Answer :
To solve the given system of equations, we need to follow several steps to find the values of \( x \) and \( y \).
### Step 1: Write the System of Equations
The system of equations given is:
[tex]\[ \frac{x-2}{3} + \frac{y+1}{6} = 2 \][/tex]
[tex]\[ \frac{x-3}{4} - \frac{2y-1}{2} = 1 \][/tex]
### Step 2: Simplify the Equations
Let's clear the fractions to make the equations easier to work with.
For the first equation, multiply every term by the least common multiple (LCM) of the denominators, which is 6:
[tex]\[ 6 \left( \frac{x-2}{3} \right) + 6 \left( \frac{y+1}{6} \right) = 6 \cdot 2 \][/tex]
[tex]\[ 2(x - 2) + (y + 1) = 12 \][/tex]
[tex]\[ 2x - 4 + y + 1 = 12 \][/tex]
[tex]\[ 2x + y - 3 = 12 \][/tex]
[tex]\[ 2x + y = 15 \quad \text{(Equation 1)} \][/tex]
For the second equation, multiply every term by the LCM of the denominators, which is 4:
[tex]\[ 4 \left( \frac{x-3}{4} \right) - 4 \left( \frac{2y-1}{2} \right) = 4 \cdot 1 \][/tex]
[tex]\[ x - 3 - 2(2y - 1) = 4 \][/tex]
[tex]\[ x - 3 - 4y + 2 = 4 \][/tex]
[tex]\[ x - 4y - 1 = 4 \][/tex]
[tex]\[ x - 4y = 5 \quad \text{(Equation 2)} \][/tex]
### Step 3: Solve the System of Linear Equations
We now have:
[tex]\[ 2x + y = 15 \quad \text{(Equation 1)} \][/tex]
[tex]\[ x - 4y = 5 \quad \text{(Equation 2)} \][/tex]
First, let's solve Equation 2 for \( x \):
[tex]\[ x = 4y + 5 \][/tex]
Now, substitute \( x = 4y + 5 \) into Equation 1:
[tex]\[ 2(4y + 5) + y = 15 \][/tex]
[tex]\[ 8y + 10 + y = 15 \][/tex]
[tex]\[ 9y + 10 = 15 \][/tex]
[tex]\[ 9y = 5 \][/tex]
[tex]\[ y = \frac{5}{9} \][/tex]
### Step 4: Substitute \( y \) back into the Equation for \( x \)
Substitute \( y = \frac{5}{9} \) back into the equation \( x = 4y + 5 \):
[tex]\[ x = 4 \left( \frac{5}{9} \right) + 5 \][/tex]
[tex]\[ x = \frac{20}{9} + 5 \][/tex]
Convert the 5 to a fraction:
[tex]\[ x = \frac{20}{9} + \frac{45}{9} \][/tex]
[tex]\[ x = \frac{65}{9} \][/tex]
### Step 5: Write the Solution
The solution to the system of equations is:
[tex]\[ x = \frac{65}{9}, \quad y = \frac{5}{9} \][/tex]
Therefore, the values of \( x \) and \( y \) are:
[tex]\[ \boxed{\left( \frac{65}{9}, \frac{5}{9} \right)} \][/tex]
### Step 1: Write the System of Equations
The system of equations given is:
[tex]\[ \frac{x-2}{3} + \frac{y+1}{6} = 2 \][/tex]
[tex]\[ \frac{x-3}{4} - \frac{2y-1}{2} = 1 \][/tex]
### Step 2: Simplify the Equations
Let's clear the fractions to make the equations easier to work with.
For the first equation, multiply every term by the least common multiple (LCM) of the denominators, which is 6:
[tex]\[ 6 \left( \frac{x-2}{3} \right) + 6 \left( \frac{y+1}{6} \right) = 6 \cdot 2 \][/tex]
[tex]\[ 2(x - 2) + (y + 1) = 12 \][/tex]
[tex]\[ 2x - 4 + y + 1 = 12 \][/tex]
[tex]\[ 2x + y - 3 = 12 \][/tex]
[tex]\[ 2x + y = 15 \quad \text{(Equation 1)} \][/tex]
For the second equation, multiply every term by the LCM of the denominators, which is 4:
[tex]\[ 4 \left( \frac{x-3}{4} \right) - 4 \left( \frac{2y-1}{2} \right) = 4 \cdot 1 \][/tex]
[tex]\[ x - 3 - 2(2y - 1) = 4 \][/tex]
[tex]\[ x - 3 - 4y + 2 = 4 \][/tex]
[tex]\[ x - 4y - 1 = 4 \][/tex]
[tex]\[ x - 4y = 5 \quad \text{(Equation 2)} \][/tex]
### Step 3: Solve the System of Linear Equations
We now have:
[tex]\[ 2x + y = 15 \quad \text{(Equation 1)} \][/tex]
[tex]\[ x - 4y = 5 \quad \text{(Equation 2)} \][/tex]
First, let's solve Equation 2 for \( x \):
[tex]\[ x = 4y + 5 \][/tex]
Now, substitute \( x = 4y + 5 \) into Equation 1:
[tex]\[ 2(4y + 5) + y = 15 \][/tex]
[tex]\[ 8y + 10 + y = 15 \][/tex]
[tex]\[ 9y + 10 = 15 \][/tex]
[tex]\[ 9y = 5 \][/tex]
[tex]\[ y = \frac{5}{9} \][/tex]
### Step 4: Substitute \( y \) back into the Equation for \( x \)
Substitute \( y = \frac{5}{9} \) back into the equation \( x = 4y + 5 \):
[tex]\[ x = 4 \left( \frac{5}{9} \right) + 5 \][/tex]
[tex]\[ x = \frac{20}{9} + 5 \][/tex]
Convert the 5 to a fraction:
[tex]\[ x = \frac{20}{9} + \frac{45}{9} \][/tex]
[tex]\[ x = \frac{65}{9} \][/tex]
### Step 5: Write the Solution
The solution to the system of equations is:
[tex]\[ x = \frac{65}{9}, \quad y = \frac{5}{9} \][/tex]
Therefore, the values of \( x \) and \( y \) are:
[tex]\[ \boxed{\left( \frac{65}{9}, \frac{5}{9} \right)} \][/tex]
