Solve {l}{3x-2y=4+3sqrt{3}}{7x-5y=-1,7sqrt{3}}

$$\left\{ \begin{array} { l } { 3 x - 2 y = 4 + 3 \sqrt { 3 } } \\ { 7 x - 5 y = - 1,7 \sqrt { 3 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{92\sqrt{3}}{5}+20\approx 51.869734859$$

$$y=\frac{261\sqrt{3}}{10}+28\approx 73.206526078$$

Steps Using Substitution

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.

$$3x-2y=3\sqrt{3}+4;7x-5y=-\frac{17\sqrt{3}}{10}$$

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

$$3x-2y=3\sqrt{3}+4$$

Add 2y to both sides of the equation.

$$3x=2y+3\sqrt{3}+4$$

Divide both sides by 3.

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

Multiply \frac{1}{3} times 2y+4+3\sqrt{3}.

$$x=\frac{2}{3}y+\sqrt{3}+\frac{4}{3}$$

Substitute \frac{2y}{3}+\frac{4}{3}+\sqrt{3} for x in the other equation, 7x-5y=-\frac{17\sqrt{3}}{10}.

$$7\left(\frac{2}{3}y+\sqrt{3}+\frac{4}{3}\right)-5y=-\frac{17\sqrt{3}}{10}$$

Multiply 7 times \frac{2y}{3}+\frac{4}{3}+\sqrt{3}.

$$\frac{14}{3}y+7\sqrt{3}+\frac{28}{3}-5y=-\frac{17\sqrt{3}}{10}$$

Add \frac{14y}{3} to -5y.

$$-\frac{1}{3}y+7\sqrt{3}+\frac{28}{3}=-\frac{17\sqrt{3}}{10}$$

Subtract \frac{28}{3}+7\sqrt{3} from both sides of the equation.

$$-\frac{1}{3}y=-\frac{87\sqrt{3}}{10}-\frac{28}{3}$$

Multiply both sides by -3.

$$y=\frac{261\sqrt{3}}{10}+28$$

Substitute \frac{261\sqrt{3}}{10}+28 for y in x=\frac{2}{3}y+\sqrt{3}+\frac{4}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2}{3}\left(\frac{261\sqrt{3}}{10}+28\right)+\sqrt{3}+\frac{4}{3}$$

Multiply \frac{2}{3} times \frac{261\sqrt{3}}{10}+28.

$$x=\frac{87\sqrt{3}}{5}+\frac{56}{3}+\sqrt{3}+\frac{4}{3}$$

Add \frac{4}{3}+\sqrt{3} to \frac{87\sqrt{3}}{5}+\frac{56}{3}.

$$x=\frac{92\sqrt{3}}{5}+20$$

The system is now solved.

$$x=\frac{92\sqrt{3}}{5}+20;y=\frac{261\sqrt{3}}{10}+28$$

Steps Using Elimination

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.

$$3x-2y=3\sqrt{3}+4;7x-5y=-\frac{17\sqrt{3}}{10}$$

To make 3x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 3.

$$7\times 3x+7\left(-2\right)y=7\left(3\sqrt{3}+4\right);3\times 7x+3\left(-5\right)y=3\left(-\frac{17\sqrt{3}}{10}\right)$$

Simplify.

$$21x-14y=21\sqrt{3}+28;21x-15y=-\frac{51\sqrt{3}}{10}$$

Subtract 21x-15y=-\frac{51\sqrt{3}}{10} from 21x-14y=21\sqrt{3}+28 by subtracting like terms on each side of the equal sign.

$$21x-21x-14y+15y=21\sqrt{3}+28+\frac{51\sqrt{3}}{10}$$

Add 21x to -21x. Terms 21x and -21x cancel out, leaving an equation with only one variable that can be solved.

$$-14y+15y=21\sqrt{3}+28+\frac{51\sqrt{3}}{10}$$

Add -14y to 15y.

$$y=21\sqrt{3}+28+\frac{51\sqrt{3}}{10}$$

Add 28+21\sqrt{3} to \frac{51\sqrt{3}}{10}.

$$y=\frac{261\sqrt{3}}{10}+28$$

Substitute 28+\frac{261\sqrt{3}}{10} for y in 7x-5y=-\frac{17\sqrt{3}}{10}. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x-5\left(\frac{261\sqrt{3}}{10}+28\right)=-\frac{17\sqrt{3}}{10}$$

Multiply -5 times 28+\frac{261\sqrt{3}}{10}.

$$7x-\frac{261\sqrt{3}}{2}-140=-\frac{17\sqrt{3}}{10}$$

Subtract -140-\frac{261\sqrt{3}}{2} from both sides of the equation.

$$7x=\frac{644\sqrt{3}}{5}+140$$

Divide both sides by 7.

$$x=\frac{92\sqrt{3}}{5}+20$$

The system is now solved.

$$x=\frac{92\sqrt{3}}{5}+20;y=\frac{261\sqrt{3}}{10}+28$$