Question

Can someone please help me with the following problem and explain how you got your answer?

5/4b + 7 = 7/8b + 19

b= ?

Answer 1

`\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides} \ \textgreater \ (5)/(4)b+7-7=(7)/(8)b+19-7 \ \textgreater \ Simplify`


`(5)/(4)b+7-7 \ \textgreater \ \mathrm{Add\:similar\:elements:}\:7-7=0 \ \textgreater \ (5)/(4)b`


`(7)/(8)b+19-7 \ \textgreater \ \mathrm{Add/Subtract\:the\:numbers:}\:19-7=12 \ \textgreater \ (7)/(8)b+12`


`(5)/(4)b=(7)/(8)b+12`


`\mathrm{Subtract\:}(7)/(8)b\mathrm{\:from\:both\:sides} \ \textgreater \ (5)/(4)b-(7)/(8)b=(7)/(8)b+12-(7)/(8)b \ \textgreater \ Simplify`


`(5)/(4)b-(7)/(8)b \ \textgreater \ \mathrm{Factor\:out\:common\:term\:}b\ \textgreater \ b\left((5)/(4)-(7)/(8)\right)`


`(5)/(4)-(7)/(8) \ \textgreater \ \mathrm{Factor\:each\:denominator\:into\:its\:primes} \ \textgreater \ 4=2^2 \ \textgreater \ 8=2^3`


`\mathrm{Find\:the\:least\:common\:denominator\:}2^3=8`


`\mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ (5\cdot \:2)/(8)-(7)/(8)`


`\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: (a)/(c)\pm (b)/(c)=(a\pm \:b)/(c)`

`(2\cdot \:5-7)/(8) \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:2\cdot \:5=10 \ \textgreater \ 10-7 \ \textgreater \ 3\ \textgreater \ (3)/(8)`

`(3)/(8)b`


`(7)/(8)b+12-(7)/(8)b \ \textgreater \ \mathrm{Add\:similar\:elements:}\:(7)/(8)b-(7)/(8)b=0 \ \textgreater \ 12 \ \textgreater \ (3)/(8)b=12`


`\mathrm{Multiply\:both\:sides\:by\:}8 \ \textgreater \ 8\cdot (3)/(8)b=12\cdot \:8 \ \textgreater \ \mathrm{Simplify} \ \textgreater \ 3b=96`


`\mathrm{Divide\:both\:sides\:by\:}3 \ \textgreater \ (3b)/(3)=(96)/(3) \ \textgreater \ Simplify \ \textgreater \ b=32`

Hope this helps!

Answer 2

First you subtract 7 minus 19 and 7/8b minus 5/4b:
5/4b+7=7/8b+19
-7/8b -7/8b

5/4b+7=7/8b+19
-7 -7
3/8b=12
Then you divide 12 by 3/8:
b=32