Solve 1+2x/x+4=3/x-1

$$1 + \frac { 2 x } { x + 4 } = \frac { 3 } { x - 1 }$$

Solve for x

$$x=-2$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using the Quadratic Formula

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

$$\left(x-1\right)\left(x+4\right)+\left(x-1\right)\times 2x=\left(x+4\right)\times 3$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{2}+3x-4+\left(x-1\right)\times 2x=\left(x+4\right)\times 3$$

Use the distributive property to multiply x-1 by 2.

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

Use the distributive property to multiply 2x-2 by x.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

Combine 3x and -2x to get x.

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

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

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

Subtract 3x from both sides.

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

Combine x and -3x to get -2x.

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

Subtract 12 from both sides.

$$3x^{2}-2x-4-12=0$$

Subtract 12 from -4 to get -16.

$$3x^{2}-2x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times 3.

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

Multiply -12 times -16.

$$x=\frac{-\left(-2\right)±\sqrt{4+192}}{2\times 3}$$

Add 4 to 192.

$$x=\frac{-\left(-2\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-\left(-2\right)±14}{2\times 3}$$

The opposite of -2 is 2.

$$x=\frac{2±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±14}{6}$$

Now solve the equation x=\frac{2±14}{6} when ± is plus. Add 2 to 14.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{2±14}{6} when ± is minus. Subtract 14 from 2.

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

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=\frac{8}{3} x=-2$$

Steps for Completing the Square

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

$$\left(x-1\right)\left(x+4\right)+\left(x-1\right)\times 2x=\left(x+4\right)\times 3$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{2}+3x-4+\left(x-1\right)\times 2x=\left(x+4\right)\times 3$$

Use the distributive property to multiply x-1 by 2.

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

Use the distributive property to multiply 2x-2 by x.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

Combine 3x and -2x to get x.

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

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

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

Subtract 3x from both sides.

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

Combine x and -3x to get -2x.

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

Add 4 to both sides.

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

Add 12 and 4 to get 16.

$$3x^{2}-2x=16$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{16}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{16}{3}+\frac{1}{9}$$

Add \frac{16}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

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

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{8}{3} x=-2$$