Solve {l}{1/x-x/x^2-6x+9+5/x^2-3x=0}{y=-3}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{1}{x} - \frac{x}{x ^ {2} - 6 x + 9} + \frac{5}{x ^ {2} - 3 x} = 0 }\\ { y = -3 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=-3$$

Solution Steps

Consider the first equation. Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right)^{2}, the least common multiple of x,x^{2}-6x+9,x^{2}-3x.

$$\left(x-3\right)^{2}-xx+\left(x-3\right)\times 5=0$$

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-x^{2}+\left(x-3\right)\times 5=0$$

Use the distributive property to multiply x-3 by 5.

$$x^{2}-6x+9-x^{2}+5x-15=0$$

Combine -6x and 5x to get -x.

$$x^{2}-x+9-x^{2}-15=0$$

Subtract 15 from 9 to get -6.

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

Combine x^{2} and -x^{2} to get 0.

$$-x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$-x=6$$

Divide both sides by -1.

$$x=-6$$

The system is now solved.

$$x=-6 y=-3 z=-3$$