Unit 4 Linear Equations And Linear Systems — Unit Plan

Equations are equivalent if values for the variables that make one equation true also make the other equation true. One way to create equivalent equations is to correctly use valid moves.

Valid moves include:

• Using the distributive property. ($2(x+6)$ is equivalent to $2x + 12$)
• Combining like terms.  ($2x + 1 - x + 5$ is equivalent to $x + 6$)
• Adding the same value to each side.
• Subtracting the same value from each side.
• Multiplying each side by the same non-zero value.
• Dividing each side by the same non-zero value.

For example, all of these equations are equivalent:

For these equations, the valid moves are used correctly, so all of the equations are equivalent. The last equation shows that 3 is the value for $x$ that makes the equation true. Because all of the equations are equivalent, 3 is the value for $x$ that makes each of these equations true.

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.

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.