Rational Equation Math Problem

Question

Solve the rational equation  6/4x^2−25 + 3/4x^2+20x+25 = 5x−8/(2x−5)(2x+5)^2.What is the value of x? Enter your answer as an integer, like this: 42 Or as a fraction, in simplest form, like this: 42/53.

Answer 1

Hi Salomon

x = -23/13

You want get the variables out of the denominator then solve for x

For

6/(4x² - 25) + 3/(4x² + 20x +25) = 5x - 8/(2x - 5)(2x + 5)²

Factor all the denominators

6/(2x - 5)(2x + 5) + 3/(2x + 5)² = 5x - 8/(2x - 5)(2x + 5)²

Multiply both sides of the equation by

(2x - 5)(2x + 5)²

To remove the denominators and give

6(2x + 5) + 3(2x - 5) = 5x - 8

12x + 30 + 6x - 15 = 5x - 8

Combine like terms

18x + 15 = 5x - 8

Add the opposites to each side of the equation to get the variables on one side the constants on the other

18x - 5x = -15 - 8

13x = -23

x = -23/13

Plug this value back into

6(2x + 5) + 3(2x - 5) = 5x - 8

6(2(-23/13) + 5) + 3(2(-23/13) - 5) = 5(-23/13) - 8

6(-46/13 + 5) + 3(-46/13 - 5) = -115/13 - 8

6(-46/13 + 65/13) + 3(-46/13 - 65/13) = -115/13 -104/13

6(19/13) + 3(-111/13) = -219/13

114/13 - 333/13 = -219/13

-219/13 = -219/13

Give it a try, check the math remember how to work with fractions

I hope this helps