Algebra 2 help pleeeeese?

Question

i did this problem like 10 times, and im not sure if im doing it correctly, can u show me the next step

1/5 (3n+5)= -(n/2 + 10)

what do u do with the - ?

does it become -5 when u try to remove the fraction or does it just make the following opposite?

(that kinda doesnt make sence so if needed ignore this just plz show me the next step or 2)

Answer 1

The negative is simply -1.  The next step I would take would be to distribute what's outside each set of parenthesis.

1/5 (3n+5)= -1 (n/2 + 10)

(3/5)n + 1 = -n/2 - 10

Multiply all terms by 10 to remove the fractions:

6n + 10 = -5n - 100

6n + 10 - 10 = -5n - 100 - 10

6n = -5n - 110

6n + 5n = -5n - 110 + 5n

11n = -110

11n/11 = -110/11

n = -10

Answer 2

You need to distribute the negative into both terms.

So, the right side will look like this: -n/2-10

Distribute the left side as well, 3/5n+1

So now you have 3/5n+1 = -n/2-10

You should be able to complete the problem from this point.

Answer 3

Hi.. you have to multiply the (-) into the bracket

1/5(3n+5) = -n/2 -10    then 10 becomes a negative because a negative wins over a positive now your equation is

1/5 times 3n + 1/5 times 5 = -n/2 -10

bring all the letters to one side and numbers on the other

1/5 times (1/5x5) = 10 = -3n-n/2

Now i dont have a calculator with a fraction button so I hope you can get it from there..  

Answer 4

it does become -5 when u remove the fraction.

i have solved it in following way:

1/5 (3n+5) = -(n/2+10),

6n+10 = -5n-100

11n = -110

therefore n= -10

hope this helps!!!!!!!!