Solve {left(frac{-9-4sqrt{3}}{3}right)}^2x+(-9-4/3/3)y=2

$${ \left( \frac{ -9-4 \sqrt{ 3 } }{ 3 } \right) }^{ 2 } x+( \frac{ -9- \frac{ 4 }{ 3 } }{ 3 } )y=2$$

Solve for x

$$x=-\frac{\left(24\sqrt{3}-43\right)\left(31y+18\right)}{363}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3.

$$3\times \left(\frac{-9-4\sqrt{3}}{3}\right)^{2}x+\left(-9-\frac{4}{3}\right)y=6$$

To raise \frac{-9-4\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(-9-4\sqrt{3}\right)^{2}}{3^{2}}x+\left(-9-\frac{4}{3}\right)y=6$$

Express 3\times \frac{\left(-9-4\sqrt{3}\right)^{2}}{3^{2}} as a single fraction.

$$\frac{3\left(-9-4\sqrt{3}\right)^{2}}{3^{2}}x+\left(-9-\frac{4}{3}\right)y=6$$

Cancel out 3 in both numerator and denominator.

$$\frac{\left(-4\sqrt{3}-9\right)^{2}}{3}x+\left(-9-\frac{4}{3}\right)y=6$$

Express \frac{\left(-4\sqrt{3}-9\right)^{2}}{3}x as a single fraction.

$$\frac{\left(-4\sqrt{3}-9\right)^{2}x}{3}+\left(-9-\frac{4}{3}\right)y=6$$

Subtract \frac{4}{3} from -9 to get -\frac{31}{3}.

$$\frac{\left(-4\sqrt{3}-9\right)^{2}x}{3}-\frac{31}{3}y=6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4\sqrt{3}-9\right)^{2}.

$$\frac{\left(16\left(\sqrt{3}\right)^{2}+72\sqrt{3}+81\right)x}{3}-\frac{31}{3}y=6$$

The square of \sqrt{3} is 3.

$$\frac{\left(16\times 3+72\sqrt{3}+81\right)x}{3}-\frac{31}{3}y=6$$

Multiply 16 and 3 to get 48.

$$\frac{\left(48+72\sqrt{3}+81\right)x}{3}-\frac{31}{3}y=6$$

Add 48 and 81 to get 129.

$$\frac{\left(129+72\sqrt{3}\right)x}{3}-\frac{31}{3}y=6$$

Add \frac{31}{3}y to both sides.

$$\frac{\left(129+72\sqrt{3}\right)x}{3}=6+\frac{31}{3}y$$

Use the distributive property to multiply 129+72\sqrt{3} by x.

$$\frac{129x+72\sqrt{3}x}{3}=6+\frac{31}{3}y$$

Multiply both sides of the equation by 3.

$$129x+72\sqrt{3}x=18+31y$$

Combine all terms containing x.

$$\left(129+72\sqrt{3}\right)x=18+31y$$

The equation is in standard form.

$$\left(72\sqrt{3}+129\right)x=31y+18$$

Divide both sides by 129+72\sqrt{3}.

$$\frac{\left(72\sqrt{3}+129\right)x}{72\sqrt{3}+129}=\frac{31y+18}{72\sqrt{3}+129}$$

Dividing by 129+72\sqrt{3} undoes the multiplication by 129+72\sqrt{3}.

$$x=\frac{31y+18}{72\sqrt{3}+129}$$

Divide 18+31y by 129+72\sqrt{3}.

$$x=-\frac{\left(24\sqrt{3}-43\right)\left(31y+18\right)}{363}$$

Solve for y

$$y=\frac{72\sqrt{3}x+129x-18}{31}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3.

$$3\times \left(\frac{-9-4\sqrt{3}}{3}\right)^{2}x+\left(-9-\frac{4}{3}\right)y=6$$

To raise \frac{-9-4\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(-9-4\sqrt{3}\right)^{2}}{3^{2}}x+\left(-9-\frac{4}{3}\right)y=6$$

Express 3\times \frac{\left(-9-4\sqrt{3}\right)^{2}}{3^{2}} as a single fraction.

$$\frac{3\left(-9-4\sqrt{3}\right)^{2}}{3^{2}}x+\left(-9-\frac{4}{3}\right)y=6$$

Cancel out 3 in both numerator and denominator.

$$\frac{\left(-4\sqrt{3}-9\right)^{2}}{3}x+\left(-9-\frac{4}{3}\right)y=6$$

Express \frac{\left(-4\sqrt{3}-9\right)^{2}}{3}x as a single fraction.

$$\frac{\left(-4\sqrt{3}-9\right)^{2}x}{3}+\left(-9-\frac{4}{3}\right)y=6$$

Subtract \frac{4}{3} from -9 to get -\frac{31}{3}.

$$\frac{\left(-4\sqrt{3}-9\right)^{2}x}{3}-\frac{31}{3}y=6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4\sqrt{3}-9\right)^{2}.

$$\frac{\left(16\left(\sqrt{3}\right)^{2}+72\sqrt{3}+81\right)x}{3}-\frac{31}{3}y=6$$

The square of \sqrt{3} is 3.

$$\frac{\left(16\times 3+72\sqrt{3}+81\right)x}{3}-\frac{31}{3}y=6$$

Multiply 16 and 3 to get 48.

$$\frac{\left(48+72\sqrt{3}+81\right)x}{3}-\frac{31}{3}y=6$$

Add 48 and 81 to get 129.

$$\frac{\left(129+72\sqrt{3}\right)x}{3}-\frac{31}{3}y=6$$

Subtract \frac{\left(129+72\sqrt{3}\right)x}{3} from both sides.

$$-\frac{31}{3}y=6-\frac{\left(129+72\sqrt{3}\right)x}{3}$$

Use the distributive property to multiply 129+72\sqrt{3} by x.

$$-\frac{31}{3}y=6-\frac{129x+72\sqrt{3}x}{3}$$

Multiply both sides of the equation by 3.

$$-31y=18-\left(129x+72\sqrt{3}x\right)$$

To find the opposite of 129x+72\sqrt{3}x, find the opposite of each term.

$$-31y=18-129x-72\sqrt{3}x$$

The equation is in standard form.

$$-31y=-72\sqrt{3}x-129x+18$$

Divide both sides by -31.

$$\frac{-31y}{-31}=\frac{-72\sqrt{3}x-129x+18}{-31}$$

Dividing by -31 undoes the multiplication by -31.

$$y=\frac{-72\sqrt{3}x-129x+18}{-31}$$

Divide 18-129x-72\sqrt{3}x by -31.

$$y=\frac{72\sqrt{3}x+129x-18}{31}$$