# College Algebra

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I posed these one's before but added parenthesis which I think made it confusing as the actually questions do not have them. So, I'm reposting them without to see if it makes sense. If not I understand.

a. -6a-16/(a+2)(a-2)^2
b. -6a+4/(a+2)(a-2)^2
c. -6a-16/(a+2)(a-2)
d. -6a+4/(a+2)^2(a-2)

2. Use the properties of exponents to simplify the expression. Write the answer using positive exponents only. So the equation is in parenthesis. (a*b^-4/c^-7)^-3

a. ab^12/c^21
b. b^12/a^3c^21
c. b^12c^7/a^3
d. ab^12c^7

3. Compute as indicated. Write your answer in lowest terms. So this one starts with 3/m-8 then at the end next to the divide line it shows -5. The answer for A start with a negative next to the divide line.

a. - 2/m-8
b. -5m-5/m-8
c. -5m-37/m-8
d. -5m+43/m-8

4. Simplify the compound rational expression. Use either method. So this equation has a +1 before the top equation and a +y before the bottom.
+1 5/y-10/ +y 25/y-10

a. 1/y-5
b. 1/y+5
c. 6/y+25
d. 1/y+1/5

5. Simplify the compound rational expression. Use either method. 2/a-1/3/4/a^2-1/9

a. a+3/3
b. a-3/3
c. 3a/6+a
d. 3a/6-a

6. Compute as indicated. Write final results in lowest terms. x^2+2x-15/2x-6 Divide by x^2-25/4x^2

a. 2x^2/x-3
b. 2x^2/x+3
c. 2x^2/x-5
d. x-5/2x^2

Thanks again for all the help its helping me!

• College Algebra -

doesn't help me. This forum doesn't have formatting.

a+3/a^2-4-a+5/a^2-4a+4

means nothing to me

Parentheses are your only option to indicate how terms and factors are formed. Maybe at least some spaces.

Judging from the answers, I'd say you mean

(a+3)/(a^2-4) - (a+5)/(a^2-4a+4)

(a+3) / (a+2)(a-2) - (a+5) / (a-2)(a-2)

If you put all over a common denominator of (a-2)(a-2)(a+2), then the numerator is

(a+3)(a-2) - (a+5)(a+2)
= (a^2+a-6) - (a^2+7a+10)
= -6a-16
(C)

2.
(a*b^-4/c^-7)^-3
(ac^7/b^4)^-3
(b^4/ac^7)^3
b^12 / a^3c^21
(B)

3.
3/(m-8) - 5
3/(m-8) - 5(m-8)/(m-8)
(3 - (5(m-8))/(m-8)
(3 - 5m + 40)/(m-8)
(43-5m)/(m-8)
(D)

4.
no clue what you mean

5.
no clue

6.
x^2+2x-15 = (x+5)(x-3)
2x-6 = 2(x-3)
so x^2+2x-15/2x-6 = (x+5)/2
and we have

(x+5)/2 * 4x^2/(x^2-25)
(x+5)/2 * 4x^2/(x+5)(x-5)
2x^2/(x-5)
(C)

• College Algebra -

You will need those brackets.
Your text book does not need them because the type was probably set as a fraction with an actual line separating numerator and denominator

e.g. your #1 should probably say, (I am guessing)

(a+3)/(a^2-4) - (a+5)/(a^2-4a+4)
which then would be
= (a+3)/((a+2)(a-2)) - (a+5)/((a-2)(a-2)

so the common denominator is (a+2)(a-2)(a-2)

= ( (a+3)(a-2) - (a+5)(a+2))/((a+2)(a-2)(a-2))
= (a^2 + a - 6 - (a^2 + 7a + 10))/((a+2)(a-2)(a-2))
= (-6a -16)/((a+2)(a-2)^2)

which is close to a) if you put the necessary brackets around your choice a)

I hope you understand why we need these brackets with the way we have to type here
a + 3/3
the way that is typed it would reduce to a + 1
but in the book it was meant to be (a+3)/3
suppose you let a = 6,

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