In Example 2.2.1 we solved a problem with two, different mixtures. In that case we knew how much of each we were adding. In other cases we know what result we want and need to figure out how much of each substance. This section presents a method for answering questions like this involving two or more constraints (equations) at the same time.
This section provides a first examples of a problem involving two, linear equations. It illustrates how we would recognize this type of problem and how to write it.
In previous dilution problems, we diluted a mixture using just diluent (water in that case). In other situations we will have two mixtures with different percents that we will combine to obtain a new mixture.
Suppose we have 16 oz of 91% isopropyl alcohol and 12 oz of 75% isopropyl alcohol. How much of each do we need to mix to produce \(10.0\) oz of 85% alcohol?
The question asks us to determine an amount for each of the two mixtures. We can use a common technique in mathematics which is to start by writing the answer. We will declare that we will use \(A\) oz of 91% alcohol and \(B\) oz of 75% alcohol. Next we can look at the constraints and use them to write equations using these variables.
The second contraint is the percent alcohol. Because our variables are in terms of amount of solution, we need to express the percent alcohol constraint in terms of ounces (not percents). We can obtain amount from percent using the definition of percent. Because the resulting solution will be 85% alcohol there will be
This will be the result of adding the amounts from each solution (just as in previous mixture problems). Because \(A\) oz of the first solution will be added and it is 91% alcohol, it will contribute \((0.91)A\) oz of alcohol. Similarly the second solution will contribute \((0.75)B\) oz of alcohol. Combined we will obtain
Based on past sales, a business determines that snack fans prefer two item trail mix consisting of cashews and dried blueberries in 6 oz portions. They are willing to pay $9/lb for it. If the cashews cost $8.50/lb and the dried blueberries cost $9.75/lb, how many oz of each should be in one 6 oz portion?
We are asked for amounts for both ingredients. Let’s call the number of ounces \(c\) and \(b\text{.}\) Before we set up equations we should convert the prices to dollars per ounce.
The next example illustrates what to do when one or more of the constraints do not use all the variables. Note, the question can be answered without a linear system (see Section 2.2); we use it here as a familiar example.
A plane has to fly 106 nm in a straight line between Merrill Field and Homer airports. If the flight from Merrill to Homer took 53 minutes, and the flight from Homer to Merrill took 45 minutes, how fast was the plane flying through the air and how fast was the head/tail wind? We will suppose the wind speed was the same the whole time.
At first it is not clear what type of calculation we will need. We can follow the advice in Example 3.6.1 to simply write down the answer. Because we are asked for plane and wind speeds, these are the variables. Call them \(p\) and \(w\text{.}\)
We also have data on two flights. These include times and distance but not speed (what we are asked to find). So we must ask ourselves how we can convert time and distance into speeds. We did this in Example 2.4.4 and Example 2.4.6.
however this has hours on the left and minutes on the right. So we must convert the time in minutes to time in hours (unit conversion is in Section 1.1). Because there are 60 minutes per hour, 53 minutes is \(53/60\) hours.
Why do we have two speeds for the same trip? The wind slows down the first flight and speeds up the second. The speeds calculated above are across the ground (nm across the ground per hour). When travelling with wind at our backs, our speed is increased by the wind speed. When travelling with a head wind, our speed is decreased by the wind speed. For example if a plane is flying at 60 nm/hr and has a 7 nm/hr head wind it will be travelling only \(60-7=53\) nm/hr over the ground. This idea finally enables us to setup equations for the constraints.
Later in this section we will learn a technique to solve these equations. However, this is a special case. The speed \(p\) must be between 120 and 141. Because the same amount (\(w\)) is subtracted and added, \(p\) is half-way between or
This section connects solving a system of two linear equations to their graphs. Graphs will help us understand why (and when) there should be a solution. The next section provides the method for solving.
Recall that a line is a relation (set of points) such that the change between any two, equally spaced points is the same. Often you have heard this described as rise over run or slope. Slope is a geometric interpretation referring to how steep the line is.
In Figure 3.6.6 there are two lines. One goes through point \(A=(1,3)\text{.}\) It rises at a slope of \(2/1\) (two up for each one over). The other line goes through the point \(B=(1,2)\) which is below \(A\text{.}\)
Vasya’s initial pay was $62,347.23. She received $5,000 raises each year. Pyotr’s initial pay was $67,242.33. He receives $3,500 raises each year. If they were both hired in 2012 in what year does Vasya first have a higher salary?
Alternatively we could note that Vasya’s raises are $1,500 more each year than Pyotr’s raises. This means she closes the gap by $1,500 each year. The difference in their initial salaries is \(67242.33-62347.23=4895.10\text{.}\) Because she gains by 1500 each year it will take \(4895.10/1500 = 3.2634 \) years. Because they receive raises once a year this result must be rounded up to 4 years. Thus she is first paid more in 2016.
Notice we can solve the first equation for \(B\text{.}\) If we do then we can substitute it into the second equation. The result will be a single equation with only one variable. We already know how to solve equations like that.
\begin{align*}
A+B & = 10.0.\\
B & = 10.0-A.
\end{align*}
If we had 7 variables instead of two, substituting would take a while. Instead we can use the following method which is more like solving as we know it, that is isolating a variable. This method is called elimination.
In the fifth line we added the two equations. Because they had opposite coefficients for \(A\text{,}\) that variable was eliminated, leaving us with just \(B\text{.}\) This can always be done with systems of linear equations.
In Figure 3.6.6 we discovered a slope that resulted in no intersection. If we were solving a pair of linear equations that represented lines like this, our calculations would not produce a solution. These are known as inconsistent systems. This section provides two examples of such systems and demonstrates how we identify them. For this book, identifying these cases and correctly describing them is all you are expected to do.
Our work is correct, but the conclusion is clearly false. You can think of this as saying, for a solution to exist 0 must equal -3. Because this is a contradiction, we call the system inconsistent. This means there are no solutions.
This time we have a true, but rather uninformative statement. We notice that after scaling (multiplying by -2) the two equations were identical. Essentially we had only one equation. Because one can be obtained from the other we call them dependent.
