Answer
$y=\dfrac{32}{7}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\dfrac{5}{16}y+\dfrac{3}{8}y=2+\dfrac{1}{4}y
,$ remove first the fraction by multiplying both sides by the $LCD.$ Then use the properties of equality to isolate the variable. Do checking of the solution.
$\bf{\text{Solution Details:}}$
The $LCD$ of the denominators, $\{
16,8,1,4
\},$ is $
16
$ since this is the least number that can be evenly divided (no remainder) by all the denominators. Multiplying both sides by the $LCD,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{5}{16}y+\dfrac{3}{8}y=2+\dfrac{1}{4}y
\\\\
16\left( \dfrac{5}{16}y+\dfrac{3}{8}y \right) =16\left( 2+\dfrac{1}{4}y \right)
\\\\
5y+6y=32+4y
.\end{array}
Using the properties of equality to isolate the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
5y+6y=32+4y
\\\\
5y+6y-4y=32
\\\\
7y=32
\\\\
y=\dfrac{32}{7}
.\end{array}
Checking: If $y=\dfrac{32}{7},$ then
\begin{array}{l}\require{cancel}
\dfrac{5}{16}y+\dfrac{3}{8}y=2+\dfrac{1}{4}y
\\\\
\dfrac{5}{16}\left( \dfrac{32}{7} \right)+\dfrac{3}{8}\left( \dfrac{32}{7} \right)=2+\dfrac{1}{4}\left( \dfrac{32}{7} \right)
\\\\
\dfrac{5}{\cancel{16}}\left( \dfrac{\cancel{16}(2)}{7} \right)+\dfrac{3}{\cancel8}\left( \dfrac{\cancel8(4)}{7} \right)=2+\dfrac{1}{\cancel4}\left( \dfrac{\cancel4(8)}{7} \right)
\\\\
\dfrac{10}{7}+\dfrac{12}{7}=2+\dfrac{8}{7}
\\\\
\dfrac{10}{7}+\dfrac{12}{7}=\dfrac{14}{7}+\dfrac{8}{7}
\\\\
\dfrac{22}{7}=\dfrac{22}{7}
\text{ (TRUE) }
.\end{array}
Hence, the solution is $
y=\dfrac{32}{7}
.$
