q: in the last problem, i would have liked to see how the n was isolated. i did not follow the math on the reciprocal to isolate n.

a: not to worry! let's take a look: 

cathy's salary is 3/7 of nora's salary and (cathy's salary) is also 5/4 of teresa's salary.

c = (3/7)n

c is also equal to (5/4)t

so we can set:

(3/7)n = (5/4)t

now, i isolate n by multiplying both sides of the equation by (7/3), the reciprocal of 3/7.

(7/3)(3/7)n = (7/3)(5/4)t

(7/3)(3/7) = 1 so (7/3)(3/7)n just equals n.

n = (7/3)(5/4)t =

(35/12)t

q: the lesson doesn't really explain why we multiply by the reciprocal to get the answer in the second problem. i am very confused.

a: happy to help! this is just an extension of one of the rules of working with algebraic expressions: when trying to isolate a variable, you can perform a number of operations to both sides of an equation, but they have to be done to both sides.

so, for example, say i have this equation:

x/2 = 4

if i want to get x alone, i multiply it by 2. but i also have to do the same to the other side:

x/2 = 4
2x/2 = 2 * 4
x = 8

now, let's try that with this equation:

2x/7 = y

first, i multiply out that 7:

(7 * 2x)/7 = 7 * y
2x = 7y

then, i divide out the 2:

2x/2 = 7y/2
x = 7y/2

look again at my original and final equations:

2x/7 = y
x = 7y/2

in order to isolate x, i multiplied y by the reciprocal of the fraction on the left side. that's just the easier, faster way to do it. it's shorthand for the same process.

in the lesson video, there were two cases of expressions like that:

300 = 2r/7
7 * 300 = 2r
7 * 300 / 2= r

and here's the other:

c = 3n/7
7c = 3n
7c/3 = n