Linear Systems

Section 3.5 Linear Systems

In Example 3.3.11 we wanted to solve a problem with two constraints. We could meet one exactly (percent alcohol) and the other in inequality (at least that volume). Many problems have more than one constraint (condition we want to meet). Here we will learn to solve one type of them.

Subsection 3.5.1 Motivation

In Example 3.3.11 we diluted a mixture using just diluent (water in that case). In other situations we will have two mixtures and want to combine them.

Example 3.5.1. Combining Mixtures.

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?

Solution.

A common technique in mathematics 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 will express our dual constraints using these answers (variables).

The first constraint is that we end up with 10 oz of solution. Thus

\begin{equation*} A+B = 10.0. \end{equation*}

The second contraint is the percent alcohol. As in Example 3.3.11 we will start by figuring out how much alcohol total will be in the resulting solution. Because it will be 85% alcohol there will be

\begin{equation*} (0.85)10.0 = 8.5 \text{ oz}\text{.} \end{equation*}

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

\begin{equation*} (0.91)A+(0.75)B = 8.5. \end{equation*}

Now we just need a way to solve this pair of equations.

Subsection 3.5.2 Crossing Lines

Our goal here is to consider what causes lines to cross. We will do this by looking at a pair of lines and seeing where they cross.

Recall that a line is a relation (set of points) such that the change between any two, equally space 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.

Example 3.5.2.

Suppose we start at a point $(1,3)\text{.}$ If we know that the line increases 2 units for each step then the points $(2,5)$ and $(3,7)$ are also on the line. This shows us that for each x value we add 2. This gives us $2x$ because that adds two for each x. But we must account for starting at $(1,3)\text{.}$ $2(1)=2$ so we need to add one to shift it up. This is why the line is $2x+1\text{.}$ This is the slope-intercept form of a line. The coefficient of x, 2 in this case, tells us how steep (how much it goes up for each incease of x). The constant, 1 in this case, shifts the line up or down.

Checkpoint 3.5.3.

In Figure 3.5.4 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 $A\text{.}$

(a)

Use the slider to set the slope of the second line to 3. Does the line cross the one through $A\text{?}$ Where (left or right of point $B$)?

(b)

Use the slider to set the slope to something bigger than 3. Does the line cross the first one? Where (left or right of point $B$)?

(c)

Use the slider to set the slope to 1. Does the line cross the first one? Where (left or right of point $B$)?

(d)

In general if one slope is steeper than the other will they cross?

(e)

Can you select a slope so that they don’t cross?

Instructions.

Use the slider to adjust the slope of the line through B.

Figure 3.5.4. Crossing Lines

Checkpoint 3.5.5.

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?

Solution 1.

We could make a table.

Year	Vasya	Pyotr
2012	$62,347.23	$67,242.33
2013	$67,347.23	$70,742.33
2014	$72,347.23	$74,242.33
2015	$77,347.23	$77,742.33
2016	$82,347.23	$81,242.33

We see that is in 2016 that Vasya is first paid more.

Solution 2.

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 it will take 4 years. Thus she is first paid more in 2016.

Subsection 3.5.3 Solving Linear Systems

We will consider two ways of solving linear systems of this type. The second method is very important for larger systems.

Example 3.5.6.

We will solve the system from Example 3.5.1. The two equations are

\begin{align*} A+B & = 10.0.\\ (0.91)A+(0.75)B & = 8.5. \end{align*}

Solution.

Notice we can solve the first equation for $B\text{,}$ then substitute it into the second.

\begin{align*} B & = 10.0-A.\\ (0.91)A+(0.75)(10.0-A) & = 8.5.\\ (0.91)A+7.5-(0.75)A & = 8.5.\\ (0.16)A & = 1.0.\\ A & = \frac{1.0}{0.16}\\ & = 6.3. \end{align*}

Now that we know that $A=6.3$ we can substitute that into $A+B=10.0.$ This gives us

\begin{align*} A+B & = 10.0.\\ 6.3+B & = 10.0.\\ B & = 3.7. \end{align*}

We can check that this works in the other equation (about percent alcohol).

\begin{align*} (0.91)A+(0.75)B & = \\ (0.91)(6.3)+(0.75)(3.7) & = \\ 5.7+2.8 & = 8.5. \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. We call it elimination.

Example 3.5.7.

We will solve the system

\begin{align*} A+B & = 10.0.\\ (0.91)A+(0.75)B & = 8.5. \end{align*}

Solution.

Note how we modify the first equation to partially match the second one.

\begin{align*} A+B & = 10.0.\\ -(0.91)(A+B) & = -(0.91)10.0.\\ -(0.91)A-(0.91)B & = -9.1.\\ (0.91)A+(0.75)B & = 8.5.\\ -(0.16)B & = -0.6.\\ B & = \frac{-0.6}{-0.16}\\ & = 3.7. \end{align*}

In the fifth line we added the two equations. Because they had opposite coefficients for $A\text{,}$ that was eliminated leaving us with just $B\text{.}$ This can always be done with systems of linear equations.

Checkpoint 3.5.8.

Solve the linear system below using substitution.

\begin{equation*} \begin{aligned} 2x+3y \amp = 17.\\ 4x+2y \amp = 14. \end{aligned} \end{equation*}

$x =$

$y =$

Answer 1.

$1$

Answer 2.

$5$

Solution.

We can solve the first equation for $x$ and substitute.

\begin{equation*} \begin{aligned} 2x+3y \amp = 17.\\ 2x \amp = 17-3y.\\ x \amp = \frac{1}{2}(17-3y). \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} 4x+2y \amp = 14.\\ 4 \left( \frac{1}{2}(17-3y) \right)+2y \amp = 14.\\ 34-6y+2y \amp = 14.\\ -4y \amp = -20.\\ y \amp = 5.\\ 2x+3(5) \amp = 17.\\ 2x+15 \amp = 17.\\ 2x \amp = 2.\\ x \amp = 1.\text{.} \end{aligned} \end{equation*}

Checkpoint 3.5.9.

Solve the linear system below using elimination.

\begin{equation*} \begin{aligned} 2x+3y \amp = 17.\\ 4x+2y \amp = 14. \end{aligned} \end{equation*}

$x =$

$y =$

Answer 1.

$1$

Answer 2.

$5$

Solution.

We note that $4/2=2$ so we should multiply the first equation by -2.

\begin{equation*} \begin{aligned} -2(2x+3y) \amp = -2(17).\\ -4x-6y \amp = -34.\\ 4x+2y \amp = 14.\\ -4y \amp = -20.\\ y \amp = 5.\\ 2x+3(5) \amp = 17.\\ 2x+15 \amp = 17.\\ 2x \amp = 2.\\ x \amp = 1. \end{aligned} \end{equation*}

Checkpoint 3.5.10.

Find the solution to this system.

\begin{equation*} \begin{aligned} 6x+8y \amp = {22}.\\ 9x+13y \amp = {35}. \end{aligned} \end{equation*}

$x=$

$y=$

Answer 1.

$1$

Answer 2.

$2$

Solution.

To eliminate the $9x$ term we need to multiply the $6x$ equation by $-9/6=-3/2$

\begin{equation*} \begin{aligned} -3/2(6x+8y) \amp = -3/2{22}\\ -9x-12y \amp = {-33}\\ 9x+13y \amp = {35}\\ y \amp = {2}\\ 6x+8({2}) \amp = {22}\\ 6x \amp = {6}\\ x \amp = {1} \end{aligned} \end{equation*}

Subsection 3.5.4 Other Cases

In Figure 3.5.4 we found a slope that caused no intersection. If we were solving a pair of linear equations that represented lines like this we would find no solution. These are known as inconsistent systems.

Example 3.5.11. Inconsistent Linear System.

Find all solutions to the system

\begin{align*} 2x + 3y & = 5.\\ 4x + 6y & = 7. \end{align*}

Solution.

We will use elimination. If we multiply -2 by the first equation we will obtain -4 (opposite of x in the second equation).

\begin{align*} 2x + 3y & = 5.\\ -2(2x+3y) & = -2(5).\\ -4x-6y & = -10.\\ 4x+6y & = 7.\\ 0 & = -3. \end{align*}

Our work is correct, but the conclusion is clearly false. This means there are no solutions. You can think of this as saying, for a solution to exist 0 must equal -3.

There is a third case.

Example 3.5.12. Dependent System.

Find all solutions to the system

\begin{align*} 2x + 3y & = 5.\\ 4x + 6y & = 10. \end{align*}

Solution.

We will use elimination. If we multiply -2 by the first equation we will obtain -4 (opposite of x in the second equation).

\begin{align*} 2x + 3y & = 5.\\ -2(2x+3y) & = -2(5).\\ -4x-6y & = -10.\\ 4x+6y & = 10.\\ 0 & = 0. \end{align*}

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.

Checkpoint 3.5.13.

Determine whether this system is inconsistent or dependent.

\begin{equation*} \begin{aligned} 6x+8y \amp = {52}.\\ 9x+12y \amp = {78}. \end{aligned} \end{equation*}

Section 3.5 Linear Systems

Subsection 3.5.1 Motivation

Example 3.5.1. Combining Mixtures.

Subsection 3.5.2 Crossing Lines

Example 3.5.2.

Checkpoint 3.5.3.

(a)

(b)

(c)

(d)

(e)

Instructions.

Checkpoint 3.5.5.

Subsection 3.5.3 Solving Linear Systems

Example 3.5.6.

Example 3.5.7.

Checkpoint 3.5.8.

Checkpoint 3.5.9.

Checkpoint 3.5.10.

Subsection 3.5.4 Other Cases

Example 3.5.11. Inconsistent Linear System.

Example 3.5.12. Dependent System.

Checkpoint 3.5.13.

Exercises 3.5.5 Exercises

1. Two Equations.

2. Two Equations.

3. Two Equations.

4. Two Equations.

5. Two Equations.

6. Two Equations.

7. Two Equations.

8. Two Equations.

9. Two Equations.

10. Linear System Application.

11. Linear System Application.

12. Linear System Application.

13. Linear System Application.

14. Linear System Application.

15. Linear System Application.

16. Linear System Application.

17. Linear System Application.