# Ratios.Given y = (x^2+5x+7)/(x^2+5x+4) prove that y = 1/(x+1) - 1/(x+4).

justaguide | Certified Educator

Given that y = (x^2+5x+7)/(x^2+5x+4) we have to prove that y = 1/(x+1) - 1/(x+4)

If y = 1/(x+1) - 1/(x+4)

=> y = [(x + 4) - (x + 1)]/(x + 1)(x + 4)

=> y = 3/(x^2 + 4x + x + 4)

=> y = 3/(x^2 + 5x + 4)

3 is not equal to x^2 + 5x + 7

We can conclude that if y = (x^2+5x+7)/(x^2+5x+4) it does not imply that y = 1/(x+1) - 1/(x+4)

giorgiana1976 | Student

To prove that (x^2+5x+7)/(x^2+5x+4) =  1/(x+1) - 1/(x+4), we'll have to solve the difference between the 2 fractions from the right side.

We notice that the ratios don't have the same denominator, so we cannot compute the subtraction before calculating their LCD.

LCD = (x+1)*(x+4)

To obtain the same LCD to both ratios, we'll multiply the first ratio by x+4 and the second ratio by x+1.

1/(x+1) - 1/(x+4) = [(x+4) - (x+1)]/(x+1)*(x+4)

We'll remove the brackets from numerator:

[(x+4) - (x+1)] = x + 4 - x - 1

We'll combine and eliminate like terms and we'll get:

[(x+4) - (x+1)] = 3

We'll multiply the brackets form denominator:

(x+1)*(x+4)  = x^2 + 4x + x + 4

We'll combine like terms and we'll get:

(x+1)*(x+4)  = x^2 + 5x + 4

1/(x+1) - 1/(x+4) = 3/((x^2 + 5x + 4)

We remark that (x^2+5x+7)/(x^2+5x+4) = 3/((x^2 + 5x + 4) is a false statement. In this case, only the denominator is the same, but numerator is different.