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Can anyone help me solve this equation? $$5.4b-14=8b+38$$

I got $b=15$ remainder $5$

but I am not sure if this is correct.

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Don't you think that a remainder is somewhat out of place in the answer to a problem in which the input coefficients are decimal fractions? 15 remainder 5 is not a number. Citing a remainder only ever makes sense if we know what number was divided by. For instance $10 \over 3$ is $3$, remainder $1$. Since we know that we divided by $3$ (it is plainly stated before the remainder) then "remainder $1$" tells us that the fractional part is $1/3$. That is, ${10\over 3} = {3 {1\over 3}}$ If we don't know what we divided by, it could mean anything. – Kaz Apr 25 '12 at 0:25

That isn't correct. I'm not sure why you expressed your answer as "$15$ remainder $5$"..

Here's one way to solve it. I suspect you made a sign error somewhere.

You start with $$ {5.4 b} -14 = {8b}+38$$ Now subtract $5.4b$ from both sides $$ {5.4 b} -14 -5.4b= {8b}+38-5.4b. $$ Simplify $$ -14 = {2.6b}+38. $$ Now subtract $38$ from both sides to obtain $$ -52=2.6b, $$ or $$ 2.6b=-52. $$ Finally, divide both sides by $2.6$ and simplify $$\eqalign{ b&={-52\over2.6}\cr &={-52\cdot 10\over 26}\cr &=-20. } $$

Note you can check your answer: $$ 5.4\cdot(-20)-14=-108-14=-122, \ \ \text{and} \ \ 8(-20)+38=-160+38=-122. $$

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$5.4 b-14 = 8 b+38$ ${}{}{}{}$ | +14 | -8b

$-2.6 b = 52$ | /(-2.6)

$b = -20$

(In case $b$ is just a normal variable. If $b$ means something else here, disregard this answer.)

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A little unsolicited advice to help you write good answers: Possibly you learned this way to keep track of solving an equation, but it seems a little nonstandard to me. I'm sure people can puzzle out what it means, but it would be most helpful for the OP if you included three more lines explicitly writing out what you mean. Cheers! – rschwieb Jun 27 '12 at 12:01

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