Solve {l}{2/x+2/3x=1/6}{3/x+frac{2}{y}=0}

$$\left. \begin{array} { l } { \frac { 2 } { x } + \frac { 2 } { 3 x } = \frac { 1 } { 6 } } \\ { \frac { 3 } { x } + \frac { 2 } { y } = 0 } \end{array} \right.$$

Solve for x, y

$$x=16$$

$$y = -\frac{32}{3} = -10\frac{2}{3} \approx -10.666666667$$

Solution Steps

Consider the first equation. Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of x,3x,6.

$$6\times 2+2\times 2=x$$

Do the multiplications.

$$12+4=x$$

Add 12 and 4 to get 16.

$$16=x$$

Swap sides so that all variable terms are on the left hand side.

$$x=16$$

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

$$\frac{3}{16}+\frac{2}{y}=0$$

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 16y, the least common multiple of 16,y.

$$y\times 3+16\times 2=0$$

Multiply 16 and 2 to get 32.

$$y\times 3+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$y\times 3=-32$$

Divide both sides by 3.

$$y=-\frac{32}{3}$$

The system is now solved.

$$x=16 y=-\frac{32}{3}$$