Solve 4/y=4-y

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

Solve for y

$$y=2$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$4=y\times 4-yy$$

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

$$4=y\times 4-y^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$y\times 4-y^{2}=4$$

Subtract 4 from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-4±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$y=\frac{-4±\sqrt{16-16}}{2\left(-1\right)}$$

Add 16 to -16.

$$y=\frac{-4±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$y=-\frac{4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=-\frac{4}{-2}$$

Divide -4 by -2.

$$y=2$$

Steps for Completing the Square

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$4=y\times 4-yy$$

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

$$4=y\times 4-y^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$y\times 4-y^{2}=4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

$$\frac{-y^{2}+4y}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{4}{-1}y=\frac{4}{-1}$$

Divide 4 by -1.

$$y^{2}-4y=\frac{4}{-1}$$

Divide 4 by -1.

$$y^{2}-4y=-4$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add -4 to 4.

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

Factor y^{2}-4y+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$y-2=0 y-2=0$$

Add 2 to both sides of the equation.

$$y=2 y=2$$

The equation is now solved. Solutions are the same.

$$y=2$$