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

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

Solve for x

$$x=-1$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-2,x-1.

$$\left(x-1\right)\times 3-\left(x-2\right)\left(5+x\right)=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-1 by 3.

$$3x-3-\left(x-2\right)\left(5+x\right)=\left(x-2\right)\left(x-1\right)$$

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

$$3x-3-\left(3x+x^{2}-10\right)=\left(x-2\right)\left(x-1\right)$$

To find the opposite of 3x+x^{2}-10, find the opposite of each term.

$$3x-3-3x-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Combine 3x and -3x to get 0.

$$-3-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Add -3 and 10 to get 7.

$$7-x^{2}=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$7-x^{2}=x^{2}-3x+2$$

Subtract x^{2} from both sides.

$$7-x^{2}-x^{2}=-3x+2$$

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

$$7-2x^{2}=-3x+2$$

Add 3x to both sides.

$$7-2x^{2}+3x=2$$

Subtract 2 from both sides.

$$7-2x^{2}+3x-2=0$$

Subtract 2 from 7 to get 5.

$$5-2x^{2}+3x=0$$

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.

$$-2x^{2}+3x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 3 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8\times 5}}{2\left(-2\right)}$$

Multiply 8 times 5.

$$x=\frac{-3±\sqrt{9+40}}{2\left(-2\right)}$$

Add 9 to 40.

$$x=\frac{-3±\sqrt{49}}{2\left(-2\right)}$$

Take the square root of 49.

$$x=\frac{-3±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±7}{-4}$$

Now solve the equation x=\frac{-3±7}{-4} when ± is plus. Add -3 to 7.

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

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-3±7}{-4} when ± is minus. Subtract 7 from -3.

$$x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-2,x-1.

$$\left(x-1\right)\times 3-\left(x-2\right)\left(5+x\right)=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-1 by 3.

$$3x-3-\left(x-2\right)\left(5+x\right)=\left(x-2\right)\left(x-1\right)$$

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

$$3x-3-\left(3x+x^{2}-10\right)=\left(x-2\right)\left(x-1\right)$$

To find the opposite of 3x+x^{2}-10, find the opposite of each term.

$$3x-3-3x-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Combine 3x and -3x to get 0.

$$-3-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Add -3 and 10 to get 7.

$$7-x^{2}=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$7-x^{2}=x^{2}-3x+2$$

Subtract x^{2} from both sides.

$$7-x^{2}-x^{2}=-3x+2$$

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

$$7-2x^{2}=-3x+2$$

Add 3x to both sides.

$$7-2x^{2}+3x=2$$

Subtract 7 from both sides.

$$-2x^{2}+3x=2-7$$

Subtract 7 from 2 to get -5.

$$-2x^{2}+3x=-5$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 3 by -2.

$$x^{2}-\frac{3}{2}x=-\frac{5}{-2}$$

Divide -5 by -2.

$$x^{2}-\frac{3}{2}x=\frac{5}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{5}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}$$

Add \frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{7}{4} x-\frac{3}{4}=-\frac{7}{4}$$

Add \frac{3}{4} to both sides of the equation.

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