How do you solve for #J# in #(CE)/2 = (3JQ)/7#?

1 Answer
Michelle
Jul 5, 2017

# J = (7CE)/(6Q)#

Explanation:

1. Cross multiply.

Multiply the numerator of the first fraction by the denominator of the second fraction, and set that equal to the numerator of the second fraction times the denominator of the first fraction. For example:

#a/b = c/d#

#ad = bc#

#(CE)/2 = (3JQ)/7#

#7(CE) = 3JQ(2)#

#7CE = 6JQ#

Commutative property:

#7CE = 6QJ#

2. To isolate J, divide both sides by 6Q.

#(7CE)/(6Q) = (cancel(6Q)J)/cancel(6Q)#

#(7CE)/(6Q) = J#

You've now solved for J. Hope this helps!

Related questions
Impact of this question
1117 views around the world
You can reuse this answer
Creative Commons License