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

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

Solve for x

$$x=\frac{5}{6}\approx 0.833333333$$

Steps Using Factoring By Grouping

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^{2}+x.

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

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

$$x^{2}+x+x\times 5x=5$$

Multiply x and x to get x^{2}.

$$x^{2}+x+x^{2}\times 5=5$$

Combine x^{2} and x^{2}\times 5 to get 6x^{2}.

$$6x^{2}+x=5$$

Subtract 5 from both sides.

$$6x^{2}+x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-5\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite 6x^{2}+x-5 as \left(6x^{2}-5x\right)+\left(6x-5\right).

$$\left(6x^{2}-5x\right)+\left(6x-5\right)$$

Factor out x in 6x^{2}-5x.

$$x\left(6x-5\right)+6x-5$$

Factor out common term 6x-5 by using distributive property.

$$\left(6x-5\right)\left(x+1\right)$$

To find equation solutions, solve 6x-5=0 and x+1=0.

$$x=\frac{5}{6} x=-1$$

Variable x cannot be equal to -1.

$$x=\frac{5}{6}$$

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^{2}+x.

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

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

$$x^{2}+x+x\times 5x=5$$

Multiply x and x to get x^{2}.

$$x^{2}+x+x^{2}\times 5=5$$

Combine x^{2} and x^{2}\times 5 to get 6x^{2}.

$$6x^{2}+x=5$$

Subtract 5 from both sides.

$$6x^{2}+x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 6\left(-5\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-1±\sqrt{1-24\left(-5\right)}}{2\times 6}$$

Multiply -24 times -5.

$$x=\frac{-1±\sqrt{1+120}}{2\times 6}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-1±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-1±11}{12}$$

Now solve the equation x=\frac{-1±11}{12} when ± is plus. Add -1 to 11.

$$x=\frac{10}{12}$$

Reduce the fraction \frac{10}{12} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{6}$$

Now solve the equation x=\frac{-1±11}{12} when ± is minus. Subtract 11 from -1.

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

Divide -12 by 12.

$$x=-1$$

The equation is now solved.

$$x=\frac{5}{6} x=-1$$

Variable x cannot be equal to -1.

$$x=\frac{5}{6}$$

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^{2}+x.

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

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

$$x^{2}+x+x\times 5x=5$$

Multiply x and x to get x^{2}.

$$x^{2}+x+x^{2}\times 5=5$$

Combine x^{2} and x^{2}\times 5 to get 6x^{2}.

$$6x^{2}+x=5$$

Divide both sides by 6.

$$\frac{6x^{2}+x}{6}=\frac{5}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{1}{6}x=\frac{5}{6}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{6}x+\left(\frac{1}{12}\right)^{2}=\frac{5}{6}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{5}{6}+\frac{1}{144}$$

Add \frac{5}{6} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{121}{144}$$

Factor x^{2}+\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{12}\right)^{2}=\frac{121}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{12}\right)^{2}}=\sqrt{\frac{121}{144}}$$

Simplify.

$$x+\frac{1}{12}=\frac{11}{12} x+\frac{1}{12}=-\frac{11}{12}$$

Subtract \frac{1}{12} from both sides of the equation.

$$x=\frac{5}{6} x=-1$$

Variable x cannot be equal to -1.

$$x=\frac{5}{6}$$