Question

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

Answer 1

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

Answer 2

$\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides} \ \textgreater \ \frac{5}{4}b+7-7=\frac{7}{8}b+19-7 \ \textgreater \ Simplify$

$\frac{5}{4}b+7-7 \ \textgreater \ \mathrm{Add\:similar\:elements:}\:7-7=0 \ \textgreater \ \frac{5}{4}b$

$\frac{7}{8}b+19-7 \ \textgreater \ \mathrm{Add/Subtract\:the\:numbers:}\:19-7=12 \ \textgreater \ \frac{7}{8}b+12$

$\frac{5}{4}b=\frac{7}{8}b+12$

$\mathrm{Subtract\:}\frac{7}{8}b\mathrm{\:from\:both\:sides} \ \textgreater \ \frac{5}{4}b-\frac{7}{8}b=\frac{7}{8}b+12-\frac{7}{8}b \ \textgreater \ Simplify$

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

$\frac{5}{4}-\frac{7}{8} \ \textgreater \ \mathrm{Factor\:each\:denominator\:into\:its\:primes} \ \textgreater \ 4=2^2 \ \textgreater \ 8=2^3$

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

$\mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{5\cdot \:2}{8}-\frac{7}{8}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{2\cdot \:5-7}{8} \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:2\cdot \:5=10 \ \textgreater \ 10-7 \ \textgreater \ 3\ \textgreater \ \frac{3}{8}$
$\frac{3}{8}b$

$\frac{7}{8}b+12-\frac{7}{8}b \ \textgreater \ \mathrm{Add\:similar\:elements:}\:\frac{7}{8}b-\frac{7}{8}b=0 \ \textgreater \ 12 \ \textgreater \ \frac{3}{8}b=12$

$\mathrm{Multiply\:both\:sides\:by\:}8 \ \textgreater \ 8\cdot \frac{3}{8}b=12\cdot \:8 \ \textgreater \ \mathrm{Simplify} \ \textgreater \ 3b=96$

$\mathrm{Divide\:both\:sides\:by\:}3 \ \textgreater \ \frac{3b}{3}=\frac{96}{3} \ \textgreater \ Simplify \ \textgreater \ b=32$

Hope this helps!