Solve {l}{9x=3y}{2x-9y=sqrt{2000}}

$$\left\{ \begin{array} { l } { 9 x = 3 y } \\ { 2 x - 9 y = \sqrt { 2000 } } \end{array} \right.$$

Solve for x, y

$$x = -\frac{4 \sqrt{5}}{5} \approx -1.788854382$$

$$y = -\frac{12 \sqrt{5}}{5} \approx -5.366563146$$

Steps Using Substitution

Consider the first equation. Subtract 3y from both sides.

$$9x-3y=0$$

Consider the second equation. Factor 2000=20^{2}\times 5. Rewrite the square root of the product \sqrt{20^{2}\times 5} as the product of square roots \sqrt{20^{2}}\sqrt{5}. Take the square root of 20^{2}.

$$2x-9y=20\sqrt{5}$$

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.

$$9x-3y=0,2x-9y=20\sqrt{5}$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$9x-3y=0$$

Add 3y to both sides of the equation.

$$9x=3y$$

Divide both sides by 9.

$$x=\frac{1}{9}\times 3y$$

Multiply \frac{1}{9} times 3y.

$$x=\frac{1}{3}y$$

Substitute \frac{y}{3} for x in the other equation, 2x-9y=20\sqrt{5}.

$$2\times \frac{1}{3}y-9y=20\sqrt{5}$$

Multiply 2 times \frac{y}{3}.

$$\frac{2}{3}y-9y=20\sqrt{5}$$

Add \frac{2y}{3} to -9y.

$$-\frac{25}{3}y=20\sqrt{5}$$

Divide both sides of the equation by -\frac{25}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{12\sqrt{5}}{5}$$

Substitute -\frac{12\sqrt{5}}{5} for y in x=\frac{1}{3}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{3}\left(-\frac{12\sqrt{5}}{5}\right)$$

Multiply \frac{1}{3} times -\frac{12\sqrt{5}}{5}.

$$x=-\frac{4\sqrt{5}}{5}$$

The system is now solved.

$$x=-\frac{4\sqrt{5}}{5},y=-\frac{12\sqrt{5}}{5}$$

Steps Using Elimination

Consider the first equation. Subtract 3y from both sides.

$$9x-3y=0$$

Consider the second equation. Factor 2000=20^{2}\times 5. Rewrite the square root of the product \sqrt{20^{2}\times 5} as the product of square roots \sqrt{20^{2}}\sqrt{5}. Take the square root of 20^{2}.

$$2x-9y=20\sqrt{5}$$

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.

$$9x-3y=0,2x-9y=20\sqrt{5}$$

To make 9x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 9.

$$2\times 9x+2\left(-3\right)y=0,9\times 2x+9\left(-9\right)y=9\times 20\sqrt{5}$$

Simplify.

$$18x-6y=0,18x-81y=180\sqrt{5}$$

Subtract 18x-81y=180\sqrt{5} from 18x-6y=0 by subtracting like terms on each side of the equal sign.

$$18x-18x-6y+81y=-180\sqrt{5}$$

Add 18x to -18x. Terms 18x and -18x cancel out, leaving an equation with only one variable that can be solved.

$$-6y+81y=-180\sqrt{5}$$

Add -6y to 81y.

$$75y=-180\sqrt{5}$$

Divide both sides by 75.

$$y=-\frac{12\sqrt{5}}{5}$$

Substitute -\frac{12\sqrt{5}}{5} for y in 2x-9y=20\sqrt{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$2x-9\left(-\frac{12\sqrt{5}}{5}\right)=20\sqrt{5}$$

Multiply -9 times -\frac{12\sqrt{5}}{5}.

$$2x+\frac{108\sqrt{5}}{5}=20\sqrt{5}$$

Subtract \frac{108\sqrt{5}}{5} from both sides of the equation.

$$2x=-\frac{8\sqrt{5}}{5}$$

Divide both sides by 2.

$$x=-\frac{4\sqrt{5}}{5}$$

The system is now solved.

$$x=-\frac{4\sqrt{5}}{5},y=-\frac{12\sqrt{5}}{5}$$