To solve the system of equations for the value of [tex]\( y \)[/tex]:
[tex]\[
\begin{array}{l}
\frac{1}{3} x + \frac{1}{4} y = 1 \\
2 x - 3 y = -30
\end{array}
\][/tex]
we will perform the following steps:
### Step 1: Clear the fractions
First, clear the fractions in the first equation by finding a common multiple of the denominators (3 and 4).
Multiplying the entire first equation by 12 (which is the least common multiple of 3 and 4):
[tex]\[
12 \left( \frac{1}{3} x + \frac{1}{4} y \right) = 12 \cdot 1
\][/tex]
This simplifies to:
[tex]\[
4x + 3y = 12
\][/tex]
Now we have the system:
[tex]\[
\begin{array}{l}
4x + 3y = 12 \\
2x - 3y = -30
\end{array}
\][/tex]
### Step 2: Add the equations
Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[
(4x + 3y) + (2x - 3y) = 12 + (-30)
\][/tex]
Simplify by combining like terms:
[tex]\[
4x + 2x + 3y - 3y = 12 - 30
\][/tex]
This reduces to:
[tex]\[
6x = -18
\][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides by 6:
[tex]\[
x = \frac{-18}{6} = -3
\][/tex]
### Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]
Substitute [tex]\( x = -3 \)[/tex] into [tex]\( 4x + 3y = 12 \)[/tex]:
[tex]\[
4(-3) + 3y = 12
\][/tex]
[tex]\[
-12 + 3y = 12
\][/tex]
Add 12 to both sides to isolate [tex]\( 3y \)[/tex]:
[tex]\[
3y = 24
\][/tex]
Divide both sides by 3:
[tex]\[
y = \frac{24}{3} = 8
\][/tex]
Therefore, the value of [tex]\( y \)[/tex] is [tex]\( \boxed{8} \)[/tex].
