Solve 8/x-3-10/x=2/1

$$\frac{ 8 }{ x-3 } - \frac{ 10 }{ x } = \frac{ 2 }{ 1 }$$

Solve for x

$$x=-3$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x-3,x.

$$x\times 8-\left(x-3\right)\times 10=x\left(x-3\right)\times 2$$

Use the distributive property to multiply x-3 by 10.

$$x\times 8-\left(10x-30\right)=x\left(x-3\right)\times 2$$

To find the opposite of 10x-30, find the opposite of each term.

$$x\times 8-10x+30=x\left(x-3\right)\times 2$$

Combine x\times 8 and -10x to get -2x.

$$-2x+30=x\left(x-3\right)\times 2$$

Use the distributive property to multiply x by x-3.

$$-2x+30=\left(x^{2}-3x\right)\times 2$$

Use the distributive property to multiply x^{2}-3x by 2.

$$-2x+30=2x^{2}-6x$$

Subtract 2x^{2} from both sides.

$$-2x+30-2x^{2}=-6x$$

Add 6x to both sides.

$$-2x+30-2x^{2}+6x=0$$

Combine -2x and 6x to get 4x.

$$4x+30-2x^{2}=0$$

Divide both sides by 2.

$$2x+15-x^{2}=0$$

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

$$-x^{2}+2x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=2 ab=-15=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=5 b=-3$$

Rewrite -x^{2}+2x+15 as \left(-x^{2}+5x\right)+\left(-3x+15\right).

$$\left(-x^{2}+5x\right)+\left(-3x+15\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-3\right)$$

To find equation solutions, solve x-5=0 and -x-3=0.

$$x=5 x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x-3,x.

$$x\times 8-\left(x-3\right)\times 10=x\left(x-3\right)\times 2$$

Use the distributive property to multiply x-3 by 10.

$$x\times 8-\left(10x-30\right)=x\left(x-3\right)\times 2$$

To find the opposite of 10x-30, find the opposite of each term.

$$x\times 8-10x+30=x\left(x-3\right)\times 2$$

Combine x\times 8 and -10x to get -2x.

$$-2x+30=x\left(x-3\right)\times 2$$

Use the distributive property to multiply x by x-3.

$$-2x+30=\left(x^{2}-3x\right)\times 2$$

Use the distributive property to multiply x^{2}-3x by 2.

$$-2x+30=2x^{2}-6x$$

Subtract 2x^{2} from both sides.

$$-2x+30-2x^{2}=-6x$$

Add 6x to both sides.

$$-2x+30-2x^{2}+6x=0$$

Combine -2x and 6x to get 4x.

$$4x+30-2x^{2}=0$$

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.

$$-2x^{2}+4x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 4 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-2\right)\times 30}}{2\left(-2\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-2\right)\times 30}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-4±\sqrt{16+8\times 30}}{2\left(-2\right)}$$

Multiply 8 times 30.

$$x=\frac{-4±\sqrt{16+240}}{2\left(-2\right)}$$

Add 16 to 240.

$$x=\frac{-4±\sqrt{256}}{2\left(-2\right)}$$

Take the square root of 256.

$$x=\frac{-4±16}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-4±16}{-4}$$

Now solve the equation x=\frac{-4±16}{-4} when ± is plus. Add -4 to 16.

$$x=\frac{12}{-4}$$

Divide 12 by -4.

$$x=-3$$

Now solve the equation x=\frac{-4±16}{-4} when ± is minus. Subtract 16 from -4.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

The equation is now solved.

$$x=-3 x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x-3,x.

$$x\times 8-\left(x-3\right)\times 10=x\left(x-3\right)\times 2$$

Use the distributive property to multiply x-3 by 10.

$$x\times 8-\left(10x-30\right)=x\left(x-3\right)\times 2$$

To find the opposite of 10x-30, find the opposite of each term.

$$x\times 8-10x+30=x\left(x-3\right)\times 2$$

Combine x\times 8 and -10x to get -2x.

$$-2x+30=x\left(x-3\right)\times 2$$

Use the distributive property to multiply x by x-3.

$$-2x+30=\left(x^{2}-3x\right)\times 2$$

Use the distributive property to multiply x^{2}-3x by 2.

$$-2x+30=2x^{2}-6x$$

Subtract 2x^{2} from both sides.

$$-2x+30-2x^{2}=-6x$$

Add 6x to both sides.

$$-2x+30-2x^{2}+6x=0$$

Combine -2x and 6x to get 4x.

$$4x+30-2x^{2}=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$4x-2x^{2}=-30$$

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.

$$-2x^{2}+4x=-30$$

Divide both sides by -2.

$$\frac{-2x^{2}+4x}{-2}=-\frac{30}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{4}{-2}x=-\frac{30}{-2}$$

Divide 4 by -2.

$$x^{2}-2x=-\frac{30}{-2}$$

Divide -30 by -2.

$$x^{2}-2x=15$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=15+1$$

Add 15 to 1.

$$x^{2}-2x+1=16$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-1=4 x-1=-4$$

Add 1 to both sides of the equation.

$$x=5 x=-3$$