Question

Bob and Patrick want to meet with thier friend Sandy. They want to meet at Sandy's treehouse which his halfway between their houses. The distance from Bob's house to Sandy's 4x + 3. The distance of Patrick's house from Sandy's is 13x - 24. How far are Bob and Patrick's houses each from Sandy's?

Answer 1

By setting the equal distances from Bob's and Patrick's houses to Sandy's treehouse, we determined that both houses are 15 units from the treehouse after solving the equation.

Step-by-step explanation:

The distance from Bob's house to Sandy's treehouse is 4x + 3, and the distance from Patrick's house to Sandy's treehouse is 13x - 24. Since Sandy's treehouse is halfway between their houses, the distances from Bob's and Patrick's houses to the treehouse should be equal. Therefore, we can set up the equation 4x + 3 = 13x - 24 to find the value of x.

Now, we'll solve for x:

1. Subtract 4x from both sides: 3 = 9x - 24
2. Add 24 to both sides: 27 = 9x
3. Divide by 9: x = 3

Substitute x = 3 back into the expressions for the distances:

• Bob's distance to the treehouse: 4(3) + 3 = 12 + 3 = 15
• Patrick's distance to the treehouse: 13(3) - 24 = 39 - 24 = 15

So, both Bob and Patrick's houses are 15 units each from Sandy's treehouse.