Can the following equation hold simultaneusly? If yes, state the values of x and y
` " " (x)/(2) +( 5y )/( 3) = 12`
` " " 0.7x - 0.3y = 1 and 1.25 x = 4 + ( y)/(6)`

Answer

To determine if the given equations can hold simultaneously and to find the values of \( x \) and \( y \), we will follow these steps:

### Step 1: Simplify the First Equation
The first equation is:
\[ \frac{x}{2} + \frac{5y}{3} = 12 \]
To eliminate the fractions, we can multiply the entire equation by 6 (the least common multiple of 2 and 3):
\[ 6 \left( \frac{x}{2} \right) + 6 \left( \frac{5y}{3} \right) = 6 \cdot 12 \]
This simplifies to:
\[ 3x + 10y = 72 \quad \text{(Equation 1)} \]

### Step 2: Simplify the Second Equation
The second equation is:
\[ 0.7x - 0.3y = 1 \]
To eliminate the decimals, we can multiply the entire equation by 10:
\[ 10(0.7x) - 10(0.3y) = 10 \cdot 1 \]
This simplifies to:
\[ 7x - 3y = 10 \quad \text{(Equation 2)} \]

### Step 3: Simplify the Third Equation
The third equation is:
\[ 1.25x = 4 + \frac{y}{6} \]
We can rewrite \( 1.25 \) as \( \frac{125}{100} \) or \( \frac{5}{4} \):
\[ \frac{5}{4}x = 4 + \frac{y}{6} \]
To eliminate the fraction, multiply the entire equation by 12 (the least common multiple of 4 and 6):
\[ 12 \left( \frac{5}{4}x \right) = 12 \cdot 4 + 12 \left( \frac{y}{6} \right) \]
This simplifies to:
\[ 15x = 48 + 2y \quad \text{(Equation 3)} \]

### Step 4: Rearranging Equation 3
Rearranging Equation 3 gives:
\[ 15x - 2y = 48 \quad \text{(Equation 3)} \]

### Step 5: Solve the System of Equations
We now have the following system of equations:
1. \( 3x + 10y = 72 \)
2. \( 7x - 3y = 10 \)
3. \( 15x - 2y = 48 \)

Let's solve Equations 1 and 2 first.

From Equation 1, express \( y \) in terms of \( x \):
\[ 10y = 72 - 3x \implies y = \frac{72 - 3x}{10} \]

Substituting this expression for \( y \) into Equation 2:
\[ 7x - 3\left(\frac{72 - 3x}{10}\right) = 10 \]
Multiply through by 10 to eliminate the fraction:
\[ 70x - 3(72 - 3x) = 100 \]
Distributing:
\[ 70x - 216 + 9x = 100 \]
Combine like terms:
\[ 79x - 216 = 100 \]
Add 216 to both sides:
\[ 79x = 316 \]
Dividing by 79 gives:
\[ x = \frac{316}{79} = 4 \]

### Step 6: Substitute \( x \) back to find \( y \)
Now substitute \( x = 4 \) back into Equation 1 to find \( y \):
\[ 3(4) + 10y = 72 \]
This simplifies to:
\[ 12 + 10y = 72 \]
Subtract 12 from both sides:
\[ 10y = 60 \]
Dividing by 10 gives:
\[ y = 6 \]

### Conclusion
The values of \( x \) and \( y \) that satisfy all three equations are:
\[ x = 4, \quad y = 6 \]

### Verification
To verify, substitute \( x = 4 \) and \( y = 6 \) back into all three original equations to ensure they hold true.