Strange Ratio problem

[Jason_04]

let b:c=7 and a:b=3. Find (a-b) = (b-c).
Came across this problem, and I'm not quite sure how to go about solving it.

 

[Romsek]

\(\displaystyle b:c=7 \text{ means the same thing as }\dfrac b c = 7\)

See if that helps any.

 

[Dr.Peterson]

let b:c=7 and a:b=3. Find (a-b) = (b-c).
Came across this problem, and I'm not quite sure how to go about solving it.

That doesn't make sense. Do you perhaps mean, Find a, b, and c if (a-b) = (b-c)?

 

[Romsek]

That doesn't make sense. Do you perhaps mean, Find a, b, and c if (a-b) = (b-c)?

OP almost certainly means find a, b, and c, which satisfy the two proportions and the one equality

[MATH]b:c = 7\\ b = 7c\\ a:b = 3\\ a = 3b = 21c\\ (a-b)= (b-c)\\ (21c - 7c) = (7c- c)\\ 14c = 6c[/MATH]
There is only one value of \(\displaystyle c\) that satisfies the above equation for c.
Then use the given proportions to solve for a and b.