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7-(-y-5)= 2(y+3)-6(y+1)
want to know if I should distributive only the neg. sign to the -y or the 7
at the beginning of the problem, I came up with 5 over 11y
but text book states -12 over 5
The - sign should be distributive only to -y &-5 and not 7
The solution goes like this : 7-(-y-5) = 2(y+3)-6(y+1)
7+y+5 = 2y+6-6y-6
y+12 = -4y+0
y+4y = -12
5y = -12 or y = -12/5
You can
put this solution on YOUR website!Question:
7-(-y-5)= 2(y+3)-6(y+1)
want to know if I should distributive only the neg. sign to the -y or the 7
at the beginning of the problem, I came up with 5 over 11y
but text book states -12 over 5
Answer:
7-(-y-5)= 2(y+3)-6(y+1)
For solving this you have to remove the parenthesis first.
Here negative sign is distributive only to -y and -5
Hope you are familiar with multiplication of signs...
here -(-y) = +y
-(-5) = +5
2(y) = 2y
2(3) = 6
-6(y) = -6y
-6(1) = -6
so 7-(-y-5)= 2(y+3)-6(y+1)
==> 7 + y + 5 = 2y + 6 - 6y - 6
==> 7 + 5 + y = 2y - 6y + 6 - 6
==> 12 + y = -4y
Add 4y on both sides.....
==> 12 + y + 4y = -4y + 4y
==> 12 + 5y = 0
Subtract 12 from both sides....
==> 12 + 5y - 12 = 0 - 12
==> 5y = -12
Divide both sides by 5
==> 5y/5 = -12/5
==>
Hope you found the explanation useful.
Regards.
Praseena.
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