Solve 3/x-4=1+5/x+4

$$\frac { 3 } { x - 4 } = 1 + \frac { 5 } { x + 4 }$$

Solve for x

$$x=-8$$

$$x=6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x-4,x+4.

$$\left(x+4\right)\times 3=\left(x-4\right)\left(x+4\right)+\left(x-4\right)\times 5$$

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

$$3x+12=\left(x-4\right)\left(x+4\right)+\left(x-4\right)\times 5$$

Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$3x+12=x^{2}-16+\left(x-4\right)\times 5$$

Use the distributive property to multiply x-4 by 5.

$$3x+12=x^{2}-16+5x-20$$

Subtract 20 from -16 to get -36.

$$3x+12=x^{2}-36+5x$$

Subtract x^{2} from both sides.

$$3x+12-x^{2}=-36+5x$$

Subtract -36 from both sides.

$$3x+12-x^{2}-\left(-36\right)=5x$$

The opposite of -36 is 36.

$$3x+12-x^{2}+36=5x$$

Subtract 5x from both sides.

$$3x+12-x^{2}+36-5x=0$$

Add 12 and 36 to get 48.

$$3x+48-x^{2}-5x=0$$

Combine 3x and -5x to get -2x.

$$-2x+48-x^{2}=0$$

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.

$$-x^{2}-2x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 48}}{2\left(-1\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 48}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 48}}{2\left(-1\right)}$$

Multiply 4 times 48.

$$x=\frac{-\left(-2\right)±\sqrt{4+192}}{2\left(-1\right)}$$

Add 4 to 192.

$$x=\frac{-\left(-2\right)±\sqrt{196}}{2\left(-1\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-2\right)±14}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±14}{-2}$$

Now solve the equation x=\frac{2±14}{-2} when ± is plus. Add 2 to 14.

$$x=\frac{16}{-2}$$

Divide 16 by -2.

$$x=-8$$

Now solve the equation x=\frac{2±14}{-2} when ± is minus. Subtract 14 from 2.

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

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=-8 x=6$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+4\right), the least common multiple of x-4,x+4.

$$\left(x+4\right)\times 3=\left(x-4\right)\left(x+4\right)+\left(x-4\right)\times 5$$

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

$$3x+12=\left(x-4\right)\left(x+4\right)+\left(x-4\right)\times 5$$

Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$3x+12=x^{2}-16+\left(x-4\right)\times 5$$

Use the distributive property to multiply x-4 by 5.

$$3x+12=x^{2}-16+5x-20$$

Subtract 20 from -16 to get -36.

$$3x+12=x^{2}-36+5x$$

Subtract x^{2} from both sides.

$$3x+12-x^{2}=-36+5x$$

Subtract 5x from both sides.

$$3x+12-x^{2}-5x=-36$$

Combine 3x and -5x to get -2x.

$$-2x+12-x^{2}=-36$$

Subtract 12 from both sides.

$$-2x-x^{2}=-36-12$$

Subtract 12 from -36 to get -48.

$$-2x-x^{2}=-48$$

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.

$$-x^{2}-2x=-48$$

Divide both sides by -1.

$$\frac{-x^{2}-2x}{-1}=-\frac{48}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{48}{-1}$$

Divide -2 by -1.

$$x^{2}+2x=-\frac{48}{-1}$$

Divide -48 by -1.

$$x^{2}+2x=48$$

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.

$$x^{2}+2x+1^{2}=48+1^{2}$$

Square 1.

$$x^{2}+2x+1=48+1$$

Add 48 to 1.

$$x^{2}+2x+1=49$$

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}.

$$\left(x+1\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+1=7 x+1=-7$$

Subtract 1 from both sides of the equation.

$$x=6 x=-8$$