Solve (x/x-1)^2-5*x/x-1+6=0

$$( \frac { x } { x - 1 } ) ^ { 2 } - 5 \cdot \frac { x } { x - 1 } + 6 = 0$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=2$$

Solution Steps

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

$$\left(x-1\right)\left(\left(\frac{x}{x-1}\right)^{2}-5\times \frac{x}{x-1}\right)+\left(x-1\right)\times 6=0$$

To raise \frac{x}{x-1} to a power, raise both numerator and denominator to the power and then divide.

$$\left(x-1\right)\left(\frac{x^{2}}{\left(x-1\right)^{2}}-5\times \frac{x}{x-1}\right)+\left(x-1\right)\times 6=0$$

Express 5\times \frac{x}{x-1} as a single fraction.

$$\left(x-1\right)\left(\frac{x^{2}}{\left(x-1\right)^{2}}-\frac{5x}{x-1}\right)+\left(x-1\right)\times 6=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-1\right)^{2} and x-1 is \left(x-1\right)^{2}. Multiply \frac{5x}{x-1} times \frac{x-1}{x-1}.

$$\left(x-1\right)\left(\frac{x^{2}}{\left(x-1\right)^{2}}-\frac{5x\left(x-1\right)}{\left(x-1\right)^{2}}\right)+\left(x-1\right)\times 6=0$$

Since \frac{x^{2}}{\left(x-1\right)^{2}} and \frac{5x\left(x-1\right)}{\left(x-1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\left(x-1\right)\times \frac{x^{2}-5x\left(x-1\right)}{\left(x-1\right)^{2}}+\left(x-1\right)\times 6=0$$

Do the multiplications in x^{2}-5x\left(x-1\right).

$$\left(x-1\right)\times \frac{x^{2}-5x^{2}+5x}{\left(x-1\right)^{2}}+\left(x-1\right)\times 6=0$$

Combine like terms in x^{2}-5x^{2}+5x.

$$\left(x-1\right)\times \frac{-4x^{2}+5x}{\left(x-1\right)^{2}}+\left(x-1\right)\times 6=0$$

Express \left(x-1\right)\times \frac{-4x^{2}+5x}{\left(x-1\right)^{2}} as a single fraction.

$$\frac{\left(x-1\right)\left(-4x^{2}+5x\right)}{\left(x-1\right)^{2}}+\left(x-1\right)\times 6=0$$

Use the distributive property to multiply x-1 by 6.

$$\frac{\left(x-1\right)\left(-4x^{2}+5x\right)}{\left(x-1\right)^{2}}+6x-6=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6x-6 times \frac{\left(x-1\right)^{2}}{\left(x-1\right)^{2}}.

$$\frac{\left(x-1\right)\left(-4x^{2}+5x\right)}{\left(x-1\right)^{2}}+\frac{\left(6x-6\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}}=0$$

Since \frac{\left(x-1\right)\left(-4x^{2}+5x\right)}{\left(x-1\right)^{2}} and \frac{\left(6x-6\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(x-1\right)\left(-4x^{2}+5x\right)+\left(6x-6\right)\left(x-1\right)^{2}}{\left(x-1\right)^{2}}=0$$

Do the multiplications in \left(x-1\right)\left(-4x^{2}+5x\right)+\left(6x-6\right)\left(x-1\right)^{2}.

$$\frac{-4x^{3}+5x^{2}+4x^{2}-5x+6x^{3}-12x^{2}+6x-6x^{2}+12x-6}{\left(x-1\right)^{2}}=0$$

Combine like terms in -4x^{3}+5x^{2}+4x^{2}-5x+6x^{3}-12x^{2}+6x-6x^{2}+12x-6.

$$\frac{2x^{3}-9x^{2}+13x-6}{\left(x-1\right)^{2}}=0$$

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

$$\frac{2x^{3}-9x^{2}+13x-6}{x^{2}-2x+1}=0$$

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{2}.

$$2x^{3}-9x^{2}+13x-6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±3,±6,±\frac{3}{2},±1,±2,±\frac{1}{2}$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-9x^{2}+13x-6 by x-1 to get 2x^{2}-7x+6. Solve the equation where the result equals to 0.

$$2x^{2}-7x+6=0$$

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 2 for a, -7 for b, and 6 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Do the calculations.

$$x=\frac{7±1}{4}$$

Solve the equation 2x^{2}-7x+6=0 when ± is plus and when ± is minus.

$$x=\frac{3}{2} x=2$$

Remove the values that the variable cannot be equal to.

$$x=2\text{ or }x=\frac{3}{2}$$

List all found solutions.

$$x=1 x=\frac{3}{2} x=2$$

Variable x cannot be equal to 1.

$$x=2 x=\frac{3}{2}$$