The solution of the simultaneous equations `x/2+y/3=4` and `x+y=10` is give by

Options

A. (6,4)
B. (4,6)
C. (-6,4)
D. (6,-4)

Answer

To solve the simultaneous equations \( \frac{x}{2} + \frac{y}{3} = 4 \) and \( x + y = 10 \), we can follow these steps:

### Step 1: Rewrite the first equation
The first equation is \( \frac{x}{2} + \frac{y}{3} = 4 \). To eliminate the fractions, we can find the least common multiple (LCM) of the denominators 2 and 3, which is 6. We multiply the entire equation by 6:

\[ 6 \left(\frac{x}{2}\right) + 6 \left(\frac{y}{3}\right) = 6 \cdot 4 \]

This simplifies to:

\[ 3x + 2y = 24 \quad \text{(Equation 1)} \]

### Step 2: Write the second equation
The second equation is already in a simple form:

\[ x + y = 10 \quad \text{(Equation 2)} \]

### Step 3: Express \( y \) in terms of \( x \)
From Equation 2, we can express \( y \) in terms of \( x \):

\[ y = 10 - x \]

### Step 4: Substitute \( y \) in Equation 1
Now, substitute \( y = 10 - x \) into Equation 1:

\[ 3x + 2(10 - x) = 24 \]

### Step 5: Simplify the equation
Expanding the equation gives:

\[ 3x + 20 - 2x = 24 \]

Combine like terms:

\[ 3x - 2x + 20 = 24 \]

This simplifies to:

\[ x + 20 = 24 \]

### Step 6: Solve for \( x \)
Now, isolate \( x \):

\[ x = 24 - 20 \]

So,

\[ x = 4 \]

### Step 7: Substitute back to find \( y \)
Now that we have \( x \), we can find \( y \) using the expression we found earlier:

\[ y = 10 - x = 10 - 4 \]

Thus,

\[ y = 6 \]

### Final Solution
The solution to the simultaneous equations is:

\[ (x, y) = (4, 6) \]

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