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

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

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=2$$

Steps Using the Quadratic Formula

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

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

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

$$12x^{2}-17x+6=\left(2x+1\right)\left(x+2\right)$$

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

$$12x^{2}-17x+6=2x^{2}+5x+2$$

Subtract 2x^{2} from both sides.

$$12x^{2}-17x+6-2x^{2}=5x+2$$

Combine 12x^{2} and -2x^{2} to get 10x^{2}.

$$10x^{2}-17x+6=5x+2$$

Subtract 5x from both sides.

$$10x^{2}-17x+6-5x=2$$

Combine -17x and -5x to get -22x.

$$10x^{2}-22x+6=2$$

Subtract 2 from both sides.

$$10x^{2}-22x+6-2=0$$

Subtract 2 from 6 to get 4.

$$10x^{2}-22x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -22 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 10\times 4}}{2\times 10}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 10\times 4}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-22\right)±\sqrt{484-40\times 4}}{2\times 10}$$

Multiply -40 times 4.

$$x=\frac{-\left(-22\right)±\sqrt{484-160}}{2\times 10}$$

Add 484 to -160.

$$x=\frac{-\left(-22\right)±\sqrt{324}}{2\times 10}$$

Take the square root of 324.

$$x=\frac{-\left(-22\right)±18}{2\times 10}$$

The opposite of -22 is 22.

$$x=\frac{22±18}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{22±18}{20}$$

Now solve the equation x=\frac{22±18}{20} when ± is plus. Add 22 to 18.

$$x=\frac{40}{20}$$

Divide 40 by 20.

$$x=2$$

Now solve the equation x=\frac{22±18}{20} when ± is minus. Subtract 18 from 22.

$$x=\frac{4}{20}$$

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

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

The equation is now solved.

$$x=2 x=\frac{1}{5}$$

Steps for Completing the Square

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

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

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

$$12x^{2}-17x+6=\left(2x+1\right)\left(x+2\right)$$

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

$$12x^{2}-17x+6=2x^{2}+5x+2$$

Subtract 2x^{2} from both sides.

$$12x^{2}-17x+6-2x^{2}=5x+2$$

Combine 12x^{2} and -2x^{2} to get 10x^{2}.

$$10x^{2}-17x+6=5x+2$$

Subtract 5x from both sides.

$$10x^{2}-17x+6-5x=2$$

Combine -17x and -5x to get -22x.

$$10x^{2}-22x+6=2$$

Subtract 6 from both sides.

$$10x^{2}-22x=2-6$$

Subtract 6 from 2 to get -4.

$$10x^{2}-22x=-4$$

Divide both sides by 10.

$$\frac{10x^{2}-22x}{10}=-\frac{4}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{22}{10}\right)x=-\frac{4}{10}$$

Reduce the fraction \frac{-22}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{5}x=-\frac{4}{10}$$

Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{5}x=-\frac{2}{5}$$

Divide -\frac{11}{5}, the coefficient of the x term, by 2 to get -\frac{11}{10}. Then add the square of -\frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{11}{10}\right)^{2}$$

Square -\frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-\frac{2}{5}+\frac{121}{100}$$

Add -\frac{2}{5} to \frac{121}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{81}{100}$$

Factor x^{2}-\frac{11}{5}x+\frac{121}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{9}{10} x-\frac{11}{10}=-\frac{9}{10}$$

Add \frac{11}{10} to both sides of the equation.

$$x=2 x=\frac{1}{5}$$