$

Divide both sides by 2.

$$y=\frac{3}{2}$$

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

$$6x+3\times \frac{3}{2}=0$$

Multiply 3 and \frac{3}{2} to get \frac{9}{2}.

$$6x+\frac{9}{2}=0$$

Subtract \frac{9}{2} from both sides. Anything subtracted from zero gives its negation.

$$6x=-\frac{9}{2}$$

Divide both sides by 6.

$$x=\frac{-\frac{9}{2}}{6}$$

Express \frac{-\frac{9}{2}}{6} as a single fraction.

$$x=\frac{-9}{2\times 6}$$

Multiply 2 and 6 to get 12.

$$x=\frac{-9}{12}$$

Reduce the fraction \frac{-9}{12} to lowest terms by extracting and canceling out 3.

$$x=-\frac{3}{4}$$

The system is now solved.

$$x=-\frac{3}{4} y=\frac{3}{2}$$