Solve 16/x-1=15/x+1

$$\frac { 16 } { x } - 1 = \frac { 15 } { x + 1 }$$

Solve for x

$$x=4$$

$$x=-4$$

Steps by Finding Square Root

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x,x+1.

$$\left(x+1\right)\times 16+x\left(x+1\right)\left(-1\right)=x\times 15$$

Use the distributive property to multiply x+1 by 16.

$$16x+16+x\left(x+1\right)\left(-1\right)=x\times 15$$

Use the distributive property to multiply x by x+1.

$$16x+16+\left(x^{2}+x\right)\left(-1\right)=x\times 15$$

Use the distributive property to multiply x^{2}+x by -1.

$$16x+16-x^{2}-x=x\times 15$$

Combine 16x and -x to get 15x.

$$15x+16-x^{2}=x\times 15$$

Subtract x\times 15 from both sides.

$$15x+16-x^{2}-x\times 15=0$$

Combine 15x and -x\times 15 to get 0.

$$16-x^{2}=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$x^{2}=\frac{-16}{-1}$$

Fraction \frac{-16}{-1} can be simplified to 16 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right), the least common multiple of x,x+1.

$$\left(x+1\right)\times 16+x\left(x+1\right)\left(-1\right)=x\times 15$$

Use the distributive property to multiply x+1 by 16.

$$16x+16+x\left(x+1\right)\left(-1\right)=x\times 15$$

Use the distributive property to multiply x by x+1.

$$16x+16+\left(x^{2}+x\right)\left(-1\right)=x\times 15$$

Use the distributive property to multiply x^{2}+x by -1.

$$16x+16-x^{2}-x=x\times 15$$

Combine 16x and -x to get 15x.

$$15x+16-x^{2}=x\times 15$$

Subtract x\times 15 from both sides.

$$15x+16-x^{2}-x\times 15=0$$

Combine 15x and -x\times 15 to get 0.

$$16-x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-x^{2}+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 16}}{2\left(-1\right)}$$

Multiply 4 times 16.

$$x=\frac{0±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±8}{-2}$$

Now solve the equation x=\frac{0±8}{-2} when ± is plus. Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{0±8}{-2} when ± is minus. Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-4 x=4$$