lesson 12: solving equations€¦ · · 2016-09-19because when adding/...

ALGEBRA I

M1 Hart Interactive – Algebra 1

Lesson 12

Lesson 12: Solving Equations

Exploratory Exercises

1. Alonzo was correct when he said the following equations had the same solution set. Discuss with your partner why Alonzo was correct.

(𝑥𝑥 − 1)(𝑥𝑥 + 3) = 17 + 𝑥𝑥 (𝑥𝑥 − 1)(𝑥𝑥 + 3) = 𝑥𝑥 + 17

2. He then said that (𝑥𝑥 − 1)(𝑥𝑥 + 3) = 17 + 𝑥𝑥 and (𝑥𝑥 − 1)(𝑥𝑥 + 3) + 500 = 517 + 𝑥𝑥 should have the same solution set. Is he correct? Explain your reasoning.

3. Finally, Alonzo said the equations (𝑥𝑥 − 1)(𝑥𝑥 + 3) = 17 + 𝑥𝑥 and 3(𝑥𝑥 − 1)(𝑥𝑥 + 3) = 51 + 3𝑥𝑥 should have the same solution set. What do you think? Why?


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Associative property: switching groups(a+b)+c=a+(b+c)Commutative property: switching order/placea+b=b+cDistributive property:distribute evenly to each terma(b+c)= ab+bcYes, correct. They're same equations because of commutative property.Yes! They have same solution set because they're equivalent equations. Because when adding/ subtracting/multiplying/dividing the same number to bothSides of equation, the solutions are the same.They have same solution set because of multiplying 3 to both sides.

ALGEBRA I


Lesson 12

4. Consider the equation 𝑥𝑥2 + 1 = 7 − 𝑥𝑥.

a. Verify that this has the solution set {−3, 2}. Draw this solution set as a graph on the number line. We will later learn how to show that these happen to be the ONLY solutions to this equation.

b. Let’s add 4 to both sides of the equation and consider the new equation 𝑥𝑥2 + 5 = 11− 𝑥𝑥. Verify 2 and −3 are still solutions.

c. Let’s now add 𝑥𝑥 to both sides of the equation and consider the new equation 𝑥𝑥2 + 5 + 𝑥𝑥 = 11. Are 2 and −3 still solutions?

d. Let’s add −5 to both sides of the equation and consider the new equation 𝑥𝑥2 + 𝑥𝑥 = 6. Are 2 and −3 still solutions?

e. Let’s go back to part (d) and add 3𝑥𝑥3 to both sides of the equation and consider the new equation 𝑥𝑥2 + 𝑥𝑥 + 3𝑥𝑥3 = 6 + 3𝑥𝑥3. Are 2 and −3 still solutions?


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ALGEBRA I


Lesson 12

From Exercise 4, whenever 𝑎𝑎 = 𝑏𝑏 is true, then 𝑎𝑎 + 𝑐𝑐 = 𝑏𝑏 + 𝑐𝑐 will also be true for all real numbers 𝑐𝑐. 5. What if 𝑎𝑎 = 𝑏𝑏 is false? Will 𝑎𝑎 + 𝑐𝑐 = 𝑏𝑏 + 𝑐𝑐 will also be false? 6. Is it also okay to subtract a number from both sides of the equation? Explain your reasoning.

7. Let’s go back to Exercise 4 and this time multiply both sides by 16 to get 𝑥𝑥2+𝑥𝑥6 = 1. Are 2 and −3 still

solutions? Whenever 𝑎𝑎 = 𝑏𝑏 is true, then 𝑎𝑎𝑐𝑐 = 𝑏𝑏𝑐𝑐 will also be true, and whenever 𝑎𝑎 = 𝑏𝑏 is false, 𝑎𝑎𝑐𝑐 = 𝑏𝑏𝑐𝑐 will also be false for all nonzero real numbers 𝑐𝑐. So, we have said earlier that applying the distributive, associative, and commutative properties does not change the solution set, and now we see that applying the additive and multiplicative properties of equality also preserves the solution set (does not change it).

8. Is 𝑥𝑥 = 5 an equation? If so, what is its solution set? This example is so simple that it may be hard to wrap your brain around, but it points out that if ever we have an equation that is this simple, we know its solution set. 9. Determine each solution set.

(a) 𝑤𝑤2 = 64 (b) 7 + 𝑃𝑃 = 5 (c) 3𝛽𝛽 = 10


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ALGEBRA I


Lesson 12

Here’s the strategy: If we are faced with the task of solving an equation, that is, finding the solution set of the equation:

Use the commutative, associative, distributive properties AND

Use the properties of equality (adding, subtracting, multiplying by non-zeros, dividing by non-zeros)

to keep rewriting the equation into one whose solution set you easily recognize. (We observed that the solution set will not change under these operations.) This usually means rewriting the equation so that all the terms with the variable appear on one side of the equation.

10. To illustrate this idea, in this exercise you and your team will solve the equation 3𝑥𝑥 + 4 = 8𝑥𝑥 − 16 starting in four different ways. Determine who will be Student 1, 2, 3 and 4. Then solve for 𝑥𝑥 using the given starting point.

3𝑥𝑥 + 4 = 8𝑥𝑥 − 16

Student 1 Student 2 Student 3 Student 4

Subtract 3𝑥𝑥 from both sides

Subtract 4 from both sides

Subtract 8𝑥𝑥 from both sides Add 16 to both sides

11. Did everyone get the same value for x? If not, check to see where a mistake was made. Remember you

must do the same thing to both sides of the equation to keep it balanced.


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ALGEBRA I


Lesson 12

Lesson Summary

Homework Problem Set

Solve the following equations, check your solutions, and then graph the solution sets.

1. −16− 6𝑣𝑣 = −2(8𝑣𝑣 − 7) 2. 2(6𝑏𝑏 + 8) = 4 + 6𝑏𝑏 3. (𝑥𝑥 − 2)(𝑥𝑥 − 2) = x2 − 8

4. 7 − 8𝑥𝑥 = 7(1 + 7𝑥𝑥) 5. 39 − 8𝑛𝑛 = −8(3 + 4𝑛𝑛) + 3𝑛𝑛 6. (𝑥𝑥 − 1)(𝑥𝑥 + 5) = 𝑥𝑥2 + 4𝑥𝑥 − 2

If 𝑥𝑥 is a solution to an equation, it will also be a solution to the new equation formed when the same number is added to (or subtracted from) each side of the original equation or when the two sides of the original equation are multiplied by (or divided by) the same nonzero number. These are referred to as the properties of equality.

If you are faced with the task of solving an equation, that is, finding the solution set of the equation:

• Use the commutative, associative, and distributive properties, AND use the properties of equality (adding, subtracting, multiplying by nonzeros, dividing by nonzeros) to keep rewriting the equation into one whose solution set you easily recognize.


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ALGEBRA I


Lesson 12

7. 𝑥𝑥2 − 7 = 𝑥𝑥2 − 6𝑥𝑥 − 7 8. −7 − 6𝑎𝑎 + 5𝑎𝑎 = 3𝑎𝑎 − 5𝑎𝑎 9. 7 − 2𝑥𝑥 = 1 − 5𝑥𝑥 + 2𝑥𝑥

10. 4(𝑥𝑥 − 2) = 8(𝑥𝑥 − 3) − 12 11. −3(1− 𝑛𝑛) = −6− 6𝑛𝑛 12. −21− 8𝑎𝑎 = −5(𝑎𝑎 + 6)

13. −11− 2𝑝𝑝 = 6𝑝𝑝 + 5(𝑝𝑝 + 3) 14. 𝑥𝑥

𝑥𝑥+2 = 4 15. 2 + 𝑥𝑥9 = 𝑥𝑥

3 − 3

16. −5(−5𝑥𝑥 − 6) = −22− 𝑥𝑥 17. 𝑥𝑥+43 = 𝑥𝑥+2

5 18. −5(2𝑟𝑟 − 0.3) + 0.5(4𝑟𝑟 + 3) = −64


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lesson 12: solving equations€¦ · · 2016-09-19because when adding/...

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