Solve 3/x-6-4/x-5=1

$$\frac{ 3 }{ x-6 } - \frac{ 4 }{ x-5 } =1$$

Solve for x

$$x=3$$

$$x=7$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 5,6 since division by zero is not defined. Multiply both sides of the equation by \left(x-6\right)\left(x-5\right), the least common multiple of x-6,x-5.

$$\left(x-5\right)\times 3-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply x-5 by 3.

$$3x-15-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply x-6 by 4.

$$3x-15-\left(4x-24\right)=\left(x-6\right)\left(x-5\right)$$

To find the opposite of 4x-24, find the opposite of each term.

$$3x-15-4x+24=\left(x-6\right)\left(x-5\right)$$

Combine 3x and -4x to get -x.

$$-x-15+24=\left(x-6\right)\left(x-5\right)$$

Add -15 and 24 to get 9.

$$-x+9=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply x-6 by x-5 and combine like terms.

$$-x+9=x^{2}-11x+30$$

Subtract x^{2} from both sides.

$$-x+9-x^{2}=-11x+30$$

Add 11x to both sides.

$$-x+9-x^{2}+11x=30$$

Combine -x and 11x to get 10x.

$$10x+9-x^{2}=30$$

Subtract 30 from both sides.

$$10x+9-x^{2}-30=0$$

Subtract 30 from 9 to get -21.

$$10x-21-x^{2}=0$$

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.

$$-x^{2}+10x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-21\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-21\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-21\right)}}{2\left(-1\right)}$$

Multiply 4 times -21.

$$x=\frac{-10±\sqrt{100-84}}{2\left(-1\right)}$$

Add 100 to -84.

$$x=\frac{-10±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$x=\frac{-10±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±4}{-2}$$

Now solve the equation x=\frac{-10±4}{-2} when ± is plus. Add -10 to 4.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-10±4}{-2} when ± is minus. Subtract 4 from -10.

$$x=-\frac{14}{-2}$$

Divide -14 by -2.

$$x=7$$

The equation is now solved.

$$x=3 x=7$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 5,6 since division by zero is not defined. Multiply both sides of the equation by \left(x-6\right)\left(x-5\right), the least common multiple of x-6,x-5.

$$\left(x-5\right)\times 3-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply x-5 by 3.

$$3x-15-\left(x-6\right)\times 4=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply x-6 by 4.

$$3x-15-\left(4x-24\right)=\left(x-6\right)\left(x-5\right)$$

To find the opposite of 4x-24, find the opposite of each term.

$$3x-15-4x+24=\left(x-6\right)\left(x-5\right)$$

Combine 3x and -4x to get -x.

$$-x-15+24=\left(x-6\right)\left(x-5\right)$$

Add -15 and 24 to get 9.

$$-x+9=\left(x-6\right)\left(x-5\right)$$

Use the distributive property to multiply x-6 by x-5 and combine like terms.

$$-x+9=x^{2}-11x+30$$

Subtract x^{2} from both sides.

$$-x+9-x^{2}=-11x+30$$

Add 11x to both sides.

$$-x+9-x^{2}+11x=30$$

Combine -x and 11x to get 10x.

$$10x+9-x^{2}=30$$

Subtract 9 from both sides.

$$10x-x^{2}=30-9$$

Subtract 9 from 30 to get 21.

$$10x-x^{2}=21$$

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.

$$-x^{2}+10x=21$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{21}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{21}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{21}{-1}$$

Divide 21 by -1.

$$x^{2}-10x=-21$$

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

$$x^{2}-10x+\left(-5\right)^{2}=-21+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-21+25$$

Add -21 to 25.

$$x^{2}-10x+25=4$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-5=2 x-5=-2$$

Add 5 to both sides of the equation.

$$x=7 x=3$$