Solve {l}{sqrt{2}x+(sqrt{3}+2)y=2(y+2)+1}{(sqrt{8}+2)x-sqrt{12}y=2(2+x)-6}

$$\left\{ \begin{array} { l } { \sqrt { 2 } x + ( \sqrt { 3 } + 2 ) y = 2 ( y + 2 ) + 1 } \\ { ( \sqrt { 8 } + 2 ) x - \sqrt { 12 } y = 2 ( 2 + x ) - 6 } \end{array} \right.$$

Solve for x, y

$$x=\sqrt{2}\approx 1.414213562$$

$$y=\sqrt{3}\approx 1.732050808$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply \sqrt{3}+2 by y.

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

Use the distributive property to multiply 2 by y+2.

$$\sqrt{2}x+\sqrt{3}y+2y=2y+4+1$$

Add 4 and 1 to get 5.

$$\sqrt{2}x+\sqrt{3}y+2y=2y+5$$

Subtract 2y from both sides.

$$\sqrt{2}x+\sqrt{3}y+2y-2y=5$$

Combine 2y and -2y to get 0.

$$\sqrt{2}x+\sqrt{3}y=5$$

Consider the second equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}+2\right)x-\sqrt{12}y=2\left(2+x\right)-6$$

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

$$2\sqrt{2}x+2x-\sqrt{12}y=2\left(2+x\right)-6$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Use the distributive property to multiply 2 by 2+x.

$$2\sqrt{2}x+2x-2\sqrt{3}y=4+2x-6$$

Subtract 6 from 4 to get -2.

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

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

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

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.

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

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

$$\sqrt{2}x+\sqrt{3}y=5$$

Subtract \sqrt{3}y from both sides of the equation.

$$\sqrt{2}x=\left(-\sqrt{3}\right)y+5$$

Divide both sides by \sqrt{2}.

$$x=\frac{\sqrt{2}}{2}\left(\left(-\sqrt{3}\right)y+5\right)$$

Multiply \frac{\sqrt{2}}{2} times -\sqrt{3}y+5.

$$x=\left(-\frac{\sqrt{6}}{2}\right)y+\frac{5\sqrt{2}}{2}$$

Substitute \frac{-\sqrt{6}y+5\sqrt{2}}{2} for x in the other equation, 2\sqrt{2}x+\left(-2\sqrt{3}\right)y=-2.

$$2\sqrt{2}\left(\left(-\frac{\sqrt{6}}{2}\right)y+\frac{5\sqrt{2}}{2}\right)+\left(-2\sqrt{3}\right)y=-2$$

Multiply 2\sqrt{2} times \frac{-\sqrt{6}y+5\sqrt{2}}{2}.

$$\left(-2\sqrt{3}\right)y+10+\left(-2\sqrt{3}\right)y=-2$$

Add -2\sqrt{3}y to -2\sqrt{3}y.

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

Subtract 10 from both sides of the equation.

$$\left(-4\sqrt{3}\right)y=-12$$

Divide both sides by -4\sqrt{3}.

$$y=\sqrt{3}$$

Substitute \sqrt{3} for y in x=\left(-\frac{\sqrt{6}}{2}\right)y+\frac{5\sqrt{2}}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\left(-\frac{\sqrt{6}}{2}\right)\sqrt{3}+\frac{5\sqrt{2}}{2}$$

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

$$x=\frac{-3\sqrt{2}+5\sqrt{2}}{2}$$

Add \frac{5\sqrt{2}}{2} to -\frac{3\sqrt{2}}{2}.

$$x=\sqrt{2}$$

The system is now solved.

$$x=\sqrt{2},y=\sqrt{3}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply \sqrt{3}+2 by y.

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

Use the distributive property to multiply 2 by y+2.

$$\sqrt{2}x+\sqrt{3}y+2y=2y+4+1$$

Add 4 and 1 to get 5.

$$\sqrt{2}x+\sqrt{3}y+2y=2y+5$$

Subtract 2y from both sides.

$$\sqrt{2}x+\sqrt{3}y+2y-2y=5$$

Combine 2y and -2y to get 0.

$$\sqrt{2}x+\sqrt{3}y=5$$

Consider the second equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}+2\right)x-\sqrt{12}y=2\left(2+x\right)-6$$

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

$$2\sqrt{2}x+2x-\sqrt{12}y=2\left(2+x\right)-6$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Use the distributive property to multiply 2 by 2+x.

$$2\sqrt{2}x+2x-2\sqrt{3}y=4+2x-6$$

Subtract 6 from 4 to get -2.

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

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

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

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.

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

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

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

Simplify.

$$4x+2\sqrt{6}y=10\sqrt{2},4x+\left(-2\sqrt{6}\right)y=-2\sqrt{2}$$

Subtract 4x+\left(-2\sqrt{6}\right)y=-2\sqrt{2} from 4x+2\sqrt{6}y=10\sqrt{2} by subtracting like terms on each side of the equal sign.

$$4x-4x+2\sqrt{6}y+2\sqrt{6}y=10\sqrt{2}+2\sqrt{2}$$

Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.

$$2\sqrt{6}y+2\sqrt{6}y=10\sqrt{2}+2\sqrt{2}$$

Add 2\sqrt{6}y to 2\sqrt{6}y.

$$4\sqrt{6}y=10\sqrt{2}+2\sqrt{2}$$

Add 10\sqrt{2} to 2\sqrt{2}.

$$4\sqrt{6}y=12\sqrt{2}$$

Divide both sides by 4\sqrt{6}.

$$y=\sqrt{3}$$

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

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

Multiply -2\sqrt{3} times \sqrt{3}.

$$2\sqrt{2}x-6=-2$$

Add 6 to both sides of the equation.

$$2\sqrt{2}x=4$$

Divide both sides by 2\sqrt{2}.

$$x=\sqrt{2}$$

The system is now solved.

$$x=\sqrt{2},y=\sqrt{3}$$