Answer :
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].
[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].
