Unit 4 Linear Equations And Linear Systems — Unit Plan

If we have equal weights on the ends of a hanger, then the hanger will be in balance. If there is more weight on one side than on the other, the hanger will tilt to the heavier side.

We can think of a balanced hanger as a representation for an equation. An equation says that the expressions on each side have equal value, just like a balanced hanger has equal weights on each side. This hanger could be represented by $a + 2b = 5b$.

If we have a balanced hanger and add or remove the same amount of weight from each side, the result will still be in balance. Here, we remove 2 triangles from each side, which is like subtracting $2b$ from each side of the equation to get $a = 3b$.

In the same way that adding or subtracting the same shapes on each side of a hanger keeps it in balance, adding or subtracting the same value to each side of an equation creates an equivalent equation.

An equation tells us that two expressions have equal value. For example, if $4x+9$ and $\text-2x-3$ have equal value, we can write the equation

$4x + 9 = \text-2x - 3$

Earlier, we used hangers to understand that if we add the same positive number to each side of the equation, the sides will still have equal value. It also works if we add negative numbers! For example, we can add -9 to each side of the equation.

Because expressions represent numbers, we can also add expressions to each side of an equation. For example, we can add $2x$ to each side and still maintain equality.

If we multiply or divide the expressions on each side of an equation by the same number, we will also maintain the equality (as long as we do not divide by zero).

or

Now we can see that $x = \text-2$ is the solution to our equation.

How do we make sure that the solution we find for an equation is correct? Accidentally adding when we meant to subtract, missing a negative when we distribute, forgetting to write an $x$ from one line to the next are some of the many possible mistakes to watch out for!

Fortunately, each valid step we take to solve an equation results in a new equation with the same solution as the original. This means that we can check our work by substituting the value of the solution into the original equation. For example, suppose we solve the following equation:

$\begin{aligned} 2x&=\text-3(x+5)\\ 2x&=\text-3x+15\\ 5x&=15\\ x&=3 \end{aligned}$

Because the last equation shows that $x$ equals 3, and because valid steps make equivalent equations, we can use the equivalence in the original equation to check that all of the steps are valid. Substituting 3 in place of $x$ into the original equation,

$\begin{aligned} 2(3) &= \text-3(3+5)\\ 6&= \text-3(8)\\ 6&=\text-24 \end{aligned}$

we get a statement that isn't true! This tells us we must have made a mistake somewhere. Checking our original steps carefully, we made a mistake when distributing -3. Fixing it, we now have

$\begin{aligned} 2x&=\text-3(x+5)\\ 2x&=\text-3x-15\\ 5x&=\text-15\\ x&=\text-3 \end{aligned}$

Substituting -3 in place of $x$ into the original equation to make sure we didn't make another mistake:

$\begin{aligned} 2(\text-3) &= \text-3(\text-3+5)\\ \text-6&= \text-3(2)\\ \text-6&=\text-6 \end{aligned}$

This equation is true, so $x=\text-3$ is the solution.

When we have an equation in one variable, there are many different ways to solve it. We generally want to make moves that get us closer to an equation that clearly shows the value that makes the equation true.

For example, $x=5$ or $t = \frac{7}{3}$ show that 5 and $\frac{7}{3}$ are solutions. Because there are many ways to do this, it helps to choose moves that leave fewer terms or factors.

If we have an equation like $3t + 5 = 7$, adding -5 to each side will leave us with fewer terms. The equation then becomes $3t = 2$.

Dividing each side of this equation by 3 results in the equivalent equation $t = \frac{2}{3}$, which is the solution.

Or, if we have an equation like $4(5 - a) = 12$, dividing each side by 4 will leave us with fewer factors on the left. The equation then becomes $5-a = 3$.

Here is a list of valid moves that can help create equivalent equations that move toward a solution:

1. Use the distributive property so that all the expressions no longer have parentheses.
2. Collect like terms on each side of the equation.
3. Add or subtract an expression on each side so that there is a variable on just one side.
4. Add or subtract an expression on each side so that there is just a number on the side without the variable.
5. Multiply or divide by a number on each side so that the variable on one side of the equation has a coefficient of 1.

For example, suppose we want to solve $9-2b + 6 =\text-3(b+5) + 4b$.

\begin{aligned} \text{Use the distributive property}&&9 - 2b + 6 &= \text-3b - 15 + 4b\\ \text{Combine like terms}&&15 - 2b &= b - 15\\ \text{Add \(2b to each side}&&15 &= 3b - 15\\ \text{Add 15 to each side}&&30 &= 3b\\ \text{Divide each side by 3}&&10 &= b\\ \end{align}\)

From lots of experience, we learn when to use different valid moves that help solve an equation.

Sometimes we are asked to solve equations with a lot of things happening on each side. For example:

$x-2(x+5)=\dfrac{3(2x-20)}{6}$

This equation has variables on each side, parentheses, and even a fraction to think about. Before we start distributing, let's take a closer look at the fraction on the right side. The expression $2x-20$ is being multiplied by 3 and divided by 6, which is the same as just dividing by 2, so we can re-write the equation as:

$x-2(x+5)=\dfrac{2x-20}{2}$

But now it’s easier to see that all the terms in the numerator of right side are divisible by 2, which means we can re-write the right side again as:

$x-2(x+5)=x-10$

At this point, we could do some distribution and then collect like terms on each side of the equation. Another choice would be to use the structure of the equation. Both the left and the right side have something being subtracted from $x$. But, if the two sides are equal, that means the things being subtracted on each side must also be equal. Thinking this way, the equation can now be rewritten with fewer terms as:

$2(x+5)=10$

Only a few steps left! But what can we tell about the solution to this problem right now? Is it positive? Negative? Zero? Well, the 2 and the 5 multiplied together are 10, so that means the 2 and the $x$ multiplied together cannot have a positive or a negative value. Finishing the steps we have:

$\begin{aligned} &&2(x+5)&=10\\ \text{Divide each side by 2}&&x+5&=5\\ \text{Subtract 5 from each side}&&x &=0\\ \end{aligned}$

Neither positive nor negative. Just as predicted.

An equation is a statement that says that two expressions have an equal value.

The equation $2x = 6$ has one and only one solution, because there is only one number that you can double to get 6.

Some equations are true no matter what the value of the variable is.

Some equations have no solutions. For example, $x=x+1$ has no solutions, because no matter what the value of $x$ is, it can’t equal 1 more than itself.

When we solve an equation, we are looking for the values of the variable that make the equation true. When we try to solve the equation, we make valid moves assuming it has a solution. Sometimes we make valid moves and get an equation like this:

$\displaystyle 8 = 7$

This statement is false, so it must be that the original equation had no solution at all.

Imagine a full 1,500 liter water tank that springs a leak, losing 2 liters per minute. We could represent the number of liters left in the tank with the expression $\text-2x+1,500$, where $x$ represents the number of minutes the tank has been leaking.

Now imagine at the same time, a second tank has 300 liters and is being filled at a rate of 6 liters per minute. We could represent the amount of water in liters in this second tank with the expression $6x+300$, where $x$ represents the number of minutes that have passed.

Since one tank is losing water and the other is gaining water, at some point they will have the same amount of water—but when? Asking when the two tanks have the same number of liters is the same as asking when $\text-2x+1,500$ (the number of liters in the first tank after $x$ minutes) is equal to $6x+300$ (the number of liters in the second tank after $x$ minutes),

$\text-2x+1,500=6x+300$

Solving for $x$ gives us $x=150$ minutes. So after 150 minutes, the number of liters of the first tank is equal to the number of liters of the second tank. We can check our answer and find the number of liters in each tank by substituting 150 for $x$ in the original expressions.

Using the expression for the first tank, we get $\text-2(150)+1,500$ which is equal to $\text-300+1,500$, or 1,200 liters.

If we use the expression for the second tank, we get $6(150)+300$, or just $900+300$, which is also 1,200 liters. That means that after 150 minutes, each tank has 1,200 liters.

We studied linear relationships in an earlier unit. We learned that values of $x$ and $y$ that make an equation true correspond to points $(x,y)$ on the graph.

For example, let’s plan the base rocks for a terrarium. We have $x$ pounds of river rocks that cost $0.80 per pound and $y$ pounds of unpolished rocks that cost $0.50 per pound, and the total cost is $9.00, so we can write an equation like this to represent the relationship between $x$ and $y:$ $0.8x + 0.5y = 9$

Because 5 pounds of river rocks cost $4.00 and 10 pounds of unpolished rocks cost $5.00, we know that $x=5$, $y=10$ is a solution to the equation, and the point $(5,10)$ is a point on the graph.

The line shown is the graph of the equation. Notice that there are 2 points shown that are not on the line. What do they mean in the context?

<p>The graph of a line in the x y plane. The line slants downward and right, crosses the y axis at 18, and passes through the point 5 comma 10. Two additional points, 9 comma 16 and 1 comma 14, are labeled on the graph.</p>  

The point $(1,14)$ means that there is 1 pound of river rock and 14 pounds of unpolished rocks. The total cost for this is $0.8 \boldcdot 1 + 0.5 \boldcdot 14$ or $7.80. Because the cost is not $9.00, this point is not on the line. Likewise, 9 pounds of river rocks and 16 pounds of unpolished rocks cost $0.8 \boldcdot 9 + 0.5 \boldcdot 16$ or $15.20, so the other point is not on the line either.

Suppose we also know that the river rocks and unpolished rocks together weigh 15 pounds. That means that $x+y=15$.

If we draw the graph of this equation on the same coordinate plane, we see it passes through 2 of the 3 labeled points:

<p>The graph of two intersecting lines in the x y plane. The first line slants downward and right, crosses the y axis at 18, and passes through the point 5 comma 10. The second line slants downward and to the right and passes through the points 1 comma 14 and 5 comma 10. An additional point, 9 comma 16, is labeled on the graph.</p>  

The point $(1,14)$ is on the graph of $x+y=15$ because $1 + 14 = 15$. Similarly, $5 + 10 = 15$. But $9 + 16 \neq 15$, so $(9, 16)$ is not on the graph of $x+y = 15$.

In general, if we have 2 lines in the coordinate plane and we have their corresponding equations,

• The coordinates of a point on a line make that equation true.
• The coordinates of a point off of a line make that equation false.
• The coordinates of a point that is the intersection of the 2 lines make both equations true.

The solutions to an equation correspond to points on its graph. For example, if Car A is traveling 75 miles per hour and passes a rest area when $t = 0$, then the distance in miles it has traveled from the rest area after $t$ hours is

$\displaystyle d = 75t$

The point $(2, 150)$ is on the graph of this equation because it makes the equation true ($150 = 75 \boldcdot 2$). This means that 2 hours after passing the rest area, the car has traveled 150 miles.

If you have 2 equations, you can ask whether there is an ordered pair that is a solution to both equations simultaneously. For example, if Car B is traveling toward the rest area, and its distance from the rest area is

$\displaystyle d = 14 - 65t$

We can ask if there is ever a time when the distance of Car A from the rest area is the same as the distance of Car B from the rest area. If the answer is yes, then the solution will correspond to a point that is on both lines.

Graph of 2 lines, origin O, with grid. Horizontal axis, time in hours, scale 0 to point 22, by point 0 2’s. Vertical axis, distance in miles, scale 0 to 14, by 2’s. One line passes through the origin and the point 0 point 1 comma 7 point 5. Another line crosses the y axis at 14 and passes through the point 0 point 1 comma 7 point 5  

Looking at the coordinates of the intersection point, we see that Car A and Car B will both be 7.5 miles from the rest area after 0.1 hours (which is 6 minutes).

Now suppose another car, Car C, also passes the rest stop at time $t=0$ and travels in the same direction as Car A, also going 75 miles per hour. It's equation is also $d=75t$. Any solution to the equation for Car A is also a solution for Car C, and any solution to the equation for Car C is also a solution for Car A. The line for Car C is on top of the line for Car A. In this case, every point on the graphed line is a solution to both equations, so there are infinitely many solutions to the question, “When are Car A and Car C the same distance from the rest stop?” This means that Car A and Car C are side by side for their whole journey.

When we have two linear equations that are equivalent to each other, like $y = 3x+2$ and $2y = 6x +4$, we get 2 lines that are right on top of each other. Any solution to one equation is also a solution to the other, so these 2 lines intersect at infinitely many points.

A system of equations is a set of 2 or more equations, where the variables represent the same unknown values. For example, suppose that two different kinds of bamboo are planted at the same time. Plant A starts at 6 ft tall and grows at a constant rate of $\frac14$ foot each day. Plant B starts at 3 ft tall and grows at a constant rate of $\frac12$ foot each day. Because Plant B grows faster than Plant A, it will eventually be taller, but when?

Solving a system of equations means to find the values of $x$ and $y$ that make both equations true at the same time. One way we have seen to find the solution to a system of equations is to graph both lines and find the intersection point. The intersection point represents the pair of $x$ and $y$ values that makes both equations true.

Here is a graph for the bamboo example:

<p>Graph of two lines, origin O, with grid. Horizontal axis, time in days, scale 0 to 13, by 1’s. Vertical axis, height in feet, scale 0 to 12, by 1’s. A line, labeled Plant A, crosses the y axis at 6. A line, labeled Plant B, crosses the y axis at 3. The lines intersect at the point 12 comma 9.</p>  

The solution to this system of equations is $(12,9)$, which means that both bamboo plants will be 9 feet tall after 12 days.

We have seen systems of equations that have no solutions, one solution, and infinitely many solutions.

• When the lines do not intersect, there is no solution. (Lines that do not intersect are parallel.)
• When the lines intersect once, there is one solution.
• When the lines are right on top of each other, there are infinitely many solutions.

$\displaystyle \begin{cases} y = \text{[some stuff]}\\ y = \text{[some other stuff]} \end{cases}$

we know that we are looking for a pair of values $(x,y)$ that makes both equations true. In particular, we know that the value for $y$ will be the same in both equations. That means that

$\displaystyle \text{[some stuff]} = \text{[some other stuff]}$

But this is only half of what we are looking for: we know the value for $x$, but we need the corresponding value for $y$.

Since both equations have the same $y$ value, we can use either equation to find the $y$-value: $2(\text-2) + 6$ or $y = \text-3(\text-2) -4$.

In both cases, we find that $y = 2$. So the solution to the system is $(\text-2,2)$. We can verify this by graphing both equations in the coordinate plane.

<p>Graph of two lines line, origin O, with grid. Horizontal axis, x, scale negative 4 to 1, by 1s. Vertical axis, y, scale negative 1 to 4, by 1’s. The lines intersect at the point negative 2 comma 2. </p>  

In general, a system of linear equations can have:

• No solutions. In this case, the lines that correspond to each equation never intersect. They have the same slope and different $y$-intercepts.
• Exactly one solution. The lines that correspond to each equation intersect in exactly one point. They have different slopes.
• An infinite number of solutions. The graphs of the two equations are the same line! They have the same slope and the same $y$-intercept.

When we have a system of linear equations where one of the equations is of the form $y = \text{[stuff]}$ or $x=\text{[stuff]}$, we can solve it algebraically by using a technique called substitution. The basic idea is to replace a variable with an expression that it is equal to (so the expression is like a substitute for the variable). For example, let's start with the system:

$\displaystyle \begin{cases} y = 5x\\2x - y = 9 \end{cases}$

Because we know that $y = 5x$, we can substitute $5x$ for $y$ in the equation $2x - y = 9$,

$\displaystyle 2x - (5x) = 9$

and then solve the equation for $x$,

$\displaystyle x =\text -3$

We can find $y$ using either equation. Using the first one, $y = 5 \boldcdot \text-3$.
So $(\text-3,\text -15)$ is the solution to this system.

We can verify this by looking at the graphs of the equations in the system:

Sure enough! They intersect at $(\text-3, \text-15)$.

<p>Graph of two lines, origin O, with grid. Horizontal axis, x, scale negative 10 to 10, by 2’s. Vertical axis, y, scale negative 30 to 30, by 10’s. One line is labeled as y equals 5 x. Another line is labeled as 2 x minus y equals 9. The lines intersect at negative 3 comma negative 15. </p>  

We didn't know it at the time, but we were actually using substitution in the last lesson as well. In that lesson, we looked at the system

$\begin{cases} y = 2x + 6 \\ y  = \text-3x - 4 \end{cases}$

We substituted $2x+6$ for $y$ into the second equation to get $2x+6=\text-3x-4$. Go back and check for yourself!