Solve x-2/x-5=m/5-xquadx/5-x+2/x

$$\frac { x - 2 } { x - 5 } = \frac { m } { 5 - x } \quad \frac { x } { 5 - x } + \frac { 2 } { x }$$

Solve for m

$$m=x-9+\frac{30}{x}-\frac{50}{x^{2}}$$

$$x\neq 5\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x\left(x-5\right), the least common multiple of x-5,5-x,x.

$$x\left(x-2\right)=-xm\times \frac{x}{5-x}+\left(x-5\right)\times 2$$

Use the distributive property to multiply x by x-2.

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

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

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

Use the distributive property to multiply x-5 by 2.

$$x^{2}-2x=-\frac{xx}{5-x}m+2x-10$$

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

$$x^{2}-2x=-\frac{x^{2}}{5-x}m+2x-10$$

Express \frac{x^{2}}{5-x}m as a single fraction.

$$x^{2}-2x=-\frac{x^{2}m}{5-x}+2x-10$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x-10 times \frac{5-x}{5-x}.

$$x^{2}-2x=-\frac{x^{2}m}{5-x}+\frac{\left(2x-10\right)\left(5-x\right)}{5-x}$$

Since -\frac{x^{2}m}{5-x} and \frac{\left(2x-10\right)\left(5-x\right)}{5-x} have the same denominator, add them by adding their numerators.

$$x^{2}-2x=\frac{-x^{2}m+\left(2x-10\right)\left(5-x\right)}{5-x}$$

Do the multiplications in -x^{2}m+\left(2x-10\right)\left(5-x\right).

$$x^{2}-2x=\frac{-x^{2}m+10x-2x^{2}-50+10x}{5-x}$$

Combine like terms in -x^{2}m+10x-2x^{2}-50+10x.

$$x^{2}-2x=\frac{20x-x^{2}m-2x^{2}-50}{5-x}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{20x-x^{2}m-2x^{2}-50}{5-x}=x^{2}-2x$$

Multiply both sides of the equation by -x+5.

$$20x-x^{2}m-2x^{2}-50=\left(-x+5\right)x^{2}-2x\left(-x+5\right)$$

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

$$20x-x^{2}m-2x^{2}-50=-x^{3}+5x^{2}-2x\left(-x+5\right)$$

Use the distributive property to multiply -2x by -x+5.

$$20x-x^{2}m-2x^{2}-50=-x^{3}+5x^{2}+2x^{2}-10x$$

Combine 5x^{2} and 2x^{2} to get 7x^{2}.

$$20x-x^{2}m-2x^{2}-50=-x^{3}+7x^{2}-10x$$

Subtract 20x from both sides.

$$-x^{2}m-2x^{2}-50=-x^{3}+7x^{2}-10x-20x$$

Combine -10x and -20x to get -30x.

$$-x^{2}m-2x^{2}-50=-x^{3}+7x^{2}-30x$$

Add 2x^{2} to both sides.

$$-x^{2}m-50=-x^{3}+7x^{2}-30x+2x^{2}$$

Combine 7x^{2} and 2x^{2} to get 9x^{2}.

$$-x^{2}m-50=-x^{3}+9x^{2}-30x$$

Add 50 to both sides.

$$-x^{2}m=-x^{3}+9x^{2}-30x+50$$

The equation is in standard form.

$$\left(-x^{2}\right)m=50-30x+9x^{2}-x^{3}$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)m}{-x^{2}}=\frac{\left(x-5\right)\left(-x^{2}+4x-10\right)}{-x^{2}}$$

Dividing by -x^{2} undoes the multiplication by -x^{2}.

$$m=\frac{\left(x-5\right)\left(-x^{2}+4x-10\right)}{-x^{2}}$$

Divide \left(-10+4x-x^{2}\right)\left(-5+x\right) by -x^{2}.

$$m=x-9+\frac{30x-50}{x^{2}}$$