"

q: what's the algebraic solution for the first practice problem in the video?

a: i'm going to say upfront that this problem is really hard to solve algebraically in a short amount of time, which is why it's included in the ""picking numbers"" video. but, let's walk through the algebra anyway :)

the vendor is selling some number of product a for $6 each and some number of product b for $21 each. we know the prices but not the respective amounts of a and b.

we need to account for the fact that a and b are different prices and may have different quantities. so we can't just work with the variables q and t. we need to introduce new variables a (number of product a sold) and b (number product b sold).

the a + b is total products sold and b/(a + b) is the percent of products sold that are b.

1) q/100 = b/(a + b)

because q is the percentage of products sold that are b

2) 21b/(6a + 21b) = t/100

because 21b is the revenue from b and 6a + 21b is the total revenue, so 21b/(6a + 21b) is the percentage of revenue from b.

we can take the inverse of both sides of 2) and solve for a/b:

(6a + 21b)/21b = 100/t

(6/21)(a/b) + 1 = 100/t

(a/b) = (7/2)(100 − t)/t = (700 − 7t)/2t

now take the inverse of equation 1)

100/q = (a + b)/b

100/q = (a/b) + 1

now substitute (700 − 7t)/2t for (a/b) and solve for q. you should (eventually) get:

q = 200t(700 − 5t) and divide everything by 5 to get:

40t(140 − t)

as you can see, plugging in is the way to go here!

q: what's the algebraic solution for the second practice problem in the video?

a: the key to this problem is that we need to use multipliers, discussed in these two videos: percent increases and decreases and sequential percent changes.   using multipliers for percent changes is a very important idea to have mastered for the test

the original price of the shoes is h and the dress is d = 5h.

shoe price increased by 50%. so the price became (1.5)h

dress price increased 40%. so the price became (1.4)5h = 7h.

so the combined price after the increase but before the discount is 1.5h + 7h = 8.5h

the discount is 30%, so roberta paid (1 − .3)(8.5)h = .7(8.5)h = 5.95h

as you can see, it's not exactly impossible (especially compared to the other practice problems) to figure this one out algebraically. but working through that series of percentage multipliers increases the chances that we're going to make a small mistake that leads us to an incorrect answer. additionally, it requires us to set up and simplify the equation in a particular way, when there are in fact many ways we could present the amount roberta paid in terms of d, h, or d and h.

"