Solve 4x-10/2x-5-2=3x-39/41

$$\frac { 4 x - 10 } { 2 x - 5 } - 2 = \frac { 3 x - 39 } { 41 }$$

Solve for x

$$x=13$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{5}{2} since division by zero is not defined. Multiply both sides of the equation by 41\left(2x-5\right), the least common multiple of 2x-5,41.

$$41\left(4x-10\right)+41\left(2x-5\right)\left(-2\right)=\left(2x-5\right)\left(3x-39\right)$$

Use the distributive property to multiply 41 by 4x-10.

$$164x-410+41\left(2x-5\right)\left(-2\right)=\left(2x-5\right)\left(3x-39\right)$$

Multiply 41 and -2 to get -82.

$$164x-410-82\left(2x-5\right)=\left(2x-5\right)\left(3x-39\right)$$

Use the distributive property to multiply -82 by 2x-5.

$$164x-410-164x+410=\left(2x-5\right)\left(3x-39\right)$$

Combine 164x and -164x to get 0.

$$-410+410=\left(2x-5\right)\left(3x-39\right)$$

Add -410 and 410 to get 0.

$$0=\left(2x-5\right)\left(3x-39\right)$$

Use the distributive property to multiply 2x-5 by 3x-39 and combine like terms.

$$0=6x^{2}-93x+195$$

Swap sides so that all variable terms are on the left hand side.

$$6x^{2}-93x+195=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -93 for b, and 195 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-93\right)±\sqrt{\left(-93\right)^{2}-4\times 6\times 195}}{2\times 6}$$

Square -93.

$$x=\frac{-\left(-93\right)±\sqrt{8649-4\times 6\times 195}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-93\right)±\sqrt{8649-24\times 195}}{2\times 6}$$

Multiply -24 times 195.

$$x=\frac{-\left(-93\right)±\sqrt{8649-4680}}{2\times 6}$$

Add 8649 to -4680.

$$x=\frac{-\left(-93\right)±\sqrt{3969}}{2\times 6}$$

Take the square root of 3969.

$$x=\frac{-\left(-93\right)±63}{2\times 6}$$

The opposite of -93 is 93.

$$x=\frac{93±63}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{93±63}{12}$$

Now solve the equation x=\frac{93±63}{12} when ± is plus. Add 93 to 63.

$$x=\frac{156}{12}$$

Divide 156 by 12.

$$x=13$$

Now solve the equation x=\frac{93±63}{12} when ± is minus. Subtract 63 from 93.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=13 x=\frac{5}{2}$$

Variable x cannot be equal to \frac{5}{2}.

$$x=13$$

Steps for Completing the Square

Variable x cannot be equal to \frac{5}{2} since division by zero is not defined. Multiply both sides of the equation by 41\left(2x-5\right), the least common multiple of 2x-5,41.

$$41\left(4x-10\right)+41\left(2x-5\right)\left(-2\right)=\left(2x-5\right)\left(3x-39\right)$$

Use the distributive property to multiply 41 by 4x-10.

$$164x-410+41\left(2x-5\right)\left(-2\right)=\left(2x-5\right)\left(3x-39\right)$$

Multiply 41 and -2 to get -82.

$$164x-410-82\left(2x-5\right)=\left(2x-5\right)\left(3x-39\right)$$

Use the distributive property to multiply -82 by 2x-5.

$$164x-410-164x+410=\left(2x-5\right)\left(3x-39\right)$$

Combine 164x and -164x to get 0.

$$-410+410=\left(2x-5\right)\left(3x-39\right)$$

Add -410 and 410 to get 0.

$$0=\left(2x-5\right)\left(3x-39\right)$$

Use the distributive property to multiply 2x-5 by 3x-39 and combine like terms.

$$0=6x^{2}-93x+195$$

Swap sides so that all variable terms are on the left hand side.

$$6x^{2}-93x+195=0$$

Subtract 195 from both sides. Anything subtracted from zero gives its negation.

$$6x^{2}-93x=-195$$

Divide both sides by 6.

$$\frac{6x^{2}-93x}{6}=-\frac{195}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{93}{6}\right)x=-\frac{195}{6}$$

Reduce the fraction \frac{-93}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{31}{2}x=-\frac{195}{6}$$

Reduce the fraction \frac{-195}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{31}{2}x=-\frac{65}{2}$$

Divide -\frac{31}{2}, the coefficient of the x term, by 2 to get -\frac{31}{4}. Then add the square of -\frac{31}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{31}{2}x+\left(-\frac{31}{4}\right)^{2}=-\frac{65}{2}+\left(-\frac{31}{4}\right)^{2}$$

Square -\frac{31}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{31}{2}x+\frac{961}{16}=-\frac{65}{2}+\frac{961}{16}$$

Add -\frac{65}{2} to \frac{961}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{31}{2}x+\frac{961}{16}=\frac{441}{16}$$

Factor x^{2}-\frac{31}{2}x+\frac{961}{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}.

$$\left(x-\frac{31}{4}\right)^{2}=\frac{441}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{31}{4}\right)^{2}}=\sqrt{\frac{441}{16}}$$

Simplify.

$$x-\frac{31}{4}=\frac{21}{4} x-\frac{31}{4}=-\frac{21}{4}$$

Add \frac{31}{4} to both sides of the equation.

$$x=13 x=\frac{5}{2}$$

Variable x cannot be equal to \frac{5}{2}.

$$x=13$$