Solve 7/x+1-4/3x-2=1

$$\frac { 7 } { x + 1 } - \frac { 4 } { 3 x - 2 } = 1$$

Solve for x

$$x=4$$

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-2\right)\left(x+1\right), the least common multiple of x+1,3x-2.

$$\left(3x-2\right)\times 7-\left(x+1\right)\times 4=\left(3x-2\right)\left(x+1\right)$$

Use the distributive property to multiply 3x-2 by 7.

$$21x-14-\left(x+1\right)\times 4=\left(3x-2\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 4.

$$21x-14-\left(4x+4\right)=\left(3x-2\right)\left(x+1\right)$$

To find the opposite of 4x+4, find the opposite of each term.

$$21x-14-4x-4=\left(3x-2\right)\left(x+1\right)$$

Combine 21x and -4x to get 17x.

$$17x-14-4=\left(3x-2\right)\left(x+1\right)$$

Subtract 4 from -14 to get -18.

$$17x-18=\left(3x-2\right)\left(x+1\right)$$

Use the distributive property to multiply 3x-2 by x+1 and combine like terms.

$$17x-18=3x^{2}+x-2$$

Subtract 3x^{2} from both sides.

$$17x-18-3x^{2}=x-2$$

Subtract x from both sides.

$$17x-18-3x^{2}-x=-2$$

Combine 17x and -x to get 16x.

$$16x-18-3x^{2}=-2$$

Add 2 to both sides.

$$16x-18-3x^{2}+2=0$$

Add -18 and 2 to get -16.

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

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.

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

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

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

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-3\right)\left(-16\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-16±\sqrt{256+12\left(-16\right)}}{2\left(-3\right)}$$

Multiply 12 times -16.

$$x=\frac{-16±\sqrt{256-192}}{2\left(-3\right)}$$

Add 256 to -192.

$$x=\frac{-16±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-16±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-16±8}{-6}$$

Now solve the equation x=\frac{-16±8}{-6} when ± is plus. Add -16 to 8.

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

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{-16±8}{-6} when ± is minus. Subtract 8 from -16.

$$x=-\frac{24}{-6}$$

Divide -24 by -6.

$$x=4$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-2\right)\left(x+1\right), the least common multiple of x+1,3x-2.

$$\left(3x-2\right)\times 7-\left(x+1\right)\times 4=\left(3x-2\right)\left(x+1\right)$$

Use the distributive property to multiply 3x-2 by 7.

$$21x-14-\left(x+1\right)\times 4=\left(3x-2\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 4.

$$21x-14-\left(4x+4\right)=\left(3x-2\right)\left(x+1\right)$$

To find the opposite of 4x+4, find the opposite of each term.

$$21x-14-4x-4=\left(3x-2\right)\left(x+1\right)$$

Combine 21x and -4x to get 17x.

$$17x-14-4=\left(3x-2\right)\left(x+1\right)$$

Subtract 4 from -14 to get -18.

$$17x-18=\left(3x-2\right)\left(x+1\right)$$

Use the distributive property to multiply 3x-2 by x+1 and combine like terms.

$$17x-18=3x^{2}+x-2$$

Subtract 3x^{2} from both sides.

$$17x-18-3x^{2}=x-2$$

Subtract x from both sides.

$$17x-18-3x^{2}-x=-2$$

Combine 17x and -x to get 16x.

$$16x-18-3x^{2}=-2$$

Add 18 to both sides.

$$16x-3x^{2}=-2+18$$

Add -2 and 18 to get 16.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 16 by -3.

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

Divide 16 by -3.

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

Divide -\frac{16}{3}, the coefficient of the x term, by 2 to get -\frac{8}{3}. Then add the square of -\frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{3}x+\left(-\frac{8}{3}\right)^{2}=-\frac{16}{3}+\left(-\frac{8}{3}\right)^{2}$$

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=-\frac{16}{3}+\frac{64}{9}$$

Add -\frac{16}{3} to \frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{16}{3}x+\frac{64}{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}.

$$\left(x-\frac{8}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{8}{3}=\frac{4}{3} x-\frac{8}{3}=-\frac{4}{3}$$

Add \frac{8}{3} to both sides of the equation.

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