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

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

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{m^{3}-3\sqrt{3}}{3\left(m^{2}-3\right)}\text{, }y=\frac{3\left(-m+\sqrt{3}\right)}{m^{2}-3}\text{, }&m\neq -\sqrt{3}\text{ and }m\neq \sqrt{3}\\x=\frac{y-\sqrt{3}}{3}\text{, }y\in \mathrm{C}\text{, }&m=\sqrt{3}\end{matrix}\right.$$

Steps Using Substitution

Consider the second equation. Reorder the terms.

$$9x-ym^{2}=-3\sqrt{3}$$

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.

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

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

$$3x-y=-m$$

Add y to both sides of the equation.

$$3x=y-m$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(y-m\right)$$

Multiply \frac{1}{3} times y-m.

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

Substitute \frac{y-m}{3} for x in the other equation, 9x+\left(-m^{2}\right)y=-3\sqrt{3}.

$$9\left(\frac{1}{3}y-\frac{m}{3}\right)+\left(-m^{2}\right)y=-3\sqrt{3}$$

Multiply 9 times \frac{y-m}{3}.

$$3y-3m+\left(-m^{2}\right)y=-3\sqrt{3}$$

Add 3y to -m^{2}y.

$$\left(3-m^{2}\right)y-3m=-3\sqrt{3}$$

Add 3m to both sides of the equation.

$$\left(3-m^{2}\right)y=3m-3\sqrt{3}$$

Divide both sides by 3-m^{2}.

$$y=-\frac{3}{m+\sqrt{3}}$$

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

$$x=\frac{1}{3}\left(-\frac{3}{m+\sqrt{3}}\right)-\frac{m}{3}$$

Multiply \frac{1}{3} times -\frac{3}{\sqrt{3}+m}.

$$x=-\frac{1}{m+\sqrt{3}}-\frac{m}{3}$$

Add -\frac{m}{3} to -\frac{1}{\sqrt{3}+m}.

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

The system is now solved.

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

Steps Using Elimination

Consider the second equation. Reorder the terms.

$$9x-ym^{2}=-3\sqrt{3}$$

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.

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

To make 3x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by 3.

$$9\times 3x+9\left(-1\right)y=9\left(-m\right),3\times 9x+3\left(-m^{2}\right)y=3\left(-3\sqrt{3}\right)$$

Simplify.

$$27x-9y=-9m,27x+\left(-3m^{2}\right)y=-9\sqrt{3}$$

Subtract 27x+\left(-3m^{2}\right)y=-9\sqrt{3} from 27x-9y=-9m by subtracting like terms on each side of the equal sign.

$$27x-27x-9y+3m^{2}y=-9m+3^{\frac{5}{2}}$$

Add 27x to -27x. Terms 27x and -27x cancel out, leaving an equation with only one variable that can be solved.

$$-9y+3m^{2}y=-9m+3^{\frac{5}{2}}$$

Add -9y to 3m^{2}y.

$$\left(3m^{2}-9\right)y=-9m+3^{\frac{5}{2}}$$

Add -9m to 3^{\frac{5}{2}}.

$$\left(3m^{2}-9\right)y=3^{\frac{5}{2}}-9m$$

Divide both sides by -9+3m^{2}.

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

Substitute \frac{3\left(-m+\sqrt{3}\right)}{-3+m^{2}} for y in 9x+\left(-m^{2}\right)y=-3\sqrt{3}. Because the resulting equation contains only one variable, you can solve for x directly.

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

Multiply -m^{2} times \frac{3\left(-m+\sqrt{3}\right)}{-3+m^{2}}.

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

Add \frac{3m^{2}\left(-m+\sqrt{3}\right)}{-3+m^{2}} to both sides of the equation.

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

Divide both sides by 9.

$$x=\frac{-\frac{m^{3}}{3}+\sqrt{3}}{m^{2}-3}$$

The system is now solved.

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

Solve for x, y

$$\left\{\begin{matrix}x=-\frac{m^{3}-3\sqrt{3}}{3\left(m^{2}-3\right)}\text{, }y=\frac{3\left(-m+\sqrt{3}\right)}{m^{2}-3}\text{, }&|m|\neq \sqrt{3}\\x=\frac{y-\sqrt{3}}{3}\text{, }y\in \mathrm{R}\text{, }&m=\sqrt{3}\end{matrix}\right.$$

Steps Using Substitution

Consider the second equation. Reorder the terms.

$$9x-ym^{2}=-3\sqrt{3}$$

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.

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

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

$$3x-y=-m$$

Add y to both sides of the equation.

$$3x=y-m$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(y-m\right)$$

Multiply \frac{1}{3} times y-m.

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

Substitute \frac{y-m}{3} for x in the other equation, 9x+\left(-m^{2}\right)y=-3\sqrt{3}.

$$9\left(\frac{1}{3}y-\frac{m}{3}\right)+\left(-m^{2}\right)y=-3\sqrt{3}$$

Multiply 9 times \frac{y-m}{3}.

$$3y-3m+\left(-m^{2}\right)y=-3\sqrt{3}$$

Add 3y to -m^{2}y.

$$\left(3-m^{2}\right)y-3m=-3\sqrt{3}$$

Add 3m to both sides of the equation.

$$\left(3-m^{2}\right)y=3m-3\sqrt{3}$$

Divide both sides by 3-m^{2}.

$$y=-\frac{3}{m+\sqrt{3}}$$

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

$$x=\frac{1}{3}\left(-\frac{3}{m+\sqrt{3}}\right)-\frac{m}{3}$$

Multiply \frac{1}{3} times -\frac{3}{\sqrt{3}+m}.

$$x=-\frac{1}{m+\sqrt{3}}-\frac{m}{3}$$

Add -\frac{m}{3} to -\frac{1}{\sqrt{3}+m}.

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

The system is now solved.

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

Steps Using Elimination

Consider the second equation. Reorder the terms.

$$9x-ym^{2}=-3\sqrt{3}$$

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.

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

To make 3x and 9x equal, multiply all terms on each side of the first equation by 9 and all terms on each side of the second by 3.

$$9\times 3x+9\left(-1\right)y=9\left(-m\right),3\times 9x+3\left(-m^{2}\right)y=3\left(-3\sqrt{3}\right)$$

Simplify.

$$27x-9y=-9m,27x+\left(-3m^{2}\right)y=-9\sqrt{3}$$

Subtract 27x+\left(-3m^{2}\right)y=-9\sqrt{3} from 27x-9y=-9m by subtracting like terms on each side of the equal sign.

$$27x-27x-9y+3m^{2}y=-9m+9\sqrt{3}$$

Add 27x to -27x. Terms 27x and -27x cancel out, leaving an equation with only one variable that can be solved.

$$-9y+3m^{2}y=-9m+9\sqrt{3}$$

Add -9y to 3m^{2}y.

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

Divide both sides by -9+3m^{2}.

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

Substitute \frac{3\left(-m+\sqrt{3}\right)}{-3+m^{2}} for y in 9x+\left(-m^{2}\right)y=-3\sqrt{3}. Because the resulting equation contains only one variable, you can solve for x directly.

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

Multiply -m^{2} times \frac{3\left(-m+\sqrt{3}\right)}{-3+m^{2}}.

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

Add \frac{3m^{2}\left(-m+\sqrt{3}\right)}{-3+m^{2}} to both sides of the equation.

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

Divide both sides by 9.

$$x=\frac{-\frac{m^{3}}{3}+\sqrt{3}}{m^{2}-3}$$

The system is now solved.

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