we will proceed to solve each case to determine the solution
we have the following expression
\(\frac{2x+3}{4}\)
case a) \(\frac{x}{2} +\frac{3}{4}\)
Multiply \(\frac{x}{2}\) by \(\frac{2}{2}\)
\(\frac{x}{2}*\frac{2}{2}=\frac{2x}{4}\)
so
\(\frac{x}{2} +\frac{3}{4}=\frac{2x}{4} +\frac{3}{4}=\frac{2x+3}{4}\)
\(\frac{2x+3}{4}=\frac{2x+3}{4}\)
therefore
the case a) is equivalent to the expression
case b) \(\frac{x}{2} -\frac{3}{4}\)
Multiply \(\frac{x}{2}\) by \(\frac{2}{2}\)
\(\frac{x}{2}*\frac{2}{2}=\frac{2x}{4}\)
so
\(\frac{x}{2} -\frac{3}{4}=\frac{2x}{4} -\frac{3}{4}=\frac{2x-3}{4}\)
\(\frac{2x-3}{4}
eq \frac{2x+3}{4}\)
therefore
the case b) is not equivalent to the expression
case c) \(\frac{3x}{4}+\frac{3}{4}\)
\(\frac{3x}{4} +\frac{3}{4}=\frac{3x+3}{4}\)
\(\frac{3x+3}{4}
eq \frac{2x+3}{4}\)
therefore
the case c) is not equivalent to the expression
case d) \(8x+24\)
Multiply \(8x+24\) by \(\frac{4}{4}\)
\((8x+24)*\frac{4}{4}=\frac{32x+96}{4}\)
\(\frac{32x+96}{4}
eq \frac{2x+3}{4}\)
therefore
the case d) is not equivalent to the expression
the answer is
The equivalent expression is \(\frac{x}{2} +\frac{3}{4}\)
