Solve 2/d+1/d-2=1

$$\frac{ 2 }{ d } + \frac{ 1 }{ d-2 } =1$$

Solve for d

$$d=1$$

$$d=4$$

Steps Using Factoring By Grouping

Variable d cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by d\left(d-2\right), the least common multiple of d,d-2.

$$\left(d-2\right)\times 2+d=d\left(d-2\right)$$

Use the distributive property to multiply d-2 by 2.

$$2d-4+d=d\left(d-2\right)$$

Combine 2d and d to get 3d.

$$3d-4=d\left(d-2\right)$$

Use the distributive property to multiply d by d-2.

$$3d-4=d^{2}-2d$$

Subtract d^{2} from both sides.

$$3d-4-d^{2}=-2d$$

Add 2d to both sides.

$$3d-4-d^{2}+2d=0$$

Combine 3d and 2d to get 5d.

$$5d-4-d^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-d^{2}+5d-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -d^{2}+ad+bd-4. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=4 b=1$$

Rewrite -d^{2}+5d-4 as \left(-d^{2}+4d\right)+\left(d-4\right).

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

Factor out -d in -d^{2}+4d.

$$-d\left(d-4\right)+d-4$$

Factor out common term d-4 by using distributive property.

$$\left(d-4\right)\left(-d+1\right)$$

To find equation solutions, solve d-4=0 and -d+1=0.

$$d=4 d=1$$

Steps Using the Quadratic Formula

Variable d cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by d\left(d-2\right), the least common multiple of d,d-2.

$$\left(d-2\right)\times 2+d=d\left(d-2\right)$$

Use the distributive property to multiply d-2 by 2.

$$2d-4+d=d\left(d-2\right)$$

Combine 2d and d to get 3d.

$$3d-4=d\left(d-2\right)$$

Use the distributive property to multiply d by d-2.

$$3d-4=d^{2}-2d$$

Subtract d^{2} from both sides.

$$3d-4-d^{2}=-2d$$

Add 2d to both sides.

$$3d-4-d^{2}+2d=0$$

Combine 3d and 2d to get 5d.

$$5d-4-d^{2}=0$$

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.

$$-d^{2}+5d-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 5.

$$d=\frac{-5±\sqrt{25-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$d=\frac{-5±\sqrt{25+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$d=\frac{-5±\sqrt{25-16}}{2\left(-1\right)}$$

Add 25 to -16.

$$d=\frac{-5±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$d=\frac{-5±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$d=\frac{-5±3}{-2}$$

Now solve the equation d=\frac{-5±3}{-2} when ± is plus. Add -5 to 3.

$$d=-\frac{2}{-2}$$

Divide -2 by -2.

$$d=1$$

Now solve the equation d=\frac{-5±3}{-2} when ± is minus. Subtract 3 from -5.

$$d=-\frac{8}{-2}$$

Divide -8 by -2.

$$d=4$$

The equation is now solved.

$$d=1 d=4$$

Steps for Completing the Square

Variable d cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by d\left(d-2\right), the least common multiple of d,d-2.

$$\left(d-2\right)\times 2+d=d\left(d-2\right)$$

Use the distributive property to multiply d-2 by 2.

$$2d-4+d=d\left(d-2\right)$$

Combine 2d and d to get 3d.

$$3d-4=d\left(d-2\right)$$

Use the distributive property to multiply d by d-2.

$$3d-4=d^{2}-2d$$

Subtract d^{2} from both sides.

$$3d-4-d^{2}=-2d$$

Add 2d to both sides.

$$3d-4-d^{2}+2d=0$$

Combine 3d and 2d to get 5d.

$$5d-4-d^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$5d-d^{2}=4$$

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.

$$-d^{2}+5d=4$$

Divide both sides by -1.

$$\frac{-d^{2}+5d}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$d^{2}+\frac{5}{-1}d=\frac{4}{-1}$$

Divide 5 by -1.

$$d^{2}-5d=\frac{4}{-1}$$

Divide 4 by -1.

$$d^{2}-5d=-4$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}-5d+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$d^{2}-5d+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$d^{2}-5d+\frac{25}{4}=\frac{9}{4}$$

Factor d^{2}-5d+\frac{25}{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}.

$$\left(d-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$d-\frac{5}{2}=\frac{3}{2} d-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$d=4 d=1$$