Solve 3x/x+1+3/x=7/x

$$\frac { 3 x } { x + 1 } + \frac { 3 } { x } = \frac { 7 } { x }$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=2$$

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 x\left(x+1\right), the least common multiple of x+1,x.

$$x\times 3x+\left(x+1\right)\times 3=\left(x+1\right)\times 7$$

Multiply x and x to get x^{2}.

$$x^{2}\times 3+\left(x+1\right)\times 3=\left(x+1\right)\times 7$$

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

$$x^{2}\times 3+3x+3=\left(x+1\right)\times 7$$

Use the distributive property to multiply x+1 by 7.

$$x^{2}\times 3+3x+3=7x+7$$

Subtract 7x from both sides.

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

Combine 3x and -7x to get -4x.

$$x^{2}\times 3-4x+3=7$$

Subtract 7 from both sides.

$$x^{2}\times 3-4x+3-7=0$$

Subtract 7 from 3 to get -4.

$$x^{2}\times 3-4x-4=0$$

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.

$$3x^{2}-4x-4=0$$

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

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

Square -4.

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

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 3}$$

Add 16 to 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$x=\frac{-\left(-4\right)±8}{2\times 3}$$

The opposite of -4 is 4.

$$x=\frac{4±8}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±8}{6}$$

Now solve the equation x=\frac{4±8}{6} when ± is plus. Add 4 to 8.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{4±8}{6} when ± is minus. Subtract 8 from 4.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

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 x\left(x+1\right), the least common multiple of x+1,x.

$$x\times 3x+\left(x+1\right)\times 3=\left(x+1\right)\times 7$$

Multiply x and x to get x^{2}.

$$x^{2}\times 3+\left(x+1\right)\times 3=\left(x+1\right)\times 7$$

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

$$x^{2}\times 3+3x+3=\left(x+1\right)\times 7$$

Use the distributive property to multiply x+1 by 7.

$$x^{2}\times 3+3x+3=7x+7$$

Subtract 7x from both sides.

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

Combine 3x and -7x to get -4x.

$$x^{2}\times 3-4x+3=7$$

Subtract 3 from both sides.

$$x^{2}\times 3-4x=7-3$$

Subtract 3 from 7 to get 4.

$$x^{2}\times 3-4x=4$$

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.

$$3x^{2}-4x=4$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

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

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{3}+\frac{4}{9}$$

Add \frac{4}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

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