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

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps for Solving Linear Equation

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

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(x^{2}-9\right)\times 2+\left(x-3\right)x=\left(x+3\right)\times 3x$$

Use the distributive property to multiply x^{2}-9 by 2.

$$2x^{2}-18+\left(x-3\right)x=\left(x+3\right)\times 3x$$

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

$$2x^{2}-18+x^{2}-3x=\left(x+3\right)\times 3x$$

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

$$3x^{2}-18-3x=\left(x+3\right)\times 3x$$

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

$$3x^{2}-18-3x=\left(3x+9\right)x$$

Use the distributive property to multiply 3x+9 by x.

$$3x^{2}-18-3x=3x^{2}+9x$$

Subtract 3x^{2} from both sides.

$$3x^{2}-18-3x-3x^{2}=9x$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-18-3x=9x$$

Subtract 9x from both sides.

$$-18-3x-9x=0$$

Combine -3x and -9x to get -12x.

$$-18-12x=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$-12x=18$$

Divide both sides by -12.

$$x=\frac{18}{-12}$$

Reduce the fraction \frac{18}{-12} to lowest terms by extracting and canceling out 6.

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