### 1.8 – Problems Involving Two Unknowns

# Problems involving Two Unknowns

In this section we will explore a method of solving equations in which there are two unknown values. In these types of problems, we will begin by assigning a variable to one of the unknown values, and expressing the other value in terms of the unknown. Doing this eliminates the need for multiple equations to describe the situation (we will do this later.) We saw this first in previous sections dealing with cutting a board into multiple pieces, or finding the values of the angles of a triangle.

Let us begin with an example:

Janet has 18 coins in her purse, all of them nickels and quarters. If she has $2.10, how many of each coin does she have? |
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We do not know how many nickels Janet has, nor do we know how many quarters she has. These are our two unknowns. We can assign the variable to represent the number of nickels she has. To avoid using a second variable, we can note that if she has 18 coins total, and of them are nickels, then the difference between 18 and is the number of quarters she has. That is: And we know that the value of nickels is the number of nickels multiplied by the value of a nickel, and similarly for the value of the quarters: And if we were to add these two amounts together, they would equal the total amount Janet has, $2.10, we can create the following equation: And from here, we simplify and solve using our regular methods: So then we have 12 nickels. Plugging 12 into the expression for quarters, we then have 18-12=6 Quarters. |

Let’s view another example:

Java Joe’s Coffee Shop sells two kinds of coffee. Mountain Aroma sells for $8 per pound, and Blue Ivy sells for $6.50 per pound. Joe wants to combine the two to sell at a rate of $7 per pound. How much of each coffee should he combine to make 30 pounds of the mix? |
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Let represent the number of pounds of Mountain Aroma coffee. Then will represent the number of pounds of Blue ivy. We can use an equation similar to the previous example: Note that the right hand side of this equation is the total cost of the mixture, since it sells at $7 per pound, and there are thirty pounds of it, then the product of the two is the total sales price of the mixture. We solve this equation: So we have ten pounds of Mountain Aroma, and thus we have 20 pounds of Blue Ivy. |

Ginger has 60 coins, all of them nickels and dimes. If she has a total of $4.45, how many nickels, and how many dimes does she have? |
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We can call our variable , and note that is the number of nickels. If there are nickels, then there are dimes. And our equation becomes: And we solve: So we have 31 nickels, and therefore, we have dimes. |

Mixture problems are very similar to the previous problems:

A scientist needs 42 gallons of a 15% mixture of saline for an experiment. He has ample 12% and 75% solutions in stock. How much of each should he mixed? |
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We identify our variable , and can say that it is the amount of 12% mixture. Then we will also have gallons of 75% mixture. Our equation is: Then we need 40 gallons of 12% mixture, and 2 gallons of 75% |

A chemist needs to use 30 Liters of a 40% solution for an experiment. He notices that in his lab, however, that he only has 25% and 50% solutions. He decides to mix the two together to get his 40% mixture. How much of each should he use? |
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The chemist wants 30 liters of the solution, but he doesn’t know how much of each to use. We can assign to represent the amount of the 25% solution to use. In doing so, then the rest of the solution is the 40% mixture. So the difference between 30 and is the amount of 40%. That is, we have: We can then create an equation using this information, and the fact that the product of a solutions volume and its strength (percentage) yields the actual amount of acid in the solution. We note that the sum of the two acids should equal the total amount of acid, and build the equation: And we can solve this equation through means discussed in earlier sections: So we see that we will use 12 liters of the 25% acid. We then will use Liters of the 50% solution. |