We are given the rational equation one over p plus one over q equals one over six. Our goal is to solve for q in terms of p. This equation involves reciprocals and is common in various mathematical contexts.
To solve for q, we need to isolate the term one over q. We do this by subtracting one over p from both sides of the equation. This maintains the balance of the equation and moves the p term to the right side.
To subtract the fractions on the right side, we need a common denominator. The common denominator for six and p is six p. We rewrite one sixth as p over six p and one over p as six over six p. Then we subtract the numerators to get p minus six over six p.
Now that we have one over q equals p minus six over six p, we can solve for q by taking the reciprocal of both sides. The reciprocal of one over q is q, and the reciprocal of the fraction is six p over p minus six. Therefore, q equals six p over p minus six.
When working with rational equations, we must consider domain restrictions. In our solution q equals six p over p minus six, the denominator p minus six cannot be zero, so p cannot equal six. Additionally, from the original equation, p cannot be zero because we have one over p. These restrictions ensure our solution is valid.
Let's verify our solution by substituting a specific value for p. We'll use p equals three. Substituting into our solution q equals six p over p minus six, we get q equals six times three over three minus six, which is eighteen over negative three, giving us q equals negative six. To verify, we substitute both values into the original equation: one over three plus one over negative six equals two sixths minus one sixth, which equals one sixth. This confirms our solution is correct.
Let's work through two more examples to solidify our understanding. First, with p equals twelve: q equals six times twelve over twelve minus six, which is seventy-two over six, giving us q equals twelve. Verification: one over twelve plus one over twelve equals two twelfths, which simplifies to one sixth. Second, with p equals negative three: q equals six times negative three over negative three minus six, which is negative eighteen over negative nine, giving us q equals two. Verification: one over negative three plus one over two equals negative two sixths plus three sixths, which equals one sixth. Both examples confirm our solution.
To summarize our solution: we found that q equals six p over p minus six. The domain restrictions are that p cannot equal zero or six. The solution method involved four key steps: first, isolating the term with q by subtracting one over p from both sides; second, finding a common denominator to combine the fractions on the right side; third, taking the reciprocal of both sides to solve for q; and fourth, checking domain restrictions to ensure our solution is valid. This approach is applicable to similar rational equations.