Solve (y+3)^2/16+x/1=1

$$\frac { ( y + 3 ) ^ { 2 } } { 16 } + \frac { x } { 1 } = 1$$

Solve for x

$$x=-\frac{\left(y-1\right)\left(y+7\right)}{16}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 16.

$$\left(y+3\right)^{2}+16x=16$$

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

$$y^{2}+6y+9+16x=16$$

Subtract y^{2} from both sides.

$$6y+9+16x=16-y^{2}$$

Subtract 6y from both sides.

$$9+16x=16-y^{2}-6y$$

Subtract 9 from both sides.

$$16x=16-y^{2}-6y-9$$

Subtract 9 from 16 to get 7.

$$16x=7-y^{2}-6y$$

The equation is in standard form.

$$16x=7-6y-y^{2}$$

Divide both sides by 16.

$$\frac{16x}{16}=\frac{\left(1-y\right)\left(y+7\right)}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x=\frac{\left(1-y\right)\left(y+7\right)}{16}$$

Solve for y (complex solution)

$$y=-4\sqrt{1-x}-3$$

$$y=4\sqrt{1-x}-3$$

Solve for y

$$y=-4\sqrt{1-x}-3$$

$$y=4\sqrt{1-x}-3\text{, }x\leq 1$$