Unit 3
Equations and Inequalities
Accelerated 7
Sign in to view assessments and invite other educators
Sign in using your existing Kendall Hunt account. If you don’t have one, create an educator account.
This week your student will be representing situations with diagrams and equations. We will look at two main types of situations.
Here is an example of the first type of situation: A standard deck of playing cards has four suits. In each suit, there are 3 face cards and \(x\) other cards. There are 52 total cards in the deck.
\(52=4(3+x)\)
The diagram and the equation both show that there are 4 groups of cards, each group contains \(x+3\) cards, and there are 52 cards in all.
Here is an example of the second type of situation: A chef makes 52 pints of spaghetti sauce. She sets aside 3 pints to take home to her family. She divides the rest of the sauce equally into 4 containers.
\(52=4x+3\)
The diagram and the equation both show that from the 52 total pints, 3 were set aside, and each of 4 containers holds \(x\) pints of sauce.
Here is a task to try with your student:
Solution:
This week your student will be solving equations. Sometimes to solve an equation, we can just think of a number that would make the equation true. For example, the solution to \(12-c=10\) is 2, because we know that \(12-2=10\). For more complicated equations that may include decimals, fractions, and negative numbers, the solution may not be so obvious.
An important method for solving equations is doing the same thing to each side. For example, let's see how we might solve \(\text-4(x-1)=24\) by doing the same thing to each side.
\(\begin{align} \text-4(x-1) &= 24 \\ \text-\tfrac14 \boldcdot \text-4(x-1) &= \text-\tfrac14 \boldcdot 24 & \text{Multiply each side by }\text-\tfrac14. \\ x-1 &= \text-6 \\ x-1+1 &= \text- 6 + 1 & \text{ Add 1 to each side.} \\ x &= \text-5 \\ \end{align}\)
Another helpful tool for solving equations is to apply the distributive property. For the above example, if we start by using the distributive property, then the solution would look like this:
\(\begin{align} \text-4(x-1) &= 24 \\ \text-4x+4 &= 24 & \text{Apply the distributive property.} \\ \text-4x+4-4 &= 24-4 &\text{Subtract 4 from each side.} \\ \text-4x &= 20 \\ \text-4x \div \text-4 &= 20\div\text-4 & \text{Divide each side by }\text-4. \\ x &= \text-5 \\ \end{align}\)
Either method shows us that the equation \(\text-4(x-1)=24\) is true when \(x=\text-5\).
Here is a task to try with your student:
Elena picks a number, adds 45 to it, and then multiplies by \(\frac12\). The result is 29. Elena says that you can find her number by solving the equation \(29=\frac12(x+45)\).
Find Elena’s number. Describe the steps you used.
Solution:
Elena’s number is 13. There are many different ways to solve her equation. Here is one example:
\(\begin{align} 29 &= \frac12(x+45) \\ 2 \boldcdot 29 &= 2 \boldcdot \frac12 (x+45) & \text{Multiply each side by }2. \\ 58 &= x+45 \\ 58-45 &= x+45-45 & \text{Subtract 45 from each side.} \\ 13 &=x \\ \end{align}\)
This week your student will be working with inequalities (expressions with the symbol \(>\) or \(<\) instead of \(=\)). We use inequalities to describe a range of numbers. For example, in many places you need to be at least 16 years old to be allowed to drive. We can represent this situation with the inequality \(a \geq 16\). We can show all the solutions to this inequality on the number line.
Here is a task to try with your student:
Noah already has \$10.50, and he earns \$3 each time he runs an errand for his neighbor. Noah wants to know how many errands he needs to run to have at least \$30, so he writes this inequality: \(\displaystyle 3e+10.50\geq30\)
We can test this inequality for different values of \(e\). For example, 4 errands is not enough for Noah to reach his goal, because \(3\boldcdot4+10.50 = 22.5\), and \$22.50 is less than \$30.
Solution: