Solve x-5/x-5+3=1/2-x

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

Solve for x

$$x = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

Steps Using the Quadratic Formula

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

$$\left(x-2\right)\left(x-5\right)+\left(x-5\right)\left(x-2\right)\times 3=5-x$$

Use the distributive property to multiply x-2 by x-5 and combine like terms.

$$x^{2}-7x+10+\left(x-5\right)\left(x-2\right)\times 3=5-x$$

Use the distributive property to multiply x-5 by x-2 and combine like terms.

$$x^{2}-7x+10+\left(x^{2}-7x+10\right)\times 3=5-x$$

Use the distributive property to multiply x^{2}-7x+10 by 3.

$$x^{2}-7x+10+3x^{2}-21x+30=5-x$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-7x+10-21x+30=5-x$$

Combine -7x and -21x to get -28x.

$$4x^{2}-28x+10+30=5-x$$

Add 10 and 30 to get 40.

$$4x^{2}-28x+40=5-x$$

Subtract 5 from both sides.

$$4x^{2}-28x+40-5=-x$$

Subtract 5 from 40 to get 35.

$$4x^{2}-28x+35=-x$$

Add x to both sides.

$$4x^{2}-28x+35+x=0$$

Combine -28x and x to get -27x.

$$4x^{2}-27x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -27 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 4\times 35}}{2\times 4}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 4\times 35}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-27\right)±\sqrt{729-16\times 35}}{2\times 4}$$

Multiply -16 times 35.

$$x=\frac{-\left(-27\right)±\sqrt{729-560}}{2\times 4}$$

Add 729 to -560.

$$x=\frac{-\left(-27\right)±\sqrt{169}}{2\times 4}$$

Take the square root of 169.

$$x=\frac{-\left(-27\right)±13}{2\times 4}$$

The opposite of -27 is 27.

$$x=\frac{27±13}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{27±13}{8}$$

Now solve the equation x=\frac{27±13}{8} when ± is plus. Add 27 to 13.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{27±13}{8} when ± is minus. Subtract 13 from 27.

$$x=\frac{14}{8}$$

Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{4}$$

The equation is now solved.

$$x=5 x=\frac{7}{4}$$

Variable x cannot be equal to 5.

$$x=\frac{7}{4}$$

Steps for Completing the Square

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

$$\left(x-2\right)\left(x-5\right)+\left(x-5\right)\left(x-2\right)\times 3=5-x$$

Use the distributive property to multiply x-2 by x-5 and combine like terms.

$$x^{2}-7x+10+\left(x-5\right)\left(x-2\right)\times 3=5-x$$

Use the distributive property to multiply x-5 by x-2 and combine like terms.

$$x^{2}-7x+10+\left(x^{2}-7x+10\right)\times 3=5-x$$

Use the distributive property to multiply x^{2}-7x+10 by 3.

$$x^{2}-7x+10+3x^{2}-21x+30=5-x$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-7x+10-21x+30=5-x$$

Combine -7x and -21x to get -28x.

$$4x^{2}-28x+10+30=5-x$$

Add 10 and 30 to get 40.

$$4x^{2}-28x+40=5-x$$

Add x to both sides.

$$4x^{2}-28x+40+x=5$$

Combine -28x and x to get -27x.

$$4x^{2}-27x+40=5$$

Subtract 40 from both sides.

$$4x^{2}-27x=5-40$$

Subtract 40 from 5 to get -35.

$$4x^{2}-27x=-35$$

Divide both sides by 4.

$$\frac{4x^{2}-27x}{4}=-\frac{35}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{27}{4}x=-\frac{35}{4}$$

Divide -\frac{27}{4}, the coefficient of the x term, by 2 to get -\frac{27}{8}. Then add the square of -\frac{27}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{27}{4}x+\left(-\frac{27}{8}\right)^{2}=-\frac{35}{4}+\left(-\frac{27}{8}\right)^{2}$$

Square -\frac{27}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{27}{4}x+\frac{729}{64}=-\frac{35}{4}+\frac{729}{64}$$

Add -\frac{35}{4} to \frac{729}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{27}{4}x+\frac{729}{64}=\frac{169}{64}$$

Factor x^{2}-\frac{27}{4}x+\frac{729}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{8}\right)^{2}=\frac{169}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{8}\right)^{2}}=\sqrt{\frac{169}{64}}$$

Simplify.

$$x-\frac{27}{8}=\frac{13}{8} x-\frac{27}{8}=-\frac{13}{8}$$

Add \frac{27}{8} to both sides of the equation.

$$x=5 x=\frac{7}{4}$$

Variable x cannot be equal to 5.

$$x=\frac{7}{4}$$