Solve x-2/3=10/3x+3

$$\frac { x - 2 } { 3 } = \frac { 10 } { 3 x + 3 }$$

Solve for x

$$x=-3$$

$$x=4$$

Steps Using the Quadratic Formula

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

$$\left(x+1\right)\left(x-2\right)=10$$

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

$$x^{2}-x-2=10$$

Subtract 10 from both sides.

$$x^{2}-x-2-10=0$$

Subtract 10 from -2 to get -12.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2}$$

Add 1 to 48.

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

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2}$$

Now solve the equation x=\frac{1±7}{2} when ± is plus. Add 1 to 7.

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

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{1±7}{2} when ± is minus. Subtract 7 from 1.

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

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=4 x=-3$$

Steps for Completing the Square

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

$$\left(x+1\right)\left(x-2\right)=10$$

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

$$x^{2}-x-2=10$$

Add 2 to both sides.

$$x^{2}-x=10+2$$

Add 10 and 2 to get 12.

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

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{7}{2} x-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=4 x=-3$$