Given that , `5=(5W+2omega)/(5+2)` and `5.1=(7W+3omega)/(7+3)`, find W and `omega`

Options

A. `W=3,omega=10`
B. `W=10,omega=3`
C. `W=-10,omega=3`
D. `W=3,omega=-10`

Answer

To solve the given simultaneous linear equations, we start with the equations provided:

1. \( 5 = \frac{5W + 2\omega}{5 + 2} \)
2. \( 5.1 = \frac{7W + 3\omega}{7 + 3} \)

### Step 1: Simplify the equations

First, simplify the denominators in both equations.

For the first equation:
\[ 5 + 2 = 7 \]
So, we rewrite the first equation as:
\[ 5 = \frac{5W + 2\omega}{7} \]

For the second equation:
\[ 7 + 3 = 10 \]
So, we rewrite the second equation as:
\[ 5.1 = \frac{7W + 3\omega}{10} \]

### Step 2: Clear the denominators

Now, we will multiply both sides of each equation by the denominators to eliminate them.

For the first equation:
\[ 5 \times 7 = 5W + 2\omega \]
This simplifies to:
\[ 35 = 5W + 2\omega \quad \text{(Equation 1)} \]

For the second equation:
\[ 5.1 \times 10 = 7W + 3\omega \]
This simplifies to:
\[ 51 = 7W + 3\omega \quad \text{(Equation 2)} \]

### Step 3: Rearrange the equations

Now we have the two equations:
1. \( 5W + 2\omega = 35 \)
2. \( 7W + 3\omega = 51 \)

### Step 4: Make coefficients of \(\omega\) equal

To eliminate \(\omega\), we can make the coefficients of \(\omega\) in both equations equal. We can multiply the first equation by 3 and the second equation by 2.

Multiplying the first equation by 3:
\[ 3(5W + 2\omega) = 3(35) \]
This gives:
\[ 15W + 6\omega = 105 \quad \text{(Equation 3)} \]

Multiplying the second equation by 2:
\[ 2(7W + 3\omega) = 2(51) \]
This gives:
\[ 14W + 6\omega = 102 \quad \text{(Equation 4)} \]

### Step 5: Subtract the equations

Now, we will subtract Equation 4 from Equation 3 to eliminate \(\omega\):
\[ (15W + 6\omega) - (14W + 6\omega) = 105 - 102 \]
This simplifies to:
\[ 15W - 14W = 3 \]
Thus, we have:
\[ W = 3 \]

### Step 6: Substitute \(W\) back to find \(\omega\)

Now that we have \(W\), we can substitute \(W = 3\) back into either Equation 1 or Equation 2 to find \(\omega\). We will use Equation 1:
\[ 5(3) + 2\omega = 35 \]
This simplifies to:
\[ 15 + 2\omega = 35 \]
Now, subtract 15 from both sides:
\[ 2\omega = 35 - 15 \]
\[ 2\omega = 20 \]
Now, divide by 2:
\[ \omega = 10 \]

### Final Solution

Thus, the values of \(W\) and \(\omega\) are:
\[ W = 3, \quad \omega = 10 \]