Three equations (I), (II) and (III) are given in each question. On the basis of these equations you have to decide the relation between 'x' , 'y' and 'z' and give answer
I. ` 7 x + 6 y = 110`
II. ` 4 x + 3y = 59`
III. ` x + z = 15`

Options

A. If ` x lt y = z `
B. If ` x le y lt z `
C. If ` x lt y gt z `
D. If ` x = y gt z `

Answer

To solve the given equations and determine the relationship between \( x \), \( y \), and \( z \), we will follow these steps:

### Step 1: Write down the equations
We have the following equations:
1. \( 7x + 6y = 110 \) (Equation I)
2. \( 4x + 3y = 59 \) (Equation II)
3. \( x + z = 15 \) (Equation III)

### Step 2: Solve for \( y \) in terms of \( x \) using Equation II
From Equation II:
\[ 4x + 3y = 59 \]
Rearranging gives:
\[ 3y = 59 - 4x \]
\[ y = \frac{59 - 4x}{3} \]

### Step 3: Substitute \( y \) in Equation I
Now, substitute \( y \) from the previous step into Equation I:
\[ 7x + 6\left(\frac{59 - 4x}{3}\right) = 110 \]
Multiply through by 3 to eliminate the fraction:
\[ 21x + 6(59 - 4x) = 330 \]
Distributing gives:
\[ 21x + 354 - 24x = 330 \]
Combining like terms:
\[ -3x + 354 = 330 \]
\[ -3x = 330 - 354 \]
\[ -3x = -24 \]
\[ x = 8 \]

### Step 4: Substitute \( x \) back to find \( y \)
Now that we have \( x = 8 \), substitute it back into the equation for \( y \):
\[ y = \frac{59 - 4(8)}{3} \]
\[ y = \frac{59 - 32}{3} \]
\[ y = \frac{27}{3} = 9 \]

### Step 5: Substitute \( x \) to find \( z \)
Now, use Equation III to find \( z \):
\[ x + z = 15 \]
Substituting \( x = 8 \):
\[ 8 + z = 15 \]
\[ z = 15 - 8 = 7 \]

### Step 6: Compare the values of \( x \), \( y \), and \( z \)
Now we have:
- \( x = 8 \)
- \( y = 9 \)
- \( z = 7 \)

### Step 7: Determine the relationship
From the values:
- \( x < y \) (since \( 8 < 9 \))
- \( y > z \) (since \( 9 > 7 \))

Thus, the relationship can be summarized as:
\[ x < y > z \]

### Final Answer
The final relationship is:
- \( x < y \) and \( y > z \)