Certainly! Let's solve the given equation step by step.
The initial equation is:
[tex]\[
\frac{2}{5}\left(\frac{1}{2} y+20\right)-\frac{4}{5}=\frac{9}{20}(2 y-1)
\][/tex]
First, we distribute on both sides. This is already done for us, resulting in:
[tex]\[
\frac{1}{5} y + 8 - \frac{4}{5} = \frac{9}{10} y - \frac{9}{20}
\][/tex]
The next step is to combine the like terms on the left side of the equation.
On the left side, combine [tex]\(\frac{1}{5} y\)[/tex] and the constants:
[tex]\[
\frac{1}{5} y + 8 - \frac{4}{5}
\][/tex]
Combine the constants:
[tex]\[
8 - \frac{4}{5}
\][/tex]
To subtract [tex]\(\frac{4}{5}\)[/tex] from 8, first express 8 as a fraction with a denominator of 5:
[tex]\[
8 = \frac{40}{5}
\][/tex]
Now:
[tex]\[
\frac{40}{5} - \frac{4}{5} = \frac{36}{5}
\][/tex]
Hence, the left side simplifies to:
[tex]\[
\frac{1}{5} y + \frac{36}{5}
\][/tex]
Now rewrite the equation with the simplified left side:
[tex]\[
\frac{1}{5} y + \frac{36}{5} = \frac{9}{10} y - \frac{9}{20}
\][/tex]
Thus, the step after using the distributive property is to combine the like terms on the left side of the equation.
