Hello. I’m Professor Von Schmohawk

and welcome to Why U. In the previous chapter

we saw how Algebra can be used to solve problems involving mixtures of solutions. In this chapter we will see how to solve problems

involving other types of mixtures. Adrienne Scienstein has taken a part-time job managing her Uncle Monopolus’ coffee shop. The shop sells two different types of roasted

coffee beans at two different price points. Their very expensive Tasmanian roasted coffee hand-picked by specially-trained wombats sells for $31 per pound. The shop also offers an inexpensive coffee roasted at the White Sands Nuclear Testing Range for $3 a pound. Adrienne thinks that a moderately-priced coffee might appeal to a broader group of customers. She decides to mix the two types of coffee beans to create a special blend costing $10 per pound. To do this, Adrienne must calculate

the quantity of each type of coffee bean needed to create a pound of coffee costing $10. As we saw in previous chapters this problem can be organized

by creating a table. In the first row of the table

we will list the price per pound of each coffee. We know that the Tasmanian coffee

costs $31 per pound the White Sands nuclear roast

costs $3 per pound and the blend we wish to create

must cost $10 per pound. In the second row

we list the weight of each coffee to mix to create one pound of coffee blend

which costs $10. If we let x represent the weight

of the Tasmanian roast then the weight of nuclear roast

must be “one minus x” since the combined weight of Tasmanian roast

plus nuclear roast must be one pound. In the third row, we will list the total cost

of each coffee given the price per pound and the weight. The cost of the Tasmanian roast

is $31 per pound times x pounds or “31x” dollars. Likewise, the cost of the Nuclear roast

is $3 per pound times “1 minus x” pounds. Since the combined cost of the two coffees

in the blend must be 10 dollars we can create an equation which says that

the cost of the Tasmanian roast plus the cost of the nuclear roast equals $10. We can now solve this equation for x. We start by distributing the 3 to the two terms in parentheses. We can then multiply 3 times 1 and eliminate the parentheses. Combining like terms

we subtract 31x minus 3x to get 28x. Subtracting 3 from both sides of the equation allows us to cancel

the positive and negative 3’s on the left leaving the equation

“28x equals 10 minus 3” or “28x equals 7”. We then divide both sides by 28 allowing us to cancel the 28’s in the numerator

and denominator of the fraction on the left. This leaves us with the equation

“x equals 7 over 28” or “x equals one-quarter”. So to create one pound

of a $10 per pound blend Adrienne must mix one-quarter pound

of Tasmanian roast with “1 minus one-quarter” or three-quarters pound of Nuclear roast. We can check our calculations by multiplying

the price per pound for each of the two coffees times their calculated weights and check to see if the total cost

for a pound of blend comes to $10. The cost of the Tasmanian roast

in one pound of Adrienne’s blend is 31 dollars times one-quarter pound and the cost of the Nuclear roast

is 3 dollars times three-quarters pound. Multiplying 31 times one-fourth we get 31 fourths. And 3 times three-fourths is nine-fourths. Adding these two fractions

we get “31 plus 9” fourths or 40 fourths which is equal to 10. Since the total cost of the calculated amounts

of the two roasts is 10 dollars our calculations appear to be correct. Our next example illustrates how algebra can

be used to solve problems involving mixtures of items

selling at different prices. The Why U Entertainment Committee

is planning a concert of the world-famous bagpipe quartet “Alasdair, Ian, Angus, and Clyde”. The concert will be performed

at the Why University concert hall which seats 1000 people. Normally tickets sell for $50 a seat. However, to be sure that the 1000 seat auditorium

is completely filled the committee has decided to offer a number

of discounted tickets at $30 a ticket. The total revenue collected from ticket sales

must be at least $45000 in order to cover the cost of the performance. The Entertainment Committee

therefore needs to know how many discounted tickets can be offered

and still collect $45000 assuming that a total of 1000 tickets are sold. Fortunately, using Algebra, we can calculate

how many discounted tickets can be sold and still collect $45000. Once again, we can organize this problem

by creating a table. In the first row of the table

we list the ticket prices for discounted and non-discounted tickets. In the second row, we list the number of

discounted and non-discounted tickets which will be sold. If we let x represent the number of

discounted tickets then the number of non-discounted tickets

will be “1000 minus x” since the total number of tickets is 1000. In the third row, we list the revenue

generated by each type of ticket given the ticket price and the number of tickets sold. The revenue produced by the discounted tickets

is 30 dollars times x tickets or “30x” dollars. Likewise, the revenue produced by the non-discounted

tickets is 50 dollars times “1000 minus x” tickets. Since the total revenue generated by both

types of tickets must be $45000 we can create an equation which says that

the revenue from discounted tickets plus the revenue from non-discounted tickets is equal to 45000. We can now solve this equation for x. We start by distributing 50 to the two terms in parentheses. We then complete the arithmetic in the parentheses

multiplying 50 times 1000 and eliminate the parentheses. Combining like terms

we subtract 30x minus 50x to get negative 20x. And subtracting 50000

from both sides of the equation allows us to cancel

the positive and negative 50000 leaving the negative 20x term

alone on the left. Completing the arithmetic on the right

we subtract 45000 minus 50000 to get negative 5000. We then eliminate the coefficient of x

by dividing both sides by negative 20 allowing us to cancel the negative 20’s in the numerator and denominator

of the fraction on the left. Finally, dividing negative 5000

by negative 20 we get “x equals 250”. So 250 discounted tickets can be sold and “1000 minus 250” or 750 non-discounted tickets can be sold. We can check our calculations by multiplying

the ticket price for each type of ticket times the number of tickets and check to see if the total revenue

comes to 45000 dollars. The revenue from the discounted tickets

is 30 dollars times 250 tickets and the revenue from the non-discounted tickets

is 50 dollars times 750 tickets. Multiplying 30 times 250 we get 7500 and 50 times 750 is 37500. Adding these two numbers we get 45000. Since the total revenue

from the calculated quantities of discounted plus non-discounted tickets

is 45000 dollars we know that our calculations are correct. In the last several lectures

we have seen how to create linear equations to solve various types of real world problems. In the next few lectures

we will explore algebraic methods which we can use to create

parallel and perpendicular lines.

Why 1-x?

I love watching this lessons!

WARBUCKS hehehe

0:51 hes a legendary super saiyan! (i need to stop)

these are great

nuclear coffee lol