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

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

Solve for x

$$x = \frac{\sqrt{229} - 7}{6} \approx 1.355457658$$

$$x=\frac{-\sqrt{229}-7}{6}\approx -3.688790992$$

Steps Using the Quadratic Formula

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

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

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

$$3x^{2}+7x-6=3\times 3$$

Multiply 3 and 3 to get 9.

$$3x^{2}+7x-6=9$$

Subtract 9 from both sides.

$$3x^{2}+7x-6-9=0$$

Subtract 9 from -6 to get -15.

$$3x^{2}+7x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$x=\frac{-7±\sqrt{49+180}}{2\times 3}$$

Add 49 to 180.

$$x=\frac{-7±\sqrt{229}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±\sqrt{229}}{6}$$

Now solve the equation x=\frac{-7±\sqrt{229}}{6} when ± is plus. Add -7 to \sqrt{229}.

$$x=\frac{\sqrt{229}-7}{6}$$

Now solve the equation x=\frac{-7±\sqrt{229}}{6} when ± is minus. Subtract \sqrt{229} from -7.

$$x=\frac{-\sqrt{229}-7}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{229}-7}{6} x=\frac{-\sqrt{229}-7}{6}$$

Steps for Completing the Square

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

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

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

$$3x^{2}+7x-6=3\times 3$$

Multiply 3 and 3 to get 9.

$$3x^{2}+7x-6=9$$

Add 6 to both sides.

$$3x^{2}+7x=9+6$$

Add 9 and 6 to get 15.

$$3x^{2}+7x=15$$

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=\frac{15}{3}$$

Divide 15 by 3.

$$x^{2}+\frac{7}{3}x=5$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=5+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=5+\frac{49}{36}$$

Add 5 to \frac{49}{36}.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{229}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{229}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{229}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{\sqrt{229}}{6} x+\frac{7}{6}=-\frac{\sqrt{229}}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=\frac{\sqrt{229}-7}{6} x=\frac{-\sqrt{229}-7}{6}$$