Question

Can someone please help me with the second problem?

Answer 1

EXPLANATION

`((x+3)^2)/(x-5) / (x^2 - 9)/(x^2 - 8x +15) \\ = (x^2 + 9)/(x-5) / (x^2 - 9)/(x^2 - 8x +15) \\ = (x + 9)/(-5) / (-9)/(-8x + 15) \\ = (x)/(-5) / (-1)/(-8x + 15) \\ = - (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

`(x + 3)/(x - 5)`

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} / ( 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.

`(x + 3)/(x - 5)`

There is your answer