How do you solve #(b-7)/4=(b+9)/7#?

1 Answer
seol
Apr 21, 2018

#85/3#

Explanation:

Common denominator: LCM of 4 and 7 is 28, or #4xx7#
Therefore, we can rewrite the equation as the following:
#(7(b-7))/28=(4(b+9))/28# which simplifies to #(7b-49)/28=(4b+36)/28#

Now let's separate the variables. Rewrite again:
#(7b)/28-49/28=(4b)/28+36/28#

Put the like terms with each other.
#(7b)/28-(4b)/28=49/28+36/28#

Take out the denominator since they're all equivalent anyway.
#7b-4b=49+36#
Simplify: #3b=85#

Divide to isolate variable: #b=85/3#

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