Solve 3x/x+1+6/2x=7/x

$$\frac { 3 x } { x + 1 } + \frac { 6 } { 2 x } = \frac { 7 } { x }$$

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 2x\left(x+1\right), the least common multiple of x+1,2x,x.

$$2x\times 3x+\left(x+1\right)\times 6=\left(2x+2\right)\times 7$$

Multiply 2 and 3 to get 6.

$$6xx+\left(x+1\right)\times 6=\left(2x+2\right)\times 7$$

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

$$6x^{2}+\left(x+1\right)\times 6=\left(2x+2\right)\times 7$$

Use the distributive property to multiply x+1 by 6.

$$6x^{2}+6x+6=\left(2x+2\right)\times 7$$

Use the distributive property to multiply 2x+2 by 7.

$$6x^{2}+6x+6=14x+14$$

Subtract 14x from both sides.

$$6x^{2}+6x+6-14x=14$$

Combine 6x and -14x to get -8x.

$$6x^{2}-8x+6=14$$

Subtract 14 from both sides.

$$6x^{2}-8x+6-14=0$$

Subtract 14 from 6 to get -8.

$$6x^{2}-8x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -8 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 6\left(-8\right)}}{2\times 6}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 6\left(-8\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-8\right)±\sqrt{64-24\left(-8\right)}}{2\times 6}$$

Multiply -24 times -8.

$$x=\frac{-\left(-8\right)±\sqrt{64+192}}{2\times 6}$$

Add 64 to 192.

$$x=\frac{-\left(-8\right)±\sqrt{256}}{2\times 6}$$

Take the square root of 256.

$$x=\frac{-\left(-8\right)±16}{2\times 6}$$

The opposite of -8 is 8.

$$x=\frac{8±16}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{8±16}{12}$$

Now solve the equation x=\frac{8±16}{12} when ± is plus. Add 8 to 16.

$$x=\frac{24}{12}$$

Divide 24 by 12.

$$x=2$$

Now solve the equation x=\frac{8±16}{12} when ± is minus. Subtract 16 from 8.

$$x=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

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

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 2x\left(x+1\right), the least common multiple of x+1,2x,x.

$$2x\times 3x+\left(x+1\right)\times 6=\left(2x+2\right)\times 7$$

Multiply 2 and 3 to get 6.

$$6xx+\left(x+1\right)\times 6=\left(2x+2\right)\times 7$$

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

$$6x^{2}+\left(x+1\right)\times 6=\left(2x+2\right)\times 7$$

Use the distributive property to multiply x+1 by 6.

$$6x^{2}+6x+6=\left(2x+2\right)\times 7$$

Use the distributive property to multiply 2x+2 by 7.

$$6x^{2}+6x+6=14x+14$$

Subtract 14x from both sides.

$$6x^{2}+6x+6-14x=14$$

Combine 6x and -14x to get -8x.

$$6x^{2}-8x+6=14$$

Subtract 6 from both sides.

$$6x^{2}-8x=14-6$$

Subtract 6 from 14 to get 8.

$$6x^{2}-8x=8$$

Divide both sides by 6.

$$\frac{6x^{2}-8x}{6}=\frac{8}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{8}{6}\right)x=\frac{8}{6}$$

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

$$x^{2}-\frac{4}{3}x=\frac{8}{6}$$

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

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

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.

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