SOLUTION: Hi, I can't figure out this problem, could you help? {{{ a/(b+(c/d))=c }}} How do you solve for c? thanks.

Algebra ->  Expressions-with-variables -> SOLUTION: Hi, I can't figure out this problem, could you help? {{{ a/(b+(c/d))=c }}} How do you solve for c? thanks.      Log On


   

Question 106453: Hi, I can't figure out this problem, could you help?
+a%2F%28b%2B%28c%2Fd%29%29=c+
How do you solve for c? thanks.
Found 2 solutions by Earlsdon, solver91311:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for c:
a%2F%28b%2B%28c%2Fd%29%29+=+c First multiply both sides by the denominator of the left side.
a+=+c%28b%2B%28c%2Fd%29%29 Now simplify the contents of the parentheses.
a+=+c%28%28bd%2Bc%29%2Fd%29 Multiply both sides by d.
ad+=+c%28bd%2Bc%29 Apply the distributive property on the right side.
ad+=+cbd%2Bc%5E2 Now rewrite this in the standard form of a quadratic equation in c.
c%5E2%2B%28bd%29c-ad+=+0 Now you can use the quadratic formula to solve for c.
c+=+%28-bd%2B-sqrt%28%28bd%29%5E2-4%28-ad%29%29%29%2F2
c+=+%28-bd%2B-sqrt%28%28bd%29%5E2%2B4ad%29%29%2F2

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
First thing is to get all of the terms containing 'c' on one side of the equation:

+a%2F%28b%2B%28c%2Fd%29%29=c+

a=c%28b%2B%28c%2Fd%29%29

Now if we multiply both sides by 'd', we can get rid of that pesky denominator on the right.

ad=c%28bd%2Bc%29

Distribute 'c'

ad+=+bdc+%2B+c%5E2

And rearrange to standard quadratic form:

c%5E2+%2B+bdc+-ad+=+0

now solve for 'c' using the quadratic formula:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

except that you would substitute 'c' for 'x'; 'bd' for 'b'; 1 for a; and 'ad' for 'c', thus:

c+=+%28-bd+%2B-+sqrt%28+%28bd%29%5E2-4%2Aad+%29%29%2F%282%29+

General hint: The variable for which you were solving existed in the numerator on one side of the equation, and in the denominator on the other side. That guaranteed that you were going to be dealing with a quadratic.