I'm not sure about what answer to these Algebra problems are. Can you help?

Question

School has just begun and i'm a little shaky at these problems. i would ask my teacher, but the assignments due when i walk in. help?

-4(2n-5)=-8n-20

4x-2(3+2x)=-6

i think the first one is no real value and the second one is all real numbers. im not quite sure. could anyone please clarify?

Answer 1

1st= -8n+20=-8n-20

=40

2nd=4x-6-4x=-6

=0

Answer 2

you're correct for both:

when you end up with something that looks like 0 = 40, there is no solution

when you end up with something tht looks like 6 = 6, the solution is all real numbers (the variable cancels)

Answer 3

-4(2n-5)=-8n-20

-8n+20=-8n-20 (multiply -4 by everything)

+8n-20=+8n-20 (get the variable to one side/number to the other)

0=40

False

4x-2(3+2x)=-6

4x-6-4x=-6 (multiply -2 by everything in parenthesis)

4x-6-4x+6=-6+6 (add 6 to both sides)

4x-4x=0

0=0

True

Answer 4

first one is false

second one indeterminate.

Answer 5

-4(2n-5)=-8n-20

-8n+20=-8n-20

0n=-40  undefined  

4x-2(3 +2x)=-6

4x-6-4x=-6

0x=0

x=0

Answer 6

-4(2n-5)=-8n-20

-8n+20=-8n-20

     +20      +20

-8n+40=-8n

+8n      +8n

40=0  (false)

4x-2(3+2x)=-6

4x-6-4x=-6

(4x-4x=0)

all thats left is..

-6=-6 (true)