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

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

Solve for x

$$x = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

Steps for Solving Linear Equation

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

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

Use the distributive property to multiply 2x+3 by 2.

$$4x+6+\left(2x-3\right)\times 2x=\left(2x-3\right)\left(2x+3\right)$$

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

$$4x+6+\left(4x-6\right)x=\left(2x-3\right)\left(2x+3\right)$$

Use the distributive property to multiply 4x-6 by x.

$$4x+6+4x^{2}-6x=\left(2x-3\right)\left(2x+3\right)$$

Combine 4x and -6x to get -2x.

$$-2x+6+4x^{2}=\left(2x-3\right)\left(2x+3\right)$$

Consider \left(2x-3\right)\left(2x+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.

$$-2x+6+4x^{2}=\left(2x\right)^{2}-9$$

Expand \left(2x\right)^{2}.

$$-2x+6+4x^{2}=2^{2}x^{2}-9$$

Calculate 2 to the power of 2 and get 4.

$$-2x+6+4x^{2}=4x^{2}-9$$

Subtract 4x^{2} from both sides.

$$-2x+6+4x^{2}-4x^{2}=-9$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-2x+6=-9$$

Subtract 6 from both sides.

$$-2x=-9-6$$

Subtract 6 from -9 to get -15.

$$-2x=-15$$

Divide both sides by -2.

$$x=\frac{-15}{-2}$$

Fraction \frac{-15}{-2} can be simplified to \frac{15}{2} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{15}{2}$$