Solve x/2x-3-1/2x=3/2(2x-3)

$$\frac { x } { 2 x - 3 } - \frac { 1 } { 2 x } = \frac { 3 } { 2 ( 2 x - 3 ) }$$

Solve for x

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x\left(2x-3\right), the least common multiple of 2x-3,2x,2\left(2x-3\right).

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

Multiply x and x to get x^{2}.

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

To find the opposite of 2x-3, find the opposite of each term.

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

Subtract x\times 3 from both sides.

$$2x^{2}-2x+3-x\times 3=0$$

Combine -2x and -x\times 3 to get -5x.

$$2x^{2}-5x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\times 3=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite 2x^{2}-5x+3 as \left(2x^{2}-3x\right)+\left(-2x+3\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(2x-3\right)-\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x-1\right)$$

To find equation solutions, solve 2x-3=0 and x-1=0.

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

Variable x cannot be equal to \frac{3}{2}.

$$x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x\left(2x-3\right), the least common multiple of 2x-3,2x,2\left(2x-3\right).

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

Multiply x and x to get x^{2}.

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

To find the opposite of 2x-3, find the opposite of each term.

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

Subtract x\times 3 from both sides.

$$2x^{2}-2x+3-x\times 3=0$$

Combine -2x and -x\times 3 to get -5x.

$$2x^{2}-5x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\times 3}}{2\times 2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2\times 2}$$

Add 25 to -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-\left(-5\right)±1}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±1}{4}$$

Now solve the equation x=\frac{5±1}{4} when ± is plus. Add 5 to 1.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{5±1}{4} when ± is minus. Subtract 1 from 5.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

The equation is now solved.

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

Variable x cannot be equal to \frac{3}{2}.

$$x=1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2x\left(2x-3\right), the least common multiple of 2x-3,2x,2\left(2x-3\right).

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

Multiply x and x to get x^{2}.

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

To find the opposite of 2x-3, find the opposite of each term.

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

Subtract x\times 3 from both sides.

$$2x^{2}-2x+3-x\times 3=0$$

Combine -2x and -x\times 3 to get -5x.

$$2x^{2}-5x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-5x=-3$$

Divide both sides by 2.

$$\frac{2x^{2}-5x}{2}=-\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{5}{2}x=-\frac{3}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{3}{2}+\frac{25}{16}$$

Add -\frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{1}{4} x-\frac{5}{4}=-\frac{1}{4}$$

Add \frac{5}{4} to both sides of the equation.

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

Variable x cannot be equal to \frac{3}{2}.

$$x=1$$