If ` 3 x + 5y + 7 z = 49 and 9 x + 8y + 21 z = 126 ` , then what is the value of y ?

Options

A. A)4
B. B)2
C. C)3
D. D)5

Answer

To solve the equations \(3x + 5y + 7z = 49\) and \(9x + 8y + 21z = 126\) for the value of \(y\), we can follow these steps:

### Step 1: Write down the equations
We have the following two equations:
1. \(3x + 5y + 7z = 49\) (Equation 1)
2. \(9x + 8y + 21z = 126\) (Equation 2)

### Step 2: Manipulate the equations
Notice that the second equation can be simplified by observing that it is a multiple of the first equation. Specifically, if we multiply the first equation by 3, we can eliminate \(x\) and \(z\) when we subtract it from the second equation.

Multiply Equation 1 by 3:
\[ 3(3x + 5y + 7z) = 3(49) \]
This gives us:
\[ 9x + 15y + 21z = 147 \quad \text{(Equation 3)} \]

### Step 3: Subtract Equation 2 from Equation 3
Now, we subtract Equation 2 from Equation 3:
\[ (9x + 15y + 21z) - (9x + 8y + 21z) = 147 - 126 \]
This simplifies to:
\[ 15y - 8y = 21 \]
Thus, we have:
\[ 7y = 21 \]

### Step 4: Solve for \(y\)
Now, divide both sides by 7:
\[ y = \frac{21}{7} = 3 \]

### Conclusion
The value of \(y\) is \(3\).