Solve 1/x-3+4/x=1

$$\frac{ 1 }{ x-3 } + \frac{ 4 }{ x } =1$$

Solve for x

$$x=2$$

$$x=6$$

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.

$$x+\left(x-3\right)\times 4=x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 4.

$$x+4x-12=x\left(x-3\right)$$

Combine x and 4x to get 5x.

$$5x-12=x\left(x-3\right)$$

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

$$5x-12=x^{2}-3x$$

Subtract x^{2} from both sides.

$$5x-12-x^{2}=-3x$$

Add 3x to both sides.

$$5x-12-x^{2}+3x=0$$

Combine 5x and 3x to get 8x.

$$8x-12-x^{2}=0$$

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

$$-x^{2}+8x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-12\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=6 b=2$$

Rewrite -x^{2}+8x-12 as \left(-x^{2}+6x\right)+\left(2x-12\right).

$$\left(-x^{2}+6x\right)+\left(2x-12\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-6\right)+2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+2\right)$$

To find equation solutions, solve x-6=0 and -x+2=0.

$$x=6 x=2$$

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.

$$x+\left(x-3\right)\times 4=x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 4.

$$x+4x-12=x\left(x-3\right)$$

Combine x and 4x to get 5x.

$$5x-12=x\left(x-3\right)$$

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

$$5x-12=x^{2}-3x$$

Subtract x^{2} from both sides.

$$5x-12-x^{2}=-3x$$

Add 3x to both sides.

$$5x-12-x^{2}+3x=0$$

Combine 5x and 3x to get 8x.

$$8x-12-x^{2}=0$$

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.

$$-x^{2}+8x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-8±\sqrt{64-48}}{2\left(-1\right)}$$

Add 64 to -48.

$$x=\frac{-8±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-8±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±4}{-2}$$

Now solve the equation x=\frac{-8±4}{-2} when ± is plus. Add -8 to 4.

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

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-8±4}{-2} when ± is minus. Subtract 4 from -8.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=2 x=6$$

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.

$$x+\left(x-3\right)\times 4=x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 4.

$$x+4x-12=x\left(x-3\right)$$

Combine x and 4x to get 5x.

$$5x-12=x\left(x-3\right)$$

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

$$5x-12=x^{2}-3x$$

Subtract x^{2} from both sides.

$$5x-12-x^{2}=-3x$$

Add 3x to both sides.

$$5x-12-x^{2}+3x=0$$

Combine 5x and 3x to get 8x.

$$8x-12-x^{2}=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$8x-x^{2}=12$$

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.

$$-x^{2}+8x=12$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{12}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}-8x=-12$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-12+16$$

Add -12 to 16.

$$x^{2}-8x+16=4$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-4=2 x-4=-2$$

Add 4 to both sides of the equation.

$$x=6 x=2$$