Unit 2 Linear Equations and Systems — Unit Plan

Suppose your class is planning a trip to a museum. The cost of admission is $7 per person,
and the cost of renting a bus for the day is $180.

• If 24 students and 3 teachers are going, we know the cost will be $7(24) + 7(3) + 180$,
  or $7(24+3) + 180$, dollars.
• If 30 students and 4 teachers are going, the cost will be $7(30+4) + 180$ dollars.

Notice that the numbers of students and teachers can vary. This means that the cost of admission and the total cost of the trip can also vary, because they depend on how many people are going.

Letters are helpful for representing quantities that vary. If $s$ represents the number of students who are going, $t$ represents the number of teachers, and $C$ represents the total cost, we can model the quantities and constraints by writing

$C = 7(s+t) + 180$

Some quantities may be fixed. In this example, the bus rental costs $180 regardless of how many students and teachers are going (assuming only one bus is needed).

Letters can also be used to represent quantities that are constant. We might do this when we don’t know what the value is, or when we want to understand the relationship between quantities (rather than the specific values).

For instance, if the bus rental is $B$ dollars, we can express the total cost of the trip as $C = 7(s + t) + B$. No matter how many teachers or students are going on the trip,
$B$ dollars need to be added to the cost of admission.

Sometimes, the relationship between two quantities is easy to see. For instance, we know that the perimeter of a square is always 4 times the side length of the square. If $P$ represents the perimeter and $s$ represents the side length, then the relationship between the two measurements (in the same unit) can be expressed as $P = 4s$, or $s = \frac{P}{4}$.

Other times, the relationship between quantities might take a bit of work to figure out—by doing calculations several times or by looking for a pattern. Here are two examples.

• A plane departed from New Orleans and is heading to San Diego. The table shows its distance from New Orleans, $x$, and its distance from San Diego, $y$, at some points along the way. 

  miles from New Orleans	miles from San Diego
  100	1,500
  300	1,300
  500	1,100
  	1,020
  900	700
  1,450
  $x$	$y$

  What is the relationship between the two distances? Do you see any patterns in how each quantity is changing? Can you find out what the missing values are?

  Notice that every time the distance from New Orleans increases by some number of miles, the distance from San Diego decreases by the same number of miles, and that the sum of the two values is always 1,600 miles.

  The relationship can be expressed with any of these equations:

  $x + y = 1,600$

  $y = 1,600 - x$

  $x = 1,600 - y$

• A company decides to donate $50,000 to charity. It will select up to 20 charitable organizations, as nominated by its employees. Each selected organization will receive an equal amount of donation.

  What is the relationship between the number of selected organizations, $n$, and the dollar amount each of them will receive, $d$?

  • If 5 organizations are selected, each one receives $10,000.
  • If 10 organizations are selected, each one receives $5,000.

  • If 20 organizations are selected, each one receives $2,500.

  Do you notice a pattern here? 10,000 is $\frac {50,000}{5}$, 5,000 is $\frac{50,000}{10}$, and 2,500 is $\frac {50,000}{20}$.

  We can generalize that the amount each organization receives is 50,000 divided by the number of selected organizations, or $d = \frac {50,000}{n}$.

An equation that contains only one unknown quantity or one quantity that can vary is called an equation in one variable.

For example, the equation $2\ell + 2w = 72$ represents the relationship between the length, $\ell$, and the width, $w$, of a rectangle that has a perimeter of 72 units. If we know that the length is 15 units, we can rewrite the equation as:

$2(15) + 2w = 72$.

This is an equation in one variable, because $w$ is the only quantity that we don't know. To solve this equation means to find a value of $w$ that makes the equation true.

In this case, 21 is the solution because substituting 21 for $w$ in the equation results in a true statement. 

$\begin{aligned}2(15) + 2w &=72\\ 2(15)+2(21) &= 72\\ 30 + 42 &=72\\ 72&=72 \end{aligned}$

An equation that contains two unknown quantities or two quantities that vary is called an equation in two variables. A solution to such an equation is a pair of numbers that makes the equation true. 

Suppose Tyler spends $45 on T-shirts and socks. A T-shirt costs $10 and a pair of socks costs $2.50. If $t$ represents the number of T-shirts and $p$ represents the number of pairs of socks that Tyler buys, we can can represent this situation with the equation:

$10t + 2.50p = 45$

This is an equation in two variables. More than one pair of values for $t$ and $p$ make the equation true.

In this situation, one constraint is that the combined cost of shirts and socks must equal $45. Solutions to the equation are pairs of $t$ and $p$ values that satisfy this constraint.

Combinations such as $t=1$ and $p = 10$ or $t=2$ and $p=7$ are not solutions because they don’t meet the constraint. When these pairs of values are substituted into the equation, they result in statements that are false.

Like an equation, a graph can give us information about the relationship between quantities and the constraints on them. 

Suppose we are buying beans and rice to feed a large gathering of people, and we plan to spend $120 on the two ingredients. Beans cost $2 a pound and rice costs $0.50 a pound. If $x$ represents pounds of beans and $y$ pounds of rice, the equation $2x + 0.50y = 120$ can represent the constraints in this situation. 

The graph of $2x + 0.50y = 120$ shows a straight line. 

Graph of a line, origin O. Horizontal axis, pounds of beans, scale is 0 to 100, by 20’s. Vertical axis, pounds of rice, scale is 0 to 280, by 40’s. Line starts at 0 comma 240, passes through 10 comma 200, 30 comma 120 and 60 comma 0. Points 20 comma 80 and 70 comma 180 are shown but not on the line.

Each point on the line is a pair of $x$- and $y$-values that makes the equation true and is, thus, a solution. It is also a pair of values that satisfy the constraints in the situation.

• The point $(10,200)$ is on the line. If we buy 10 pounds of beans and 200 pounds of rice, the cost will be $2(10) + 0.50(200)$, which equals 120. 
• The points $(60,0)$ and $(45,60)$ are also on the line. If we buy only beans—60 pounds of them—and no rice, we will spend $120. If we buy 45 pounds of beans and 60 pounds of rice, we will also spend $120. 

What about points that are not on the line? They are not solutions because they don't satisfy the constraints, but they still have meaning in the situation.

• The point $(20, 80)$ is not on the line. Buying 20 pounds of beans and 80 pounds of rice costs $2(20) + 0.50(80)$, or 80, which does not equal 120. This combination costs less than what we intend to spend.
• The point $(70,180)$ means that we buy 70 pounds of beans and 180 pounds of rice. It will cost $2(70)+0.50(180)$, or 230, which is over our budget of 120.

Suppose we bought 2 packs of markers and a $0.50 glue stick for $6.10. If $p$ is the dollar cost of 1 pack of markers, the equation $2p +0.50 = 6.10$ represents this purchase. The solution to this equation is 2.80.

Now suppose a friend bought 6 of the same packs of markers and 3 $0.50 glue sticks, and paid $18.30. The equation $6p + 1.50 = 18.30$ represents this purchase. The solution to this equation is also 2.80.

We can say that $2p + 0.50= 6.10$ and $6p + 1.50 = 18.30$ are equivalent equations because they have exactly the same solution. Besides 2.80, no other values of $p$ make either equation true. Only the price of $2.80 per pack of markers satisfies the constraint in each purchase.

In this example, the second equation, $6p + 1.50 =18.30$, is a result of multiplying each side of the first equation by 3. Buying 3 times as many markers and glue sticks means paying 3 times as much money. The unit price of the markers hasn't changed.

Here are some other equations that are equivalent to $2p + 0.50= 6.10$, along with the moves that led to these equations.

In each case:

• The move is acceptable because it doesn't change the equality of the two sides of the equation. If $2p + 0.50$ has the same value as 6.10, then multiplying $2p + 0.50$ by $\frac12$ and multiplying 6.10 by $\frac12$ keep the two sides equal.
• Only $p=2.80$ makes the equation true. Any value of $p$ that makes an equation false also makes the other equivalent equations false. (Try it!)

These moves—applying the distributive property, adding the same amount to both sides, dividing each side by the same number, and so on—might be familiar because we have performed them when solving equations. Solving an equation essentially involves writing a series of equivalent equations that eventually isolates the variable on one side.

Not all moves that we make on an equation would create equivalent equations, however!

For example, if we subtract 0.50 from the left side but add 0.50 to the right side, the result is $2p = 6.60$. The solution to this equation is 3.30, not 2.80. This means that $2p = 6.60$ is not equivalent to $2p + 0.50 =6.10$.

A relationship between quantities can be described in more than one way. Some ways are more helpful than others, depending on what we want to find out. Let’s look at the angles of an isosceles triangle, for example.

The two angles near the horizontal side have equal measurement in degrees, $a$. 

The sum of angles in a triangle is $180^{\circ}$, so the relationship between the angles can be expressed as:

 $a + a+ b=180$

Suppose we want to find $a$ when $b$ is $20^{\circ}$.  

Now suppose the bottom two angles are $34^\circ$ each. How many degrees is the top angle?

Notice that when $b$ is given, we did the same calculation repeatedly to find $a$: We substituted $b$ into the first equation, subtracted $b$ from 180, and then divided the result by 2. 

Instead of taking these steps over and over whenever we know $b$ and want to find $a$, we can rearrange the equation to isolate $a$:

$\begin{aligned} a + a + b &= 180\\ 2a + b&=180\\2a &=180-b\\ a &=\dfrac{180-b}{2} \end{aligned}$

This equation is equivalent to the first one. To find $a$, we can now simply substitute any value of $b$ into this equation and evaluate the expression on the right side.

Likewise, we can write an equivalent equation to make it easier to find $b$ when we know $a$:

$\begin{aligned} a + a + b &= 180\\ 2a + b&=180\\b &=180-2a \end{aligned}$

Rearranging an equation to isolate one variable is called solving for a variable. In this example, we have solved for $a$ and for $b$. All three equations are equivalent. Depending on what information we have and what we are interested in, we can choose a particular equation to use. 

Linear equations can be written in different forms. Some forms allow us to better see the relationship between quantities or to predict the graph of the equation.

Suppose a person wishes to travel 7,000 meters a day by running and swimming. The person runs at a speed of 130 meters per minute and swims at a speed of 54 meters per minute.

Let $x$ represents the number of minutes of running and $y$ the number of minutes of swimming. To represent the combination of running and swimming that would allow the person to travel 7,000 meters, we can write:

$130x+54y=7,000$

We can reason that the more minutes the person runs, the fewer minutes the person has to swim to meet the goal. In other words, as $x$ increases, $y$ decreases. If we graph the equation, the line will slant down from left to right.

To determine how many minutes the person would need to swim after running for 15 minutes, 20 minutes, or 30 minutes, substitute each of these values for $x$ in the equation and find $y$. Or, first solve the equation for $y$:

$\begin{aligned} 130x+54y=7,000 \\ 54y = 7,000-130x\\ y = \frac{7,000-130x}{54}\\ y = 129.63 - 2.41x  \end{aligned}$

Notice that $y=129.63-2.41x$, or $y=\text-2.41x+129.63$, is written in slope-intercept form.

• The coefficient of $x$, -2.41, is the slope of the graph. It means that as $x$ increases by 1, $y$ falls by 2.41. For every additional minute of running, the person can swim 2.41 fewer minutes.
• The constant term, 129.63, tells us where the graph intersects the $y$-axis. It tells us the number minutes the person would need to swim if they do no running.

The first equation we wrote, $130x+54y=7,000$, is a linear equation in standard form. ​In general, it is expressed as $Ax + By = C$, where $x$ and $y$ are variables, and $A, B$, and $C$ are numbers.

The two equations, $130x+54y=7,000$ and $y=\text-2.41x+129.63$, are equivalent, so they have the same solutions and the same graph.

A costume designer needs some silver and gold thread for the costumes for a school play. She needs a total of 240 yards. At a store that sells thread by the yard, silver thread costs $0.04 a yard and gold thread costs $0.07 a yard. The designer has $15 to spend on the thread.

How many of each color should she get if she is buying exactly what is needed and spending all of her budget?

This situation involves two quantities and two constraints—length and cost. Answering the question means finding a pair of values that meets both constraints simultaneously. To do so, we can write two equations and graph them on the same coordinate plane.

Let $x$ represents yards of silver thread and $y$ yards of gold thread. 

• The length constraint: $x + y = 240$
• The cost constraint: $0.04x + 0.07y = 15$

Every point on the graph of $x+y=240$ is a pair of values that meets the length constraint.

Every point on the graph of $0.04x + 0.07y = 15$ is a pair of values that meets the cost constraint.

The point where the two graphs intersect gives the pair of values that meets both constraints. 

Graph of 2 intersecting lines, origin O. Horizontal axis 0 to 280, by 20’s, labeled, yards of silver thread. Vertical from 0 to 280, by 20’s, labeled yards of gold thread. Line 1 starts at 240 comma 0, passes through 60 comma 180, ends at 240 comma 0. Line 2 starts at 0 comma 214 point 2 9, passes through 60 comma 180, ends at 280 comma 54 point 2 9.  

​​​​

That point is $(60, 180)$, which represents 60 yards of silver thread and 180 yards of gold thread.

If we substitute 60 for $x$ and 180 for $y$ in each equation, we find that these values make the equation true. $(60,180)$ is a solution to both equations simultaneously. 

Two or more equations that represent the constraints in the same situation form a system of equations. A curly bracket is often used to indicate a system.

$\begin {cases} x + y = 240\\0.04x + 0.07y = 15 \end {cases}$

The solution to a system of equations is a pair of values that makes all of the equations in the system true. Graphing the equations is one way to find the solution to a system of equations.

The solution to a system can usually be found by graphing, but graphing may not always be the most precise or the most efficient way to solve a system. 

Here is a system of equations:

$\begin {cases} 3p + q = 71\\2p - q = 30 \end {cases}$

The graphs of the equations show an intersection at approximately 20 for $p$ and approximately 10 for $q$.

Without technology, however, it is not easy to tell what the exact values are.

Instead of solving by graphing, we can solve the system algebraically. Here is one way.

If we subtract $3p$ from each side of the first equation, $3p + q = 71$, we get an equivalent equation: $q= 71 - 3p$. Rewriting the original equation this way allows us to isolate the variable $q$. 

Because $q$ is equal to $71-3p$, we can substitute the expression $71-3p$ in the place of $q$ in the second equation. Doing this gives us an equation with only one variable, $p$, and makes it possible to find $p$.

$\begin{aligned} 2p - q &= 30 &\quad& \text {original equation} \\ 2p - (71 - 3p) &=30 &\quad& \text {substitute }71-3p \text{ for }q\\ 2p - 71 + 3p &=30 &\quad& \text {apply distributive property}\\ 5p - 71 &= 30 &\quad& \text {combine like terms}\\ 5p &= 101 &\quad& \text {add 71 to both sides}\\ p &= \dfrac{101}{5} &\quad& \text {divide both sides by 5} \\ p&=20.2 \end{aligned}$

Now that we know the value of $p$, we can find the value of $q$ by substituting 20.2 for $p$ in either of the original equations and solving the equation.

The solution to the system is the pair $p=20.2$ and $q=10.4$, or the point $(20.2, 10.4)$ on the graph. 

This method of solving a system of equations is called solving by substitution, because we substituted an expression for $q$ into the second equation.

Another way to solve systems of equations algebraically is by elimination. Just like in substitution, the idea is to eliminate one variable so that we can solve for the other. This is done by adding or subtracting equations in the system. Let’s look at an example.

$\begin {cases} \begin{aligned}5x+7y=64\\ 0.5x - 7y = \text-9 \end{aligned} \end {cases}$

Now that we know $x = 10$, we can substitute 10 for $x$ in either of the equations and find $y$:

In this system, the coefficient of $y$ in the first equation happens to be the opposite of the coefficient of $y$ in the second equation. The sum of the terms with $y$-variables is 0.

Adding or subtracting the equations in a system creates a new equation. How do we know the new equation shares a solution with the original system?

If we graph the original equations in the system and the new equation, we can see that all three lines intersect at the same point, but why do they?

Three intersecting lines on coordinate plane, origin O. X axis from negative 4 to 14 by 2’s. Y axis from negative 4 to 12 by 2’s. Black line through negative 4 comma 12 and 10 comma 2. Blue line passes through negative 4 comma 1 and 10 comma 2. Red vertical line passes through 10 comma 2. All lines intersect at 10 comma 2.

In future lessons, we will investigate why this strategy works. 

We now have two algebraic strategies for solving systems of equations: by substitution and by elimination. In some systems, the equations may give us a clue as to which strategy to use. For example:

In other systems, which strategy to use is less straightforward, either because no variables are isolated, or because no variables have equal or opposite coefficients. For example:

$\begin{cases} \begin{aligned} 2x+3y&=15 \quad&\text{Equation A}\\ 3x-9y&=18 \quad&\text{Equation B} \\ \end{aligned}\end{cases}$

To solve this system by elimination, we first need to rewrite one or both equations so that one variable can be eliminated. To do that, we can multiply both sides of an equation by the same factor. Remember that doing this doesn't change the equality of the two sides of the equation, so the $x$- and $y$-values that make the first equation true also make the new equation true.

There are different ways to eliminate a variable with this approach. For instance, we could:

Each multiple of an original equation is equivalent to the original equation. So each new pair of equations is equivalent to the original system and has the same solution.

When we solve a system by elimination, we are essentially writing a series of equivalent systems, or systems with the same solution. Each equivalent system gets us closer and closer to the solution of the original system.