If `x(x+y+z)=9, y(x+y+z)=16 and z(x+y+z)=144` then what is the value of x?

Options

A. `(9)/(5)`
B. `(9)/(7)`
C. `(9)/(13)`
D. `(16)/(3)`

Answer

To solve the problem, we have the following equations:

1. \( x(x + y + z) = 9 \)
2. \( y(x + y + z) = 16 \)
3. \( z(x + y + z) = 144 \)

Let's denote \( S = x + y + z \). Then we can rewrite the equations as:

1. \( xS = 9 \)
2. \( yS = 16 \)
3. \( zS = 144 \)

### Step 1: Express \( S \) in terms of \( x, y, z \)

From the first equation, we can express \( S \):

\[ S = \frac{9}{x} \]

From the second equation:

\[ S = \frac{16}{y} \]

From the third equation:

\[ S = \frac{144}{z} \]

### Step 2: Set the equations for \( S \) equal to each other

Since all three expressions are equal to \( S \), we can set them equal to each other:

\[ \frac{9}{x} = \frac{16}{y} = \frac{144}{z} \]

### Step 3: Cross-multiply to find relationships between \( x, y, z \)

From \( \frac{9}{x} = \frac{16}{y} \):

\[ 9y = 16x \quad \text{(1)} \]

From \( \frac{16}{y} = \frac{144}{z} \):

\[ 16z = 144y \quad \text{(2)} \]

From \( \frac{9}{x} = \frac{144}{z} \):

\[ 9z = 144x \quad \text{(3)} \]

### Step 4: Solve the equations

From equation (1):

\[ y = \frac{16}{9}x \]

Substituting \( y \) in equation (2):

\[ 16z = 144 \left( \frac{16}{9}x \right) \]
\[ 16z = \frac{2304}{9}x \]
\[ z = \frac{144}{9}x = 16x \]

Now we have \( y \) and \( z \) in terms of \( x \):

\[ y = \frac{16}{9}x, \quad z = 16x \]

### Step 5: Substitute \( y \) and \( z \) back into \( S \)

Now, substitute \( y \) and \( z \) into \( S \):

\[ S = x + y + z = x + \frac{16}{9}x + 16x \]
\[ S = x + \frac{16}{9}x + \frac{144}{9}x = x + \frac{160}{9}x = \frac{169}{9}x \]

### Step 6: Substitute \( S \) back into one of the original equations

Using \( S = \frac{9}{x} \):

\[ \frac{169}{9}x = \frac{9}{x} \]

Cross-multiplying gives:

\[ 169x^2 = 81 \]

### Step 7: Solve for \( x \)

\[ x^2 = \frac{81}{169} \]
\[ x = \frac{9}{13} \]

### Conclusion

Thus, the value of \( x \) is:

\[ \boxed{\frac{9}{13}} \]