Solve 1+5x/x+1=2+x/x

$$\frac { 1 + 5 x } { x + 1 } = \frac { 2 + x } { x }$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=1$$

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+1,x.

$$x\left(1+5x\right)=\left(x+1\right)\left(2+x\right)$$

Use the distributive property to multiply x by 1+5x.

$$x+5x^{2}=\left(x+1\right)\left(2+x\right)$$

Use the distributive property to multiply x+1 by 2+x and combine like terms.

$$x+5x^{2}=3x+x^{2}+2$$

Subtract 3x from both sides.

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

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

$$-2x+5x^{2}=x^{2}+2$$

Subtract x^{2} from both sides.

$$-2x+5x^{2}-x^{2}=2$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

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

Subtract 2 from both sides.

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

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.

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

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

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

Square -2.

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

Multiply -4 times 4.

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

Multiply -16 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 4}$$

Add 4 to 32.

$$x=\frac{-\left(-2\right)±\sqrt{36}}{2\times 4}$$

Take the square root of 36.

$$x=\frac{-\left(-2\right)±6}{2\times 4}$$

The opposite of -2 is 2.

$$x=\frac{2±6}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{2±6}{8}$$

Now solve the equation x=\frac{2±6}{8} when ± is plus. Add 2 to 6.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{2±6}{8} when ± is minus. Subtract 6 from 2.

$$x=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

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+1,x.

$$x\left(1+5x\right)=\left(x+1\right)\left(2+x\right)$$

Use the distributive property to multiply x by 1+5x.

$$x+5x^{2}=\left(x+1\right)\left(2+x\right)$$

Use the distributive property to multiply x+1 by 2+x and combine like terms.

$$x+5x^{2}=3x+x^{2}+2$$

Subtract 3x from both sides.

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

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

$$-2x+5x^{2}=x^{2}+2$$

Subtract x^{2} from both sides.

$$-2x+5x^{2}-x^{2}=2$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

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

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.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Add \frac{1}{4} to both sides of the equation.

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