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Math 7 | Unit 1 2 3 4 5 6 7 8 9


Representing Situations of the Form PX+Q=R and P(X+Q)=R

In this unit, your student will be representing situations with diagrams and equations. There are two main categories of situations with associated diagrams and equations.

Here is an example of the first type: A standard deck of playing cards has four suits. In each suit, there are 3 face cards andĀ xĀ other cards. There are 52 total cards in the deck. A diagram we might use to represent this situation is:

and its associated equation could beĀ 52=4(3+x). There are 4 groups of cards, each group containsĀ x+3Ā cards, and there are 52 cards in all.

Here is an example of the second type: A chef makes 52 pints of spaghetti sauce. She reserves 3 pints to take home to her family, and divides the remaining sauce equally into 4 containers. A diagram we might use to represent this situation is:

and its associated equation could beĀ 52=4x+3. From the 52 pints of sauce, 3 were set aside, and each of 4 containers holdsĀ xĀ pints of sauce.

Here is a task to try with your student:

  1. Draw a diagram to represent the equationĀ 3x+6=39
  2. Draw a diagram to represent the equationĀ 39=3(y+6)
  3. Decide which story goes with which equation-diagram pair:
    • Three friends went cherry picking and each picked the same amount of cherries, in pounds. Before they left the cherry farm, someone gave them an additional 6 pounds of cherries. Altogether, they had 39 pounds of cherries.
    • One of the friends made three cherry tarts. She put the same number of cherries in each tart, and then added 6 more cherries to each tart. Altogether, the three tarts contained 39 cherries.
    Ā 

Solution:

Diagram A representsĀ 3x+6=39Ā and the story about cherry picking. Diagram B representsĀ 3(y+6)=39Ā and the story about making cherry tarts.Ā 

Solving Equations of the Form PX+Q=R and P(X+Q)=R and Problems that Lead to Those Questions

Your student is studying efficient methods to solve equations and working to understand why these methods work. Sometimes to solve an equation, we can just think of a number that would make the equation true. For example, the solution toĀ 12āˆ’c=10Ā is 2, because we know thatĀ 12āˆ’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 isĀ doing the same thing to each side. For example, let's show how we might solveĀ -4(xāˆ’1)=20Ā by doing the same thing to each side.

-4(xāˆ’1)-14ā‹…-4(xāˆ’1)xāˆ’1xāˆ’1+1x=24=-14ā‹…24=-6=-6+1=-5multiply each side byĀ -14Ā add 1 to each side

Another helpful tool for solving equations is to apply the distributive property. In the example above, instead of multiplying each side byĀ -14, you could apply the distributive property toĀ -4(xāˆ’1)Ā and replace it withĀ -4x+4. Your solution would look like this:

-4(xāˆ’1)-4x+4-4x+4āˆ’4-4x-4xĆ·-4x=24=24=24āˆ’4=20=20Ć·-4=-5apply the distributive propertysubtract 4 from each sidedivide each side byĀ -4

Here is a task to try with your student:

Elena picks a number, adds 45 to it, and then multiplies byĀ 12. The result is 29. Elena says that you can find her number by solving the equationĀ 29=12(x+45).

Find Elenaā€™s number. Describe the steps you used.

Solution:

Elenaā€™s number was 13. There are many different ways to solve her equation. Here is one example:

292ā‹…295858āˆ’4513=12(x+45)=2ā‹…12(x+45)=x+45=x+45āˆ’45=xmultiply each side byĀ 2subtract 45 from each side

Inequalities

This week your student will be working with inequalities (expressions withĀ >orĀ <Ā instead ofĀ =). 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 inequalityĀ aā‰„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 hasĀ $10.50, and he earnsĀ $3 each time he runs an errand for his neighbor. Noah wants to know how many errands he needs to run to have at leastĀ $30, so he writes this inequality:

3e+10.50ā‰„30

Ā 

We can test this inequality for different values ofĀ e. For example, 4 errands is not enough for Noah to reach his goal, becauseĀ 3ā‹…4+10.50=22.5, andĀ $22.50 is less thanĀ $30.

  1. Will Noah reach his goal if he runs:
    1. 8 errands?
    2. 9 errands?
    Ā 
  2. What value ofĀ eĀ makes the equationĀ 3e+10.50=30Ā true?
  3. What does this tell you about all the solutions to the inequalityĀ 3e+10.50ā‰„30?
  4. What does this mean for Noahā€™s situation?

SolutionsĀ 

  1. Ā 
    1. Yes, if Noah runs 8 errands, he will haveĀ 3ā‹…8+10.50, orĀ $34.50.
    2. Yes, since 9 is more than 8, and 8 errands was enough, so 9 will also be enough.
    Ā 
  2. The equation is true whenĀ e=6.5. We can rewrite the equation asĀ 3e=30āˆ’10.50, orĀ 3e=19.50. Then we can rewrite this asĀ e=19.50Ć·3, orĀ e=6.5.
  3. This means that whenĀ eā‰„6.5Ā then Noahā€™s inequality is true.
  4. Noah canā€™t really run 6.5 errands, but he could run 7 or more errands, and then he would have more thanĀ $30.

Writing Equivalent Expressions

This week your student will be working with equivalent expressions (expressions that are always equal, for any value of the variable). For example,Ā 2x+7+4xĀ andĀ 6x+10āˆ’3Ā are equivalent expressions. We can see that these expressions are equal when we try different values forĀ x.

Row 1

Ā 
2x+7+4x
6x+10āˆ’3
Row 2 whenĀ xĀ is 5
2ā‹…5+7+4ā‹…5
10+7+20
37
6ā‹…5+10āˆ’3
30+10āˆ’3
37
Row 3 whenĀ xĀ is -1
2ā‹…-1+7+4ā‹…-1
-2+7+-4
1

Ā 

6ā‹…-1+10āˆ’3
-6+10āˆ’3
1

Ā 

We can also use properties of operations to see why these expressions have to be equivalentā€”they are each equivalent to the expressionĀ 6x+7.

Here is a task to try with your student:

Match each expression with an equivalent expression from the list below. One expression in the list will be left over.Ā 

  1. 5x+8āˆ’2x+1
  2. 6(4xāˆ’3)
  3. (5x+8)āˆ’(2x+1)
  4. -12x+9

List:

  • 3x+7
  • 3x+9
  • -3(4xāˆ’3)
  • 24x+3
  • 24xāˆ’18

Solution

  1. 3x+9Ā is equivalent toĀ 5x+8āˆ’2x+1, becauseĀ 5x+-2x=3xĀ andĀ 8+1=9.
  2. 24xāˆ’18Ā is equivalent toĀ 6(4xāˆ’3), becauseĀ 6ā‹…4x=24xĀ andĀ 6ā‹…-3=-18.
  3. 3x+7Ā is equivalent toĀ (5x+8)āˆ’(2x+1), becauseĀ 5xāˆ’2x=3xĀ andĀ 8āˆ’1=7.
  4. -3(4xāˆ’3)Ā is equivalent toĀ -12x+9, becauseĀ -3ā‹…4x=-12xĀ andĀ -3ā‹…-3=9.

IM 6ā€“8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6ā€“8 Math Curriculum is available at https://openupresources.org/math-curriculum/.


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