3 College Algebra problems.?

Question

Solve and write as improper fraction if needed.
1. x+6=2(5x-5)
2. 8=4(8-9)-7x
Simplify.
3. k^5m^3
----------
n
---------------
k^4m^2
---------
n

4. 3(x-2)+5= -5+x-22

(Sorry. 4 Questions! lol) Need answers ASAP! Honest answers too, no fake ones please!

Answer 1

1.)
x + 6 = 2(5x - 5)
x + 6 = 10x - 10
-9x = -16
x = (16/9)

-------------------------------

2.)
8 = 4(8 - 9) - 7x
8 = 4(-1) - 7x
8 = -4 - 7x
12 = -7x
x = (-12/7)

You probably wrote something wrong on this one.

--------------------------------

3.)

((k^5m^3)/n)/((k^4m^2)/n)
((k^5m^3)/n)*(n/(k^4m^2))
(k^5m^3)/(k^4m^2)

k^(5 - 4) * m^(3 - 2)
k^1 * m^1

ANS : km

------------------------------------

4.)
3(x - 2) + 5 = -5 + x - 22
3(x - 2) + 5 = x - 27
3(x - 2) = x - 32
3x - 6 = x - 32
2x = -26
x = -13

Answer 2

x+6=2(5x-5)
x+6=10x-10
10x -x = 6+10
9x=16
x= 16/9

8=4(8-9)-7x
8=4(-1)-7x
7x= -4-8
7x=-12
x=-12/7

(k^5m^3/n)/(k^4m^2/n)
(k^5m^3/n)*(n/k^4m^2)
km
when you divide compound fractions, you invert the second fraction and multiply them. You cancel like terms. The "n"s disappear, leaving:
(k^5m^3)*(1/k^4m^2)
that equals (k^5m^3)/(k^4m^2)
when you divide terms with powers you subtract the bottom powers from the top powers, leaving:
km as your final answer.

3(x-2)+5 = -5+x-22
3x-6+5 = x -22 -5
3x-1 = x -27
3x -x = -27 +1
2x = -26
x = -13

I hope this helps, and you're not using this site to merely do your homework. I am also in a college algebra class and I do this to strengthen my knowledge and skill. Maybe you'll be answering questions soon.

Answer 3

1. x + 6 = 2(5x - 5)
x + 6 = 10x - 10 -------distributive property
x -10x = -10 - 6 -------transitivity
-9x = -16 --------subtraction property
x = -16/-9 --------cross multiplication
x = 16/9 --------a negative over a negative is equal to positive

2. 8 = 4(8 - 9) - 7x
8 = 32 -36 - 7x -------distributive property
7x = 32 - 36 - 8 --------transitivity
7x = -12 --------subtraction property
x = -12/7 --------cross multiplication

3. k^5m^3
----------
n
-----------------
k^4m^2
-----------
n

k^5m^3 * n
= ------------------------------- ---------division to multiplication
n * k^4m^2

n(k^5m^3)
= -------------- ---------multiplication property for fractions
n(k^4m^2)

= k^(5-4)m^(3-2) ---------dividing variables with exponents

= km

4. 3(x - 2) + 5 = -5 + x - 22
3x - 6 + 5 = x - 27 ---------distributive property
3x - x = -27 + 6 - 5 --------transitivity
2x = -32 + 6 ---------subtraction property
2x = -26 ---------addition property
x = -26/2 --------cross multiplication
x = -13 --------a negative over a positive is equal to a negative

e-mail me for clarifications

Answer 4

¤equation a million ----- X - 7 = 0 ¤equation 2 ----- X^2 + 12 = 7x ¤equation 3 ----- X + 5 = 5 + X ¤equation 4 ----- X + 2 = X + 3 ¤equation 5 ----- X/X = a million ¤equation 6 ----- 3/(X + 4) = 2/(3X - 2) ~Conditional Equation " If there is one variety interior the area of the variable that isn't interior the answer set." ~Equation a million, 2, 4, and six are conditional equations. because; @ equation a million, if x is replaced through 7, the consequent reality is authentic, yet when x is replaced through a form except 7, the consequent reality is fake. accordingly the in worry-free words answer is 7, and the answer set is {7}. @ equation 2, if x is replaced through both 3 or 4, we get a real reality, and if x is replaced through a form except 3 and four, we get a pretend reality. for this reason, the equation has 2 innovations, 3 and four, and the answer set is {3, 4} @ equation 4, if any authentic variety is substituted for x, we get a pretend reality. accordingly the answer set of equation 4 is null. @ equation 6, the answer set of equation set of equation 6 isn't glaring through inspection; even if, we prepare a thanks to outline it in party 2. ¤party 2; locate the answer set of the equation 3/(X + 4) = 2/(3X - 2) a million. locate the liquid crystal exhibit liquid crystal exhibit = (X + 4)(3X - 2) 2. Multiply each and every area of the equation through the liquid crystal exhibit. 3(3X - 2) = 2(X + 4) 9X - 6 = 2X + 8 9X - 2X = 6 + 8 7X = 14 X = 2, answer ~ Identities " If the answer set of any equation in a unmarried variable is an similar because the area of the variable." - equation 3 and 5 are identities. @ equation 3, if any authentic variety is substituted for x, we get a real reality. accordingly the answer set is R. @ equation 5, if any variety except 0 is substituted for x, we get a real reality. for this reason, the answer set is {x|x E R, X isn't equivalent to 0}.

Answer 5

Giving you the answers wouldn't do you any good - you need to learn how to do them - all you're doing is solving for x on the first one- first multiply out the parathetical - then group the Xs on one side, and the numbers on the other - it's easy to solve for 1 X.....Q 2 is the same, as is 4...can't tell what 3 is.

Answer 6

1) x+ 6 = 2(5x-5)

x+6= 10x-10
16=9x

16/9 = x

1 7/9 =x


4) 3(x-2)+5= -5+x-22
3x-6+5=-27+x

3x-1=-27+x
2x=-26
x=-13

Answer 7

1. x=16/9
2. If you copied the problem in right, the answer is x= -12/7
3. I don't understand your symbology.
4. x= -13

Answer 8

1. x=1 7/9
2. x=-1 5/7
3. idk
4 x=-13

Answer 9

well someone will probably answer it for you eventually anyway, i think it's :

1. x= 16/9
2. x = -12/7
3. km
4. x= -13

But it's been awhile so don't know if it's right... maybe you should figure out how I got those answers. Remember, you won't have yahoo answers for the test. :/

Answer 10

1. 16/9 = x
2. x = -12/7
3. km
4. x = -13