Question

Can someone please help me with the second problem?

Answer 1

EXPLANATION

$\frac{(x+3)^2}{x-5} \div \frac{x^2 - 9}{x^2 - 8x +15} \\ = \frac{x^2 + 9}{x-5} \div \frac{x^2 - 9}{x^2 - 8x +15} \\ = \frac{x + 9}{-5} \div \frac{-9}{-8x + 15} \\ = \frac{x}{-5} \div \frac{-1}{-8x + 15} \\ = - \frac{x(8x-15)}{5}$

Note : You only need to substitute, divide and eliminate the numbers to simplify.

Hope it helps.

ANSWER DETAILS

Subject : Mathematics

Class : 9th Grade

Chapter : -

Categorization Code : -

Keywords : Substitution, Algebra.

Answer 2

Answer:

$\frac{x + 3}{x - 5}$

Step-by-step explanation:

The first order of business is to get the denominator the same on both fractions. To do this change the 2nd terms denominator into intercept form. See below.

${x}^{2} - 8x + 15 = (x - 5)(x - 3)$

Do this for the numerator on the 2nd term also.

${x}^{2} - 9 = (x - 3)(x + 3)$

Now we can exclude the (x-3) from both the numerator and denominator of the second term. Now we have the equation below.

$\frac{(x + 3 {)}^{2} }{x - 5} \div \frac{ x + 3}{x - 5}$

Now the denominators are the same. Now we need to exclude x+3 from both numerators. Notice that in the firt term, you do not exclude both x+3, only 1.

$\frac{x + 3}{x - 5}$

There is your answer