Solve {l}{-sqrt{3}=-sqrt{2}k_{1}+frac{k_{2}}{-sqrt{2}}}{19/6=1/3k_{1}+3k_{2}}

$$\left\{ \begin{array} { l } { - \sqrt { 3 } = - \sqrt { 2 } k _ { 1 } + \frac { k _ { 2 } } { - \sqrt { 2 } } } \\ { \frac { 19 } { 6 } = \frac { 1 } { 3 } k _ { 1 } + 3 k _ { 2 } } \end{array} \right.$$

Solve for k_1, k_2

$$k_{1}=\frac{9\sqrt{6}}{17}-\frac{19}{34}\approx 0.737965158$$

$$k_{2}=\frac{19-\sqrt{6}}{17}\approx 0.973559427$$

Steps Using Substitution

Consider the first equation. To add or subtract expressions, expand them to make their denominators the same. Multiply \left(-\sqrt{2}\right)k_{1} times \frac{-\sqrt{2}}{-\sqrt{2}}.

$$-\sqrt{3}=\frac{\left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)}{-\sqrt{2}}+\frac{k_{2}}{-\sqrt{2}}$$

Since \frac{\left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)}{-\sqrt{2}} and \frac{k_{2}}{-\sqrt{2}} have the same denominator, add them by adding their numerators.

$$-\sqrt{3}=\frac{\left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)+k_{2}}{-\sqrt{2}}$$

Do the multiplications in \left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)+k_{2}.

$$-\sqrt{3}=\frac{2k_{1}+k_{2}}{-\sqrt{2}}$$

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

$$\frac{2k_{1}+k_{2}}{-\sqrt{2}}=-\sqrt{3}$$

Rationalize the denominator of \frac{2k_{1}+k_{2}}{-\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(2k_{1}+k_{2}\right)\sqrt{2}}{-\left(\sqrt{2}\right)^{2}}=-\sqrt{3}$$

The square of \sqrt{2} is 2.

$$\frac{\left(2k_{1}+k_{2}\right)\sqrt{2}}{-2}=-\sqrt{3}$$

Use the distributive property to multiply 2k_{1}+k_{2} by \sqrt{2}.

$$\frac{2k_{1}\sqrt{2}+k_{2}\sqrt{2}}{-2}=-\sqrt{3}$$

Multiply both sides of the equation by -2.

$$2k_{1}\sqrt{2}+k_{2}\sqrt{2}=2\sqrt{3}$$

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

$$\frac{1}{3}k_{1}+3k_{2}=\frac{19}{6}$$

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.

$$2\sqrt{2}k_{1}+\sqrt{2}k_{2}=2\sqrt{3},\frac{1}{3}k_{1}+3k_{2}=\frac{19}{6}$$

Choose one of the equations and solve it for k_{1} by isolating k_{1} on the left hand side of the equal sign.

$$2\sqrt{2}k_{1}+\sqrt{2}k_{2}=2\sqrt{3}$$

Subtract \sqrt{2}k_{2} from both sides of the equation.

$$2\sqrt{2}k_{1}=\left(-\sqrt{2}\right)k_{2}+2\sqrt{3}$$

Divide both sides by 2\sqrt{2}.

$$k_{1}=\frac{\sqrt{2}}{4}\left(\left(-\sqrt{2}\right)k_{2}+2\sqrt{3}\right)$$

Multiply \frac{\sqrt{2}}{4} times -\sqrt{2}k_{2}+2\sqrt{3}.

$$k_{1}=-\frac{1}{2}k_{2}+\frac{\sqrt{6}}{2}$$

Substitute \frac{-k_{2}+\sqrt{6}}{2} for k_{1} in the other equation, \frac{1}{3}k_{1}+3k_{2}=\frac{19}{6}.

$$\frac{1}{3}\left(-\frac{1}{2}k_{2}+\frac{\sqrt{6}}{2}\right)+3k_{2}=\frac{19}{6}$$

Multiply \frac{1}{3} times \frac{-k_{2}+\sqrt{6}}{2}.

$$-\frac{1}{6}k_{2}+\frac{\sqrt{6}}{6}+3k_{2}=\frac{19}{6}$$

Add -\frac{k_{2}}{6} to 3k_{2}.

$$\frac{17}{6}k_{2}+\frac{\sqrt{6}}{6}=\frac{19}{6}$$

Subtract \frac{\sqrt{6}}{6} from both sides of the equation.

$$\frac{17}{6}k_{2}=\frac{19-\sqrt{6}}{6}$$

Divide both sides of the equation by \frac{17}{6}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$k_{2}=\frac{19-\sqrt{6}}{17}$$

Substitute \frac{19-\sqrt{6}}{17} for k_{2} in k_{1}=-\frac{1}{2}k_{2}+\frac{\sqrt{6}}{2}. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$k_{1}=-\frac{1}{2}\times \frac{19-\sqrt{6}}{17}+\frac{\sqrt{6}}{2}$$

Multiply -\frac{1}{2} times \frac{19-\sqrt{6}}{17}.

$$k_{1}=\frac{\sqrt{6}-19}{34}+\frac{\sqrt{6}}{2}$$

Add \frac{\sqrt{6}}{2} to \frac{-19+\sqrt{6}}{34}.

$$k_{1}=\frac{9\sqrt{6}}{17}-\frac{19}{34}$$

The system is now solved.

$$k_{1}=\frac{9\sqrt{6}}{17}-\frac{19}{34},k_{2}=\frac{19-\sqrt{6}}{17}$$

Steps Using Elimination

Consider the first equation. To add or subtract expressions, expand them to make their denominators the same. Multiply \left(-\sqrt{2}\right)k_{1} times \frac{-\sqrt{2}}{-\sqrt{2}}.

$$-\sqrt{3}=\frac{\left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)}{-\sqrt{2}}+\frac{k_{2}}{-\sqrt{2}}$$

Since \frac{\left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)}{-\sqrt{2}} and \frac{k_{2}}{-\sqrt{2}} have the same denominator, add them by adding their numerators.

$$-\sqrt{3}=\frac{\left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)+k_{2}}{-\sqrt{2}}$$

Do the multiplications in \left(-\sqrt{2}\right)k_{1}\left(-\sqrt{2}\right)+k_{2}.

$$-\sqrt{3}=\frac{2k_{1}+k_{2}}{-\sqrt{2}}$$

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

$$\frac{2k_{1}+k_{2}}{-\sqrt{2}}=-\sqrt{3}$$

Rationalize the denominator of \frac{2k_{1}+k_{2}}{-\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(2k_{1}+k_{2}\right)\sqrt{2}}{-\left(\sqrt{2}\right)^{2}}=-\sqrt{3}$$

The square of \sqrt{2} is 2.

$$\frac{\left(2k_{1}+k_{2}\right)\sqrt{2}}{-2}=-\sqrt{3}$$

Use the distributive property to multiply 2k_{1}+k_{2} by \sqrt{2}.

$$\frac{2k_{1}\sqrt{2}+k_{2}\sqrt{2}}{-2}=-\sqrt{3}$$

Multiply both sides of the equation by -2.

$$2k_{1}\sqrt{2}+k_{2}\sqrt{2}=2\sqrt{3}$$

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

$$\frac{1}{3}k_{1}+3k_{2}=\frac{19}{6}$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$2\sqrt{2}k_{1}+\sqrt{2}k_{2}=2\sqrt{3},\frac{1}{3}k_{1}+3k_{2}=\frac{19}{6}$$

To make 2k_{1}\sqrt{2} and \frac{k_{1}}{3} equal, multiply all terms on each side of the first equation by \frac{1}{3} and all terms on each side of the second by 2\sqrt{2}.

$$\frac{1}{3}\times 2\sqrt{2}k_{1}+\frac{1}{3}\sqrt{2}k_{2}=\frac{1}{3}\times 2\sqrt{3},2\sqrt{2}\times \frac{1}{3}k_{1}+2\sqrt{2}\times 3k_{2}=2\sqrt{2}\times \frac{19}{6}$$

Simplify.

$$\frac{2\sqrt{2}}{3}k_{1}+\frac{\sqrt{2}}{3}k_{2}=\frac{2\sqrt{3}}{3},\frac{2\sqrt{2}}{3}k_{1}+6\sqrt{2}k_{2}=\frac{19\sqrt{2}}{3}$$

Subtract \frac{2\sqrt{2}}{3}k_{1}+6\sqrt{2}k_{2}=\frac{19\sqrt{2}}{3} from \frac{2\sqrt{2}}{3}k_{1}+\frac{\sqrt{2}}{3}k_{2}=\frac{2\sqrt{3}}{3} by subtracting like terms on each side of the equal sign.

$$\frac{2\sqrt{2}}{3}k_{1}+\left(-\frac{2\sqrt{2}}{3}\right)k_{1}+\frac{\sqrt{2}}{3}k_{2}+\left(-6\sqrt{2}\right)k_{2}=\frac{2\sqrt{3}-19\sqrt{2}}{3}$$

Add \frac{2\sqrt{2}k_{1}}{3} to -\frac{2\sqrt{2}k_{1}}{3}. Terms \frac{2\sqrt{2}k_{1}}{3} and -\frac{2\sqrt{2}k_{1}}{3} cancel out, leaving an equation with only one variable that can be solved.

$$\frac{\sqrt{2}}{3}k_{2}+\left(-6\sqrt{2}\right)k_{2}=\frac{2\sqrt{3}-19\sqrt{2}}{3}$$

Add \frac{\sqrt{2}k_{2}}{3} to -6\sqrt{2}k_{2}.

$$\left(-\frac{17\sqrt{2}}{3}\right)k_{2}=\frac{2\sqrt{3}-19\sqrt{2}}{3}$$

Divide both sides by -\frac{17\sqrt{2}}{3}.

$$k_{2}=\frac{19-\sqrt{6}}{17}$$

Substitute \frac{-\sqrt{6}+19}{17} for k_{2} in \frac{1}{3}k_{1}+3k_{2}=\frac{19}{6}. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$\frac{1}{3}k_{1}+3\times \frac{19-\sqrt{6}}{17}=\frac{19}{6}$$

Multiply 3 times \frac{-\sqrt{6}+19}{17}.

$$\frac{1}{3}k_{1}+\frac{57-3\sqrt{6}}{17}=\frac{19}{6}$$

Subtract \frac{-3\sqrt{6}+57}{17} from both sides of the equation.

$$\frac{1}{3}k_{1}=\frac{3\sqrt{6}}{17}-\frac{19}{102}$$

Multiply both sides by 3.

$$k_{1}=\frac{9\sqrt{6}}{17}-\frac{19}{34}$$

The system is now solved.

$$k_{1}=\frac{9\sqrt{6}}{17}-\frac{19}{34},k_{2}=\frac{19-\sqrt{6}}{17}$$