Solve 3/x+3/x-5=3x-12/x-5

$$\frac { 3 } { x } + \frac { 3 } { x - 5 } = \frac { 3 x - 12 } { x - 5 }$$

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,5 since division by zero is not defined. Multiply both sides of the equation by x\left(x-5\right), the least common multiple of x,x-5.

$$\left(x-5\right)\times 3+x\times 3=x\left(3x-12\right)$$

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

$$3x-15+x\times 3=x\left(3x-12\right)$$

Combine 3x and x\times 3 to get 6x.

$$6x-15=x\left(3x-12\right)$$

Use the distributive property to multiply x by 3x-12.

$$6x-15=3x^{2}-12x$$

Subtract 3x^{2} from both sides.

$$6x-15-3x^{2}=-12x$$

Add 12x to both sides.

$$6x-15-3x^{2}+12x=0$$

Combine 6x and 12x to get 18x.

$$18x-15-3x^{2}=0$$

Divide both sides by 3.

$$6x-5-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+6x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-5\right)=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=5 b=1$$

Rewrite -x^{2}+6x-5 as \left(-x^{2}+5x\right)+\left(x-5\right).

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

Factor out -x in -x^{2}+5x.

$$-x\left(x-5\right)+x-5$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x+1\right)$$

To find equation solutions, solve x-5=0 and -x+1=0.

$$x=5 x=1$$

Variable x cannot be equal to 5.

$$x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,5 since division by zero is not defined. Multiply both sides of the equation by x\left(x-5\right), the least common multiple of x,x-5.

$$\left(x-5\right)\times 3+x\times 3=x\left(3x-12\right)$$

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

$$3x-15+x\times 3=x\left(3x-12\right)$$

Combine 3x and x\times 3 to get 6x.

$$6x-15=x\left(3x-12\right)$$

Use the distributive property to multiply x by 3x-12.

$$6x-15=3x^{2}-12x$$

Subtract 3x^{2} from both sides.

$$6x-15-3x^{2}=-12x$$

Add 12x to both sides.

$$6x-15-3x^{2}+12x=0$$

Combine 6x and 12x to get 18x.

$$18x-15-3x^{2}=0$$

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.

$$-3x^{2}+18x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 18 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-3\right)\left(-15\right)}}{2\left(-3\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-3\right)\left(-15\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-18±\sqrt{324+12\left(-15\right)}}{2\left(-3\right)}$$

Multiply 12 times -15.

$$x=\frac{-18±\sqrt{324-180}}{2\left(-3\right)}$$

Add 324 to -180.

$$x=\frac{-18±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$x=\frac{-18±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-18±12}{-6}$$

Now solve the equation x=\frac{-18±12}{-6} when ± is plus. Add -18 to 12.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-18±12}{-6} when ± is minus. Subtract 12 from -18.

$$x=-\frac{30}{-6}$$

Divide -30 by -6.

$$x=5$$

The equation is now solved.

$$x=1 x=5$$

Variable x cannot be equal to 5.

$$x=1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,5 since division by zero is not defined. Multiply both sides of the equation by x\left(x-5\right), the least common multiple of x,x-5.

$$\left(x-5\right)\times 3+x\times 3=x\left(3x-12\right)$$

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

$$3x-15+x\times 3=x\left(3x-12\right)$$

Combine 3x and x\times 3 to get 6x.

$$6x-15=x\left(3x-12\right)$$

Use the distributive property to multiply x by 3x-12.

$$6x-15=3x^{2}-12x$$

Subtract 3x^{2} from both sides.

$$6x-15-3x^{2}=-12x$$

Add 12x to both sides.

$$6x-15-3x^{2}+12x=0$$

Combine 6x and 12x to get 18x.

$$18x-15-3x^{2}=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$18x-3x^{2}=15$$

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.

$$-3x^{2}+18x=15$$

Divide both sides by -3.

$$\frac{-3x^{2}+18x}{-3}=\frac{15}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{18}{-3}x=\frac{15}{-3}$$

Divide 18 by -3.

$$x^{2}-6x=\frac{15}{-3}$$

Divide 15 by -3.

$$x^{2}-6x=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$

Variable x cannot be equal to 5.

$$x=1$$