`6x+py=21` ltbgt `px+5y=7`
If the above system of equations has infinitely many solutions, what is the value of `(p)/(q)` ?

Answer

To solve the problem, we need to determine the values of \( p \) and \( q \) such that the system of equations has infinitely many solutions. This occurs when the two equations represent the same line, meaning they are proportional.

Given the equations:

1. \( 6x + py = 21 \) (Equation 1)
2. \( px + 5y = 7 \) (Equation 2)

**Step 1: Rewrite the equations in a comparable form.**

We can express both equations in the form \( Ax + By = C \).

Equation 1 is already in that form:
- \( A_1 = 6 \), \( B_1 = p \), \( C_1 = 21 \)

For Equation 2:
- \( A_2 = p \), \( B_2 = 5 \), \( C_2 = 7 \)

**Step 2: Set up the proportionality condition.**

For the two equations to represent the same line, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

Substituting the values we have:
\[ \frac{6}{p} = \frac{p}{5} = \frac{21}{7} \]

**Step 3: Simplify the ratios.**

The ratio \( \frac{21}{7} = 3 \). Therefore, we can set up the equations:
1. \( \frac{6}{p} = 3 \)
2. \( \frac{p}{5} = 3 \)

**Step 4: Solve for \( p \) using the first equation.**

From \( \frac{6}{p} = 3 \):
\[ 6 = 3p \implies p = \frac{6}{3} = 2 \]

**Step 5: Solve for \( p \) using the second equation.**

From \( \frac{p}{5} = 3 \):
\[ p = 3 \times 5 = 15 \]

**Step 6: Find \( q \) using the value of \( p \).**

We can now find \( q \) using the proportionality condition:
From \( \frac{6}{p} = 3 \):
\[ \frac{6}{2} = 3 \implies q = 2 \]

**Step 7: Calculate \( \frac{p}{q} \).**

Now we have \( p = 15 \) and \( q = 2 \):
\[ \frac{p}{q} = \frac{15}{2} \]

Thus, the final answer is:
\[ \frac{p}{q} = 7.5 \]