How do you solve #2/3(x-7)=4/5#?

2 Answers
Burglar
Jul 5, 2016

The solution is #x=41/5#.

Explanation:

We can first apply the multiplication on the left side of the #=#

#2/3(x-7)=4/5#

#2/3*x-2/3*7=4/5#

We can continue calculating #2/3*7=14/3#

#2/3x-2/3*7=4/5#

#2/3x-14/3=4/5#

Now we want to isolate the term with the #x# and to do this we can add on both sides #14/3# in this way we can cancel the #-14/3# from the side where there is the term with the #x#

#2/3x-cancel(14/3)+cancel(14/3)=4/5+14/3#

#2/3x=4/5+14/3#

On the right side we can calculate #4/5+14/3#

#2/3x=4/5+14/3#

#2/3x=(3*4+14*5)/(3*5)#

#2/3x=(12+70)/(15)#

#2/3x=82/15#.

Finally we want to remove the coefficient in front of the #x#. We have that #x# is multiplied by #2# and divided by #3#, so we will divide by #2# and multiply by #3# both side of the equation

#2/3x=82/15#

#cancel(2/3)*cancel(3/2)x=82/15*3/2#

#x=82/15*3/2#

we can simplify #82/2=41# and #3/15=1/5#

#x=82/15*3/2=41/5#.

Tazwar Sikder
Sep 4, 2016

#x = (41) / (5)#

Explanation:

We have: #(2) / (3) (x - 7) = (4) / (5)#

Let's cross-multiply:

#=> 10 (x - 7) = 12#

Then, let's divide both sides of the equation by #2#:

#=> 5 (x - 7) = 6#

#=> 5 x - 35 = 6#

Adding #35# to both sides:

#=> 5 x = 41#

Finally, let's solve for #x# by dividing both sides by #5#:

#=> x = (41) / (5)#

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