Let $A$ be the number of days it takes Andrew to paint the house alone, and let $B$ be the number of days it takes Bailey to paint the house alone.

From the problem, we know that Andrew can paint the house 6 times as fast as Bailey. This means: $A = \frac{B}{6}$

When Andrew and Bailey work together, the combined rate of work is the sum of their individual rates. Andrew's rate is $\frac{1}{A}$ of the house per day, and Bailey's rate is $\frac{1}{B}$ of the house per day.

Together, they complete $\frac{1}{A} + \frac{1}{B}$ of the house per day, and it took them 7 days to finish the house, so their combined rate is: $\frac{1}{A} + \frac{1}{B} = \frac{1}{7}$

Now, substitute $A = \frac{B}{6}$ into the equation: $\frac{1}{\frac{B}{6}} + \frac{1}{B} = \frac{1}{7}$ This simplifies to: $\frac{6}{B} + \frac{1}{B} = \frac{1}{7}$ $\frac{7}{B} = \frac{1}{7}$

Now, solve for $B$: $B = 49$

So, it would take Bailey 49 days to paint the house alone.

Now, find $A$ using $A = \frac{B}{6}$: $A = \frac{49}{6} \approx 8.17 \text{ days}$

Thus, it would take Andrew approximately 8.17 days to paint the house alone.

Summary:

• Bailey would take 49 days to paint the house alone.
• Andrew would take about 8.17 days to paint the house alone.

Do you want more details or have any questions?

Here are 5 related questions:

1. How would the time change if Andrew painted 5 times as fast as Bailey instead of 6?
2. If Bailey improved and could paint in 40 days, how long would it take them together?
3. What if Andrew and Bailey alternated days instead of working together every day?
4. How long would it take if another person joined them and painted at the same speed as Bailey?
5. If Andrew painted 50% faster than before, how would it affect their combined time?

Tip: When combining rates, always sum the fractions of work completed per unit of time for each individual.