Solve {l}{2=-frac{sqrt{3}}{2}k+6}{1=frac{sqrt{3}}{2}k+b}

$$\left. \begin{array} { l } { 2 = - \frac { \sqrt { 3 } } { 2 } k + 6 } \\ { 1 = \frac { \sqrt { 3 } } { 2 } k + b } \end{array} \right.$$

Solve for k, b

$$k = \frac{8 \sqrt{3}}{3} \approx 4.618802154$$

$$b=-3$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 2.

$$4=2\left(-\frac{\sqrt{3}}{2}\right)k+12$$

Express 2\left(-\frac{\sqrt{3}}{2}\right) as a single fraction.

$$4=\frac{-2\sqrt{3}}{2}k+12$$

Cancel out 2 and 2.

$$4=-\sqrt{3}k+12$$

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

$$-\sqrt{3}k+12=4$$

Subtract 12 from both sides.

$$-\sqrt{3}k=4-12$$

Subtract 12 from 4 to get -8.

$$-\sqrt{3}k=-8$$

Consider the second equation. Multiply both sides of the equation by 2.

$$2=\sqrt{3}k+2b$$

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

$$\sqrt{3}k+2b=2$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$\left(-\sqrt{3}\right)k=-8,\sqrt{3}k+2b=2$$

Pick one of the two equations which is more simple to solve for k by isolating k on the left hand side of the equal sign.

$$\left(-\sqrt{3}\right)k=-8$$

Divide both sides by -\sqrt{3}.

$$k=\frac{8\sqrt{3}}{3}$$

Substitute \frac{8\sqrt{3}}{3} for k in the other equation, \sqrt{3}k+2b=2.

$$\sqrt{3}\times \frac{8\sqrt{3}}{3}+2b=2$$

Multiply \sqrt{3} times \frac{8\sqrt{3}}{3}.

$$8+2b=2$$

Subtract 8 from both sides of the equation.

$$2b=-6$$

Divide both sides by 2.

$$b=-3$$

The system is now solved.

$$k=\frac{8\sqrt{3}}{3},b=-3$$