Solve x-8-6x-19/x^2-5x+6-1/x-3=0

$$x-8- \frac{ 6x-19 }{ { x }^{ 2 } -5x+6 } - \frac{ 1 }{ x-3 } =0$$

Solve for x

$$x=1$$

$$x=9$$

Solution Steps

Variable x cannot be equal to any of the values 2,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-2\right), the least common multiple of x^{2}-5x+6,x-3.

$$\left(x-3\right)\left(x-2\right)x+\left(x-3\right)\left(x-2\right)\left(-8\right)-\left(6x-19\right)-\left(x-2\right)=0$$

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

$$\left(x^{2}-5x+6\right)x+\left(x-3\right)\left(x-2\right)\left(-8\right)-\left(6x-19\right)-\left(x-2\right)=0$$

Use the distributive property to multiply x^{2}-5x+6 by x.

$$x^{3}-5x^{2}+6x+\left(x-3\right)\left(x-2\right)\left(-8\right)-\left(6x-19\right)-\left(x-2\right)=0$$

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

$$x^{3}-5x^{2}+6x+\left(x^{2}-5x+6\right)\left(-8\right)-\left(6x-19\right)-\left(x-2\right)=0$$

Use the distributive property to multiply x^{2}-5x+6 by -8.

$$x^{3}-5x^{2}+6x-8x^{2}+40x-48-\left(6x-19\right)-\left(x-2\right)=0$$

Combine -5x^{2} and -8x^{2} to get -13x^{2}.

$$x^{3}-13x^{2}+6x+40x-48-\left(6x-19\right)-\left(x-2\right)=0$$

Combine 6x and 40x to get 46x.

$$x^{3}-13x^{2}+46x-48-\left(6x-19\right)-\left(x-2\right)=0$$

To find the opposite of 6x-19, find the opposite of each term.

$$x^{3}-13x^{2}+46x-48-6x+19-\left(x-2\right)=0$$

Combine 46x and -6x to get 40x.

$$x^{3}-13x^{2}+40x-48+19-\left(x-2\right)=0$$

Add -48 and 19 to get -29.

$$x^{3}-13x^{2}+40x-29-\left(x-2\right)=0$$

To find the opposite of x-2, find the opposite of each term.

$$x^{3}-13x^{2}+40x-29-x+2=0$$

Combine 40x and -x to get 39x.

$$x^{3}-13x^{2}+39x-29+2=0$$

Add -29 and 2 to get -27.

$$x^{3}-13x^{2}+39x-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-13x^{2}+39x-27 by x-1 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.

$$x=1\text{ or }x=9$$

List all found solutions.

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

Variable x cannot be equal to 3.

$$x=9 x=1$$