If z varies as px + y, and if z = 3 when x = 1 and y = 2. Also if z = 5 then x =2 and y = 3, find the value of p.

Answer

To solve the problem step by step, we start by interpreting the given information and setting up equations based on the relationships provided.

### Step 1: Set up the relationship
We know that \( z \) varies as \( px + y \). This can be expressed mathematically as:
\[ z = k(px + y) \]
where \( k \) is a constant.

### Step 2: Use the first condition
According to the problem, when \( x = 1 \) and \( y = 2 \), \( z = 3 \). Plugging these values into the equation gives:
\[ 3 = k(p \cdot 1 + 2) \]
This simplifies to:
\[ 3 = k(p + 2) \quad \text{(Equation 1)} \]

### Step 3: Use the second condition
Next, we use the second condition where \( z = 5 \) when \( x = 2 \) and \( y = 3 \):
\[ 5 = k(p \cdot 2 + 3) \]
This simplifies to:
\[ 5 = k(2p + 3) \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( 3 = k(p + 2) \)
2. \( 5 = k(2p + 3) \)

We can express \( k \) from both equations:
From Equation 1:
\[ k = \frac{3}{p + 2} \]

From Equation 2:
\[ k = \frac{5}{2p + 3} \]

### Step 5: Set the equations for \( k \) equal to each other
Now we can set the two expressions for \( k \) equal to each other:
\[ \frac{3}{p + 2} = \frac{5}{2p + 3} \]

### Step 6: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 3(2p + 3) = 5(p + 2) \]

### Step 7: Expand both sides
Expanding both sides results in:
\[ 6p + 9 = 5p + 10 \]

### Step 8: Rearrange the equation
Now, rearranging the equation to isolate \( p \):
\[ 6p - 5p = 10 - 9 \]
\[ p = 1 \]

### Conclusion
Thus, the value of \( p \) is:
\[ \boxed{1} \]