Solve 6-4x/4x-2=2x+1/2x+3

$$\frac { 6 - 4 x } { 4 x - 2 } = \frac { 2 x + 1 } { 2 x + 3 }$$

Solve for x

$$x = \frac{\sqrt{5}}{2} \approx 1.118033989$$

$$x = -\frac{\sqrt{5}}{2} \approx -1.118033989$$

Steps by Finding Square Root

Variable x cannot be equal to any of the values -\frac{3}{2},\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2\left(2x-1\right)\left(2x+3\right), the least common multiple of 4x-2,2x+3.

$$\left(2x+3\right)\left(6-4x\right)=\left(4x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2x+3 by 6-4x and combine like terms.

$$-8x^{2}+18=\left(4x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 4x-2 by 2x+1 and combine like terms.

$$-8x^{2}+18=8x^{2}-2$$

Subtract 8x^{2} from both sides.

$$-8x^{2}+18-8x^{2}=-2$$

Combine -8x^{2} and -8x^{2} to get -16x^{2}.

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

Subtract 18 from both sides.

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

Subtract 18 from -2 to get -20.

$$-16x^{2}=-20$$

Divide both sides by -16.

$$x^{2}=\frac{-20}{-16}$$

Reduce the fraction \frac{-20}{-16} to lowest terms by extracting and canceling out -4.

$$x^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{5}}{2} x=-\frac{\sqrt{5}}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{3}{2},\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2\left(2x-1\right)\left(2x+3\right), the least common multiple of 4x-2,2x+3.

$$\left(2x+3\right)\left(6-4x\right)=\left(4x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2x+3 by 6-4x and combine like terms.

$$-8x^{2}+18=\left(4x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 4x-2 by 2x+1 and combine like terms.

$$-8x^{2}+18=8x^{2}-2$$

Subtract 8x^{2} from both sides.

$$-8x^{2}+18-8x^{2}=-2$$

Combine -8x^{2} and -8x^{2} to get -16x^{2}.

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

Add 2 to both sides.

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

Add 18 and 2 to get 20.

$$-16x^{2}+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 0 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-16\right)\times 20}}{2\left(-16\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-16\right)\times 20}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{0±\sqrt{64\times 20}}{2\left(-16\right)}$$

Multiply 64 times 20.

$$x=\frac{0±\sqrt{1280}}{2\left(-16\right)}$$

Take the square root of 1280.

$$x=\frac{0±16\sqrt{5}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{0±16\sqrt{5}}{-32}$$

Now solve the equation x=\frac{0±16\sqrt{5}}{-32} when ± is plus.

$$x=-\frac{\sqrt{5}}{2}$$

Now solve the equation x=\frac{0±16\sqrt{5}}{-32} when ± is minus.

$$x=\frac{\sqrt{5}}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{5}}{2} x=\frac{\sqrt{5}}{2}$$