1. ## New Inverse Variation (check this please)

When a seesaw is perfectly balanced, each person’s distance, “d,” from the fulcrum varies inversely as his weight, “w.” If a 120-pound person sits 5 feet from the seesaw fulcrum, where must a 150-pound person sit to balance the seesaw?

w = k/d

120 = k/5

120*5 = k

600 = k

... find d when w = 150...

d = 600/w
d = 600/150
d= 4

Thanks a lot!

2. Originally Posted by bobbyboy1111
When a seesaw is perfectly balanced, each person’s distance, “d,” from the fulcrum varies inversely as his weight, “w.” If a 120-pound person sits 5 feet from the seesaw fulcrum, where must a 150-pound person sit to balance the seesaw?

w = k/d <=== This is incorrect.

120 = k/5

120*5 = k

600 = k

... find d when w = 150...

d = 600/w
d = 600/150
d= 4

Thanks a lot!
Hi bobbyboy1111,

Your original equation is incorrect. However, it didn't affect the value of k.

It should have been $d=\frac{k}{w}$

I noticed that you fixed it when you solved for the distance a 150 lb person needs to sit from the fulcrum. Your answer is correct.