Algebra 32 – Solving Mixture Problems with Linear Equations

Algebra 32 – Solving Mixture Problems with Linear Equations


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.

6 thoughts on “Algebra 32 – Solving Mixture Problems with Linear Equations

  • August 2, 2014 at 3:12 pm
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    Why 1-x?

    Reply
  • August 4, 2014 at 4:53 am
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    I love watching this lessons!

    Reply
  • March 16, 2015 at 4:41 am
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    WARBUCKS hehehe

    Reply
  • February 13, 2017 at 12:18 am
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    0:51 hes a legendary super saiyan! (i need to stop)

    Reply
  • May 31, 2017 at 6:31 am
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    these are great

    Reply
  • August 7, 2017 at 3:13 pm
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    nuclear coffee lol

    Reply

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