Solve (x+2)/9+(y-3)^2/4=1

$$\frac { ( x + 2 ) } { 9 } + \frac { ( y - 3 ) ^ { 2 } } { 4 } = 1$$

Solve for x

$$x=-\frac{9y^{2}}{4}+\frac{27y}{2}-\frac{53}{4}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 36, the least common multiple of 9,4.

$$4\left(x+2\right)+9\left(y-3\right)^{2}=36$$

Use the distributive property to multiply 4 by x+2.

$$4x+8+9\left(y-3\right)^{2}=36$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-3\right)^{2}.

$$4x+8+9\left(y^{2}-6y+9\right)=36$$

Use the distributive property to multiply 9 by y^{2}-6y+9.

$$4x+8+9y^{2}-54y+81=36$$

Add 8 and 81 to get 89.

$$4x+89+9y^{2}-54y=36$$

Subtract 89 from both sides.

$$4x+9y^{2}-54y=36-89$$

Subtract 89 from 36 to get -53.

$$4x+9y^{2}-54y=-53$$

Subtract 9y^{2} from both sides.

$$4x-54y=-53-9y^{2}$$

Add 54y to both sides.

$$4x=-53-9y^{2}+54y$$

The equation is in standard form.

$$4x=-9y^{2}+54y-53$$

Divide both sides by 4.

$$\frac{4x}{4}=\frac{-9y^{2}+54y-53}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x=\frac{-9y^{2}+54y-53}{4}$$

Divide -53-9y^{2}+54y by 4.

$$x=-\frac{9y^{2}}{4}+\frac{27y}{2}-\frac{53}{4}$$

Solve for y (complex solution)

$$y=-\frac{2\sqrt{7-x}}{3}+3$$

$$y=\frac{2\sqrt{7-x}}{3}+3$$

Solve for y

$$y=-\frac{2\sqrt{7-x}}{3}+3$$

$$y=\frac{2\sqrt{7-x}}{3}+3\text{, }x\leq 7$$