Solve {l}{(x-2)^{2}+(y-1)(y+1)=x^2+y^2+3}{(x-3y)(x+3y)-x^2+3y=4-9y^2-2x}

$$\left\{ \begin{array} { l } { ( x - 2 ) ^ { 2 } + ( y - 1 ) ( y + 1 ) = x ^ { 2 } + y ^ { 2 } + 3 } \\ { ( x - 3 y ) ( x + 3 y ) - x ^ { 2 } + 3 y = 4 - 9 y ^ { 2 } - 2 x } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Solution Steps

Consider the first equation. Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Consider \left(y-1\right)\left(y+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-4x+4+y^{2}-1=x^{2}+y^{2}+3$$

Subtract 1 from 4 to get 3.

$$x^{2}-4x+3+y^{2}=x^{2}+y^{2}+3$$

Subtract x^{2} from both sides.

$$x^{2}-4x+3+y^{2}-x^{2}=y^{2}+3$$

Combine x^{2} and -x^{2} to get 0.

$$-4x+3+y^{2}=y^{2}+3$$

Subtract y^{2} from both sides.

$$-4x+3+y^{2}-y^{2}=3$$

Combine y^{2} and -y^{2} to get 0.

$$-4x+3=3$$

Subtract 3 from both sides.

$$-4x=3-3$$

Subtract 3 from 3 to get 0.

$$-4x=0$$

Divide both sides by -4. Zero divided by any non-zero number gives zero.

$$x=0$$

Consider the second equation. Insert the known values of variables into the equation.

$$\left(0-3y\right)\left(0+3y\right)-0^{2}+3y=4-9y^{2}-2\times 0$$

Anything plus zero gives itself.

$$-3y\times 3y-0^{2}+3y=4-9y^{2}-2\times 0$$

Multiply -3 and 3 to get -9.

$$-9yy-0^{2}+3y=4-9y^{2}-2\times 0$$

Multiply y and y to get y^{2}.

$$-9y^{2}-0^{2}+3y=4-9y^{2}-2\times 0$$

Calculate 0 to the power of 2 and get 0.

$$-9y^{2}-0+3y=4-9y^{2}-2\times 0$$

Multiply -1 and 0 to get 0.

$$-9y^{2}+0+3y=4-9y^{2}-2\times 0$$

Anything plus zero gives itself.

$$-9y^{2}+3y=4-9y^{2}-2\times 0$$

Multiply -2 and 0 to get 0.

$$-9y^{2}+3y=4-9y^{2}+0$$

Add 4 and 0 to get 4.

$$-9y^{2}+3y=4-9y^{2}$$

Add 9y^{2} to both sides.

$$-9y^{2}+3y+9y^{2}=4$$

Combine -9y^{2} and 9y^{2} to get 0.

$$3y=4$$

Divide both sides by 3.

$$y=\frac{4}{3}$$

The system is now solved.

$$x=0 y=\frac{4}{3}$$