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

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 4x\left(x+1\right), the least common multiple of x,4,x+1.

$$\left(4x+4\right)\times 3=4x\left(x+1\right)\times \frac{3}{4}+4x$$

Use the distributive property to multiply 4x+4 by 3.

$$12x+12=4x\left(x+1\right)\times \frac{3}{4}+4x$$

Multiply 4 and \frac{3}{4} to get 3.

$$12x+12=3x\left(x+1\right)+4x$$

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

$$12x+12=3x^{2}+3x+4x$$

Combine 3x and 4x to get 7x.

$$12x+12=3x^{2}+7x$$

Subtract 3x^{2} from both sides.

$$12x+12-3x^{2}=7x$$

Subtract 7x from both sides.

$$12x+12-3x^{2}-7x=0$$

Combine 12x and -7x to get 5x.

$$5x+12-3x^{2}=0$$

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

$$-3x^{2}+5x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=5 ab=-3\times 12=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=9 b=-4$$

Rewrite -3x^{2}+5x+12 as \left(-3x^{2}+9x\right)+\left(-4x+12\right).

$$\left(-3x^{2}+9x\right)+\left(-4x+12\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(-x+3\right)+4\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(3x+4\right)$$

To find equation solutions, solve -x+3=0 and 3x+4=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 4x\left(x+1\right), the least common multiple of x,4,x+1.

$$\left(4x+4\right)\times 3=4x\left(x+1\right)\times \frac{3}{4}+4x$$

Use the distributive property to multiply 4x+4 by 3.

$$12x+12=4x\left(x+1\right)\times \frac{3}{4}+4x$$

Multiply 4 and \frac{3}{4} to get 3.

$$12x+12=3x\left(x+1\right)+4x$$

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

$$12x+12=3x^{2}+3x+4x$$

Combine 3x and 4x to get 7x.

$$12x+12=3x^{2}+7x$$

Subtract 3x^{2} from both sides.

$$12x+12-3x^{2}=7x$$

Subtract 7x from both sides.

$$12x+12-3x^{2}-7x=0$$

Combine 12x and -7x to get 5x.

$$5x+12-3x^{2}=0$$

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.

$$-3x^{2}+5x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 5 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-3\right)\times 12}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-5±\sqrt{25+12\times 12}}{2\left(-3\right)}$$

Multiply 12 times 12.

$$x=\frac{-5±\sqrt{25+144}}{2\left(-3\right)}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2\left(-3\right)}$$

Take the square root of 169.

$$x=\frac{-5±13}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-5±13}{-6}$$

Now solve the equation x=\frac{-5±13}{-6} when ± is plus. Add -5 to 13.

$$x=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-5±13}{-6} when ± is minus. Subtract 13 from -5.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by 4x\left(x+1\right), the least common multiple of x,4,x+1.

$$\left(4x+4\right)\times 3=4x\left(x+1\right)\times \frac{3}{4}+4x$$

Use the distributive property to multiply 4x+4 by 3.

$$12x+12=4x\left(x+1\right)\times \frac{3}{4}+4x$$

Multiply 4 and \frac{3}{4} to get 3.

$$12x+12=3x\left(x+1\right)+4x$$

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

$$12x+12=3x^{2}+3x+4x$$

Combine 3x and 4x to get 7x.

$$12x+12=3x^{2}+7x$$

Subtract 3x^{2} from both sides.

$$12x+12-3x^{2}=7x$$

Subtract 7x from both sides.

$$12x+12-3x^{2}-7x=0$$

Combine 12x and -7x to get 5x.

$$5x+12-3x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$5x-3x^{2}=-12$$

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.

$$-3x^{2}+5x=-12$$

Divide both sides by -3.

$$\frac{-3x^{2}+5x}{-3}=-\frac{12}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{5}{-3}x=-\frac{12}{-3}$$

Divide 5 by -3.

$$x^{2}-\frac{5}{3}x=-\frac{12}{-3}$$

Divide -12 by -3.

$$x^{2}-\frac{5}{3}x=4$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=4+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=4+\frac{25}{36}$$

Add 4 to \frac{25}{36}.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{169}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{13}{6} x-\frac{5}{6}=-\frac{13}{6}$$

Add \frac{5}{6} to both sides of the equation.

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