Solve 1/x-1+1/x-4=5/4

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

Solve for x

$$x=5$$

$$x = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 1,4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right)\left(x-1\right), the least common multiple of x-1,x-4,4.

$$4x-16+4x-4=5\left(x-4\right)\left(x-1\right)$$

Combine 4x and 4x to get 8x.

$$8x-16-4=5\left(x-4\right)\left(x-1\right)$$

Subtract 4 from -16 to get -20.

$$8x-20=5\left(x-4\right)\left(x-1\right)$$

Use the distributive property to multiply 5 by x-4.

$$8x-20=\left(5x-20\right)\left(x-1\right)$$

Use the distributive property to multiply 5x-20 by x-1 and combine like terms.

$$8x-20=5x^{2}-25x+20$$

Subtract 5x^{2} from both sides.

$$8x-20-5x^{2}=-25x+20$$

Add 25x to both sides.

$$8x-20-5x^{2}+25x=20$$

Combine 8x and 25x to get 33x.

$$33x-20-5x^{2}=20$$

Subtract 20 from both sides.

$$33x-20-5x^{2}-20=0$$

Subtract 20 from -20 to get -40.

$$33x-40-5x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5x^{2}+33x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 33 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-5\right)\left(-40\right)}}{2\left(-5\right)}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\left(-5\right)\left(-40\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-33±\sqrt{1089+20\left(-40\right)}}{2\left(-5\right)}$$

Multiply 20 times -40.

$$x=\frac{-33±\sqrt{1089-800}}{2\left(-5\right)}$$

Add 1089 to -800.

$$x=\frac{-33±\sqrt{289}}{2\left(-5\right)}$$

Take the square root of 289.

$$x=\frac{-33±17}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-33±17}{-10}$$

Now solve the equation x=\frac{-33±17}{-10} when ± is plus. Add -33 to 17.

$$x=-\frac{16}{-10}$$

Reduce the fraction \frac{-16}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{5}$$

Now solve the equation x=\frac{-33±17}{-10} when ± is minus. Subtract 17 from -33.

$$x=-\frac{50}{-10}$$

Divide -50 by -10.

$$x=5$$

The equation is now solved.

$$x=\frac{8}{5} x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 1,4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right)\left(x-1\right), the least common multiple of x-1,x-4,4.

$$4x-16+4x-4=5\left(x-4\right)\left(x-1\right)$$

Combine 4x and 4x to get 8x.

$$8x-16-4=5\left(x-4\right)\left(x-1\right)$$

Subtract 4 from -16 to get -20.

$$8x-20=5\left(x-4\right)\left(x-1\right)$$

Use the distributive property to multiply 5 by x-4.

$$8x-20=\left(5x-20\right)\left(x-1\right)$$

Use the distributive property to multiply 5x-20 by x-1 and combine like terms.

$$8x-20=5x^{2}-25x+20$$

Subtract 5x^{2} from both sides.

$$8x-20-5x^{2}=-25x+20$$

Add 25x to both sides.

$$8x-20-5x^{2}+25x=20$$

Combine 8x and 25x to get 33x.

$$33x-20-5x^{2}=20$$

Add 20 to both sides.

$$33x-5x^{2}=20+20$$

Add 20 and 20 to get 40.

$$33x-5x^{2}=40$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5x^{2}+33x=40$$

Divide both sides by -5.

$$\frac{-5x^{2}+33x}{-5}=\frac{40}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{33}{-5}x=\frac{40}{-5}$$

Divide 33 by -5.

$$x^{2}-\frac{33}{5}x=\frac{40}{-5}$$

Divide 40 by -5.

$$x^{2}-\frac{33}{5}x=-8$$

Divide -\frac{33}{5}, the coefficient of the x term, by 2 to get -\frac{33}{10}. Then add the square of -\frac{33}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{33}{5}x+\left(-\frac{33}{10}\right)^{2}=-8+\left(-\frac{33}{10}\right)^{2}$$

Square -\frac{33}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{33}{5}x+\frac{1089}{100}=-8+\frac{1089}{100}$$

Add -8 to \frac{1089}{100}.

$$x^{2}-\frac{33}{5}x+\frac{1089}{100}=\frac{289}{100}$$

Factor x^{2}-\frac{33}{5}x+\frac{1089}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{33}{10}\right)^{2}=\frac{289}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{33}{10}\right)^{2}}=\sqrt{\frac{289}{100}}$$

Simplify.

$$x-\frac{33}{10}=\frac{17}{10} x-\frac{33}{10}=-\frac{17}{10}$$

Add \frac{33}{10} to both sides of the equation.

$$x=5 x=\frac{8}{5}$$