Solve x/x-3-2/x^2-4x+3/frac{5{x-1}+5/x-3}=1

$$\frac { \frac { x } { x - 3 } - \frac { 2 } { x ^ { 2 } - 4 x + 3 } } { \frac { 5 } { x - 1 } + \frac { 5 } { x - 3 } } = 1$$

Solve for x

$$x=9$$

Solution Steps

Factor x^{2}-4x+3.

$$\frac{\frac{x}{x-3}-\frac{2}{\left(x-3\right)\left(x-1\right)}}{\frac{5}{x-1}+\frac{5}{x-3}}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-3 and \left(x-3\right)\left(x-1\right) is \left(x-3\right)\left(x-1\right). Multiply \frac{x}{x-3} times \frac{x-1}{x-1}.

$$\frac{\frac{x\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\frac{2}{\left(x-3\right)\left(x-1\right)}}{\frac{5}{x-1}+\frac{5}{x-3}}=1$$

Since \frac{x\left(x-1\right)}{\left(x-3\right)\left(x-1\right)} and \frac{2}{\left(x-3\right)\left(x-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{x\left(x-1\right)-2}{\left(x-3\right)\left(x-1\right)}}{\frac{5}{x-1}+\frac{5}{x-3}}=1$$

Do the multiplications in x\left(x-1\right)-2.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-1 and x-3 is \left(x-3\right)\left(x-1\right). Multiply \frac{5}{x-1} times \frac{x-3}{x-3}. Multiply \frac{5}{x-3} times \frac{x-1}{x-1}.

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

Since \frac{5\left(x-3\right)}{\left(x-3\right)\left(x-1\right)} and \frac{5\left(x-1\right)}{\left(x-3\right)\left(x-1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 5\left(x-3\right)+5\left(x-1\right).

$$\frac{\frac{x^{2}-x-2}{\left(x-3\right)\left(x-1\right)}}{\frac{5x-15+5x-5}{\left(x-3\right)\left(x-1\right)}}=1$$

Combine like terms in 5x-15+5x-5.

$$\frac{\frac{x^{2}-x-2}{\left(x-3\right)\left(x-1\right)}}{\frac{10x-20}{\left(x-3\right)\left(x-1\right)}}=1$$

Variable x cannot be equal to any of the values 1,3 since division by zero is not defined. Divide \frac{x^{2}-x-2}{\left(x-3\right)\left(x-1\right)} by \frac{10x-20}{\left(x-3\right)\left(x-1\right)} by multiplying \frac{x^{2}-x-2}{\left(x-3\right)\left(x-1\right)} by the reciprocal of \frac{10x-20}{\left(x-3\right)\left(x-1\right)}.

$$\frac{\left(x^{2}-x-2\right)\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)\left(10x-20\right)}=1$$

Use the distributive property to multiply x^{2}-x-2 by x-3 and combine like terms.

$$\frac{\left(x^{3}-4x^{2}+x+6\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)\left(10x-20\right)}=1$$

Use the distributive property to multiply x^{3}-4x^{2}+x+6 by x-1 and combine like terms.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6}{\left(x-3\right)\left(x-1\right)\left(10x-20\right)}=1$$

Use the distributive property to multiply x-3 by x-1 and combine like terms.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6}{\left(x^{2}-4x+3\right)\left(10x-20\right)}=1$$

Use the distributive property to multiply x^{2}-4x+3 by 10x-20 and combine like terms.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6}{10x^{3}-60x^{2}+110x-60}=1$$

Subtract 1 from both sides.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6}{10x^{3}-60x^{2}+110x-60}-1=0$$

Factor 10x^{3}-60x^{2}+110x-60.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}-1=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}-\frac{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}=0$$

Since \frac{x^{4}-5x^{3}+5x^{2}+5x-6}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)} and \frac{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6-10\left(x-3\right)\left(x-2\right)\left(x-1\right)}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}=0$$

Do the multiplications in x^{4}-5x^{3}+5x^{2}+5x-6-10\left(x-3\right)\left(x-2\right)\left(x-1\right).

$$\frac{x^{4}-5x^{3}+5x^{2}+5x-6-10x^{3}+30x^{2}-20x+30x^{2}-90x+60}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}=0$$

Combine like terms in x^{4}-5x^{3}+5x^{2}+5x-6-10x^{3}+30x^{2}-20x+30x^{2}-90x+60.

$$\frac{x^{4}-15x^{3}+65x^{2}-105x+54}{10\left(x-3\right)\left(x-2\right)\left(x-1\right)}=0$$

Variable x cannot be equal to any of the values 1,2,3 since division by zero is not defined. Multiply both sides of the equation by 10\left(x-3\right)\left(x-2\right)\left(x-1\right).

$$x^{4}-15x^{3}+65x^{2}-105x+54=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 54 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±54,±27,±18,±9,±6,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-15x^{3}+65x^{2}-105x+54 by x-1 to get x^{3}-14x^{2}+51x-54. Solve the equation where the result equals to 0.

$$x^{3}-14x^{2}+51x-54=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -54 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±54,±27,±18,±9,±6,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-14x^{2}+51x-54 by x-2 to get x^{2}-12x+27. Solve the equation where the result equals to 0.

$$x^{2}-12x+27=0$$

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 1 for a, -12 for b, and 27 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 1\times 27}}{2}$$

Do the calculations.

$$x=\frac{12±6}{2}$$

Solve the equation x^{2}-12x+27=0 when ± is plus and when ± is minus.

$$x=3 x=9$$

Remove the values that the variable cannot be equal to.

$$x=9$$

List all found solutions.

$$x=1 x=2 x=3 x=9$$

Variable x cannot be equal to any of the values 1,2,3.

$$x=9$$