How do you solve #-32- 4n = 5( n - 1)#?

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Answer 1 · 2018-05-14T18:51:00

#n = -3#

#-32 - 4n = 5(n - 1)#

First, distribute 5 to (n -1), per PEMDAS. You should now have:

#-32 - 4n = 5n - 5#

We want to negate the lowest variable in order to solve for n. Add 4n to each side to negate -4n. You should now have:

#-32 = 9n - 5#

Add 5 to each side to negate -5.

#-27 = 9n#

Divide by 9 to isolate for n.

#-27/9# = #-3# = #n#

#n# = #-3#

Answer 2 · 2018-05-14T18:57:57

#n = -3#

To solve for the variable #n# in the equation #-32-4n=5(n-1)

Begin by using the distributive property to eliminate the parenthesis.

#-32 -4n =5(n-1)

#-32 - 4n = 5n - 5#

Now use the additive inverse to place the variable terms on the same side of the equation.

#-32 - 4n -5n = cancel(5n) - 5 cancel (-5n)#

#-32 -9n = -5#

Now use the additive inverse to place the numeric terms on the same side of the equation.

#cancel(-32) -9n cancel (+32) = -5 +32#

#-9n = 27#

Use the multiplicative inverse to isolate the variable.

#((cancel-9)n)/(cancel(-9)) = 27/-9#

#n = -3#