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

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

Solve for x

$$x=-1$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$3x+1=2xx+2x\times 2$$

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

$$3x+1=2x^{2}+2x\times 2$$

Multiply 2 and 2 to get 4.

$$3x+1=2x^{2}+4x$$

Subtract 2x^{2} from both sides.

$$3x+1-2x^{2}=4x$$

Subtract 4x from both sides.

$$3x+1-2x^{2}-4x=0$$

Combine 3x and -4x to get -x.

$$-x+1-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}-x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-2$$

Rewrite -2x^{2}-x+1 as \left(-2x^{2}+x\right)+\left(-2x+1\right).

$$\left(-2x^{2}+x\right)+\left(-2x+1\right)$$

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

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

Factor out common term 2x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$3x+1=2xx+2x\times 2$$

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

$$3x+1=2x^{2}+2x\times 2$$

Multiply 2 and 2 to get 4.

$$3x+1=2x^{2}+4x$$

Subtract 2x^{2} from both sides.

$$3x+1-2x^{2}=4x$$

Subtract 4x from both sides.

$$3x+1-2x^{2}-4x=0$$

Combine 3x and -4x to get -x.

$$-x+1-2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}-x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -1 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2\left(-2\right)}$$

Add 1 to 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of 9.

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

The opposite of -1 is 1.

$$x=\frac{1±3}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±3}{-4}$$

Now solve the equation x=\frac{1±3}{-4} when ± is plus. Add 1 to 3.

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

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{1±3}{-4} when ± is minus. Subtract 3 from 1.

$$x=-\frac{2}{-4}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$3x+1=2xx+2x\times 2$$

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

$$3x+1=2x^{2}+2x\times 2$$

Multiply 2 and 2 to get 4.

$$3x+1=2x^{2}+4x$$

Subtract 2x^{2} from both sides.

$$3x+1-2x^{2}=4x$$

Subtract 4x from both sides.

$$3x+1-2x^{2}-4x=0$$

Combine 3x and -4x to get -x.

$$-x+1-2x^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-x-2x^{2}=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}-x=-1$$

Divide both sides by -2.

$$\frac{-2x^{2}-x}{-2}=-\frac{1}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{1}{-2}\right)x=-\frac{1}{-2}$$

Divide -1 by -2.

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

Divide -1 by -2.

$$x^{2}+\frac{1}{2}x=\frac{1}{2}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{1}{2}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

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

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

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