Sure! Let's solve the equation step-by-step:
Given equation:
[tex]\[ 2(3y + 5) = 3\left(5y + \frac{1}{3}\right) \][/tex]
1. Distribute the constants on both sides of the equation:
On the left side:
[tex]\[ 2 \cdot (3y + 5) = 2 \cdot 3y + 2 \cdot 5 = 6y + 10 \][/tex]
On the right side:
[tex]\[ 3 \cdot \left(5y + \frac{1}{3}\right) = 3 \cdot 5y + 3 \cdot \frac{1}{3} = 15y + 1 \][/tex]
The equation now becomes:
[tex]\[ 6y + 10 = 15y + 1 \][/tex]
2. Move all the terms involving [tex]\( y \)[/tex] to one side and the constant terms to the other side:
Subtract [tex]\( 15y \)[/tex] from both sides:
[tex]\[ 6y + 10 - 15y = 15y + 1 - 15y \][/tex]
[tex]\[ 6y - 15y + 10 = 1 \][/tex]
[tex]\[ -9y + 10 = 1 \][/tex]
Now, subtract 10 from both sides:
[tex]\[ -9y + 10 - 10 = 1 - 10 \][/tex]
[tex]\[ -9y = -9 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
Divide both sides of the equation by [tex]\(-9\)[/tex]:
[tex]\[ y = \frac{-9}{-9} \][/tex]
[tex]\[ y = 1 \][/tex]
Therefore, the solution is [tex]\(y = 1\)[/tex].
The solution is [tex]\( \boxed{1} \)[/tex].
