Question #fb844

1 Answer
Sep 22, 2017

#n=1#

Explanation:

First, we will convert all the mixed numbers (#1 2/3#) to improper fractions (#12/3#).

#-43/24+2 2/3n=-7/4(-3/2n+1)# to

#-43/24+8/3n=-7/4(-3/2n+1)#

Let's focus on the right side of the equation (#-7/4(-3/2n+1)#)

We will distribute parentheses/brackets with: #x(y+z)=xy+xz#

In this case, #x# represents #-7/4#, #y# represents #-3/2n# and #z# represents #1#. Therefore: #-7/4*(-3/2n)cancel+ -7/4*1#

We cancel out the #+# as #+ -# is equal to #-# so the #+# is redundant.
We will apply this to #-7/4*(-3/2n)-7/4*1#:

#-(-x)=x# and #+(-x)=-x#. Therefore, the equation will now be:

#7/4*3/2n-7/4*1#.

Now we will focus on #7/4*3/2n#, which is the same as #(7*3)/(4*2)n#.

#7*3=21# and #4*2=8# so therefore the equation will be #21/8n#.

The right side is just #7/4*1# which is already equal to #7/4#.
Combine the left and right side (and the left side of the whole equation) and you will get #-43/24+8/3n=21/8n-7/4#.

Add #43/24# to both sides.
#-43/24+8/3ncolor(blue)+color(blue)43/color(blue)24=21/8n-7/4color(blue)+color(blue)43/color(blue)24#
The left side would be #8/3n#.

Let's solve the right side: #21/8n-7/4+43/24#
First, we leave #21/8n# which is #(21n)/# aside for now.
Then, change the denominator of #-7/4# and #43/24# to #24# (Make sure to multiply #6# for the numerator of #-7/4#)

#-(7*6)/24+43/24# which is #(-6*7+43)/24#

#-6*7=-42# and #-42+43=1#.

Therefore, the whole equation would be equal to #8/3n=1/24+(21n)/8#

Now we subtract #(21n)/8# from both sides
#8/3n color(blue)-(color(blue)21color(blue)n)/color(blue)8=1/24+(21n)/8color(blue)-(color(blue)21color(blue)n)/color(blue)8#

which is equal to #n/24=1/24# so #n=1#.