Question

Please answer fast!!!!!!

Answer 1

Answer:

8 and 10

Step-by-step explanation:

Given

$\frac{1}{x}$ + $\frac{1}{x+2}$ = $\frac{9}{40}$ ← combine the fractions on the left side

$\frac{x+2+x}{x(x+2)}$ = $\frac{9}{40}$

$\frac{2x+2}{x^2+2x}$ = $\frac{9}{40}$ ( cross- multiply )

9(x² + 2x) = 40(2x + 2) ← distribute both sides

9x² + 18x = 80x + 80 ( subtract 80x + 80 from both sides )

9x² - 62x - 80 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 9 × - 80 = - 720 and sum = - 62

The factors are - 72 and + 10

Use these factors to split the x- term

9x² - 72x + 10x - 80 = 0 ( factor the first/second and third/fourth terms )

9x(x - 8) + 10(x - 8) = 0 ← factor out (x - 8) from each term

(x - 8)(9x + 10) = 0

Equate each factor to zero and solve for x

x - 8 = 0 ⇒ x = 8

9x + 10 = 0 ⇒ 9x = - 10 ⇒ x = - $\frac{10}{9}$

x is an integer , then x = 8 and x + 2 = 8 + 2 = 10

The integers are 8 and 10