Solve 2/x-3+1/x-4=2

$$\frac { 2 } { x - 3 } + \frac { 1 } { x - 4 } = 2$$

Solve for x

$$x=5$$

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 3,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-3\right), the least common multiple of x-3,x-4.

$$\left(x-4\right)\times 2+x-3=2\left(x-4\right)\left(x-3\right)$$

Use the distributive property to multiply x-4 by 2.

$$2x-8+x-3=2\left(x-4\right)\left(x-3\right)$$

Combine 2x and x to get 3x.

$$3x-8-3=2\left(x-4\right)\left(x-3\right)$$

Subtract 3 from -8 to get -11.

$$3x-11=2\left(x-4\right)\left(x-3\right)$$

Use the distributive property to multiply 2 by x-4.

$$3x-11=\left(2x-8\right)\left(x-3\right)$$

Use the distributive property to multiply 2x-8 by x-3 and combine like terms.

$$3x-11=2x^{2}-14x+24$$

Subtract 2x^{2} from both sides.

$$3x-11-2x^{2}=-14x+24$$

Add 14x to both sides.

$$3x-11-2x^{2}+14x=24$$

Combine 3x and 14x to get 17x.

$$17x-11-2x^{2}=24$$

Subtract 24 from both sides.

$$17x-11-2x^{2}-24=0$$

Subtract 24 from -11 to get -35.

$$17x-35-2x^{2}=0$$

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.

$$-2x^{2}+17x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 17 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\left(-2\right)\left(-35\right)}}{2\left(-2\right)}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\left(-2\right)\left(-35\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-17±\sqrt{289+8\left(-35\right)}}{2\left(-2\right)}$$

Multiply 8 times -35.

$$x=\frac{-17±\sqrt{289-280}}{2\left(-2\right)}$$

Add 289 to -280.

$$x=\frac{-17±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of 9.

$$x=\frac{-17±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-17±3}{-4}$$

Now solve the equation x=\frac{-17±3}{-4} when ± is plus. Add -17 to 3.

$$x=-\frac{14}{-4}$$

Reduce the fraction \frac{-14}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{-17±3}{-4} when ± is minus. Subtract 3 from -17.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

The equation is now solved.

$$x=\frac{7}{2} x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 3,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-3\right), the least common multiple of x-3,x-4.

$$\left(x-4\right)\times 2+x-3=2\left(x-4\right)\left(x-3\right)$$

Use the distributive property to multiply x-4 by 2.

$$2x-8+x-3=2\left(x-4\right)\left(x-3\right)$$

Combine 2x and x to get 3x.

$$3x-8-3=2\left(x-4\right)\left(x-3\right)$$

Subtract 3 from -8 to get -11.

$$3x-11=2\left(x-4\right)\left(x-3\right)$$

Use the distributive property to multiply 2 by x-4.

$$3x-11=\left(2x-8\right)\left(x-3\right)$$

Use the distributive property to multiply 2x-8 by x-3 and combine like terms.

$$3x-11=2x^{2}-14x+24$$

Subtract 2x^{2} from both sides.

$$3x-11-2x^{2}=-14x+24$$

Add 14x to both sides.

$$3x-11-2x^{2}+14x=24$$

Combine 3x and 14x to get 17x.

$$17x-11-2x^{2}=24$$

Add 11 to both sides.

$$17x-2x^{2}=24+11$$

Add 24 and 11 to get 35.

$$17x-2x^{2}=35$$

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.

$$-2x^{2}+17x=35$$

Divide both sides by -2.

$$\frac{-2x^{2}+17x}{-2}=\frac{35}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{17}{-2}x=\frac{35}{-2}$$

Divide 17 by -2.

$$x^{2}-\frac{17}{2}x=\frac{35}{-2}$$

Divide 35 by -2.

$$x^{2}-\frac{17}{2}x=-\frac{35}{2}$$

Divide -\frac{17}{2}, the coefficient of the x term, by 2 to get -\frac{17}{4}. Then add the square of -\frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{2}x+\left(-\frac{17}{4}\right)^{2}=-\frac{35}{2}+\left(-\frac{17}{4}\right)^{2}$$

Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=-\frac{35}{2}+\frac{289}{16}$$

Add -\frac{35}{2} to \frac{289}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{17}{2}x+\frac{289}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{17}{4}=\frac{3}{4} x-\frac{17}{4}=-\frac{3}{4}$$

Add \frac{17}{4} to both sides of the equation.

$$x=5 x=\frac{7}{2}$$