Solve m+1/2m*m-1/m+1=-1

$$\frac { m + 1 } { 2 m } \times \frac { m - 1 } { m + 1 } = - 1$$

Solve for m

$$m=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Variable m cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 2m\left(m+1\right), the least common multiple of 2m,m+1.

$$\left(m+1\right)\left(m+1\right)\times \frac{m-1}{m+1}=-2m\left(m+1\right)$$

Multiply m+1 and m+1 to get \left(m+1\right)^{2}.

$$\left(m+1\right)^{2}\times \frac{m-1}{m+1}=-2m\left(m+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+1\right)^{2}.

$$\left(m^{2}+2m+1\right)\times \frac{m-1}{m+1}=-2m\left(m+1\right)$$

Express \left(m^{2}+2m+1\right)\times \frac{m-1}{m+1} as a single fraction.

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

Use the distributive property to multiply -2m by m+1.

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

Use the distributive property to multiply m^{2}+2m+1 by m-1 and combine like terms.

$$\frac{m^{3}+m^{2}-m-1}{m+1}=-2m^{2}-2m$$

Add 2m^{2} to both sides.

$$\frac{m^{3}+m^{2}-m-1}{m+1}+2m^{2}=-2m$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2m^{2} times \frac{m+1}{m+1}.

$$\frac{m^{3}+m^{2}-m-1}{m+1}+\frac{2m^{2}\left(m+1\right)}{m+1}=-2m$$

Since \frac{m^{3}+m^{2}-m-1}{m+1} and \frac{2m^{2}\left(m+1\right)}{m+1} have the same denominator, add them by adding their numerators.

$$\frac{m^{3}+m^{2}-m-1+2m^{2}\left(m+1\right)}{m+1}=-2m$$

Do the multiplications in m^{3}+m^{2}-m-1+2m^{2}\left(m+1\right).

$$\frac{m^{3}+m^{2}-m-1+2m^{3}+2m^{2}}{m+1}=-2m$$

Combine like terms in m^{3}+m^{2}-m-1+2m^{3}+2m^{2}.

$$\frac{3m^{3}+3m^{2}-m-1}{m+1}=-2m$$

Add 2m to both sides.

$$\frac{3m^{3}+3m^{2}-m-1}{m+1}+2m=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2m times \frac{m+1}{m+1}.

$$\frac{3m^{3}+3m^{2}-m-1}{m+1}+\frac{2m\left(m+1\right)}{m+1}=0$$

Since \frac{3m^{3}+3m^{2}-m-1}{m+1} and \frac{2m\left(m+1\right)}{m+1} have the same denominator, add them by adding their numerators.

$$\frac{3m^{3}+3m^{2}-m-1+2m\left(m+1\right)}{m+1}=0$$

Do the multiplications in 3m^{3}+3m^{2}-m-1+2m\left(m+1\right).

$$\frac{3m^{3}+3m^{2}-m-1+2m^{2}+2m}{m+1}=0$$

Combine like terms in 3m^{3}+3m^{2}-m-1+2m^{2}+2m.

$$\frac{3m^{3}+5m^{2}+m-1}{m+1}=0$$

Variable m cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by m+1.

$$3m^{3}+5m^{2}+m-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$m=-1$$

By Factor theorem, m-k is a factor of the polynomial for each root k. Divide 3m^{3}+5m^{2}+m-1 by m+1 to get 3m^{2}+2m-1. Solve the equation where the result equals to 0.

$$3m^{2}+2m-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, 2 for b, and -1 for c in the quadratic formula.

$$m=\frac{-2±\sqrt{2^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Do the calculations.

$$m=\frac{-2±4}{6}$$

Solve the equation 3m^{2}+2m-1=0 when ± is plus and when ± is minus.

$$m=-1 m=\frac{1}{3}$$

Remove the values that the variable cannot be equal to.

$$m=\frac{1}{3}$$

List all found solutions.

$$m=-1 m=\frac{1}{3}$$

Variable m cannot be equal to -1.

$$m=\frac{1}{3}$$