Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y...

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Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1. 3x – 2 when x = 5 2. 7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12 5. 9m + 3n when m = –1 and n = 2 13 19 –7 1 –3 2 p 4 1

Transcript of Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y...

Page 1: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Warm-Up 10.28.15Evaluate each expression for the given value(s).

1. 3x – 2 when x = 5

2. 7 – 6y when y = –2

3. – 12 when p = 10

4. d + (–2) when d = 12

5. 9m + 3n when m = –1 and n = 2

2

p

4

1

13

19

–7

1

–3

Page 2: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Solving One-Step Equations

Use inverse operations to solve one-step equations.

Lesson 4.2

Page 3: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

The Properties of Equality

For any numbers a, b and c:

Subtraction Property of Equality If a = b, then a – c = b – c

Addition Property of EqualityIf a = b, then a + c = b + c

Multiplication Property of Equality If a = b, then ac = bc

Division Property of Equality If a = b, then c

b

c

a

Page 4: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Explore! Introduction to Equation Mats

Step 1 If you do not have an equation mat, draw one like the one seen below on a blank sheet of paper.

Step 2 On your equation mat, place a variable cube on one side with 3 negative integer chips. On the other side of the mat, place 5 positive integer chips. This represents the equation x − 3 = 5.

Page 5: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Explore! Introduction to Equation Mats

Step 3 In order to get the variable by itself, you must cancel out the 3 negative integer chips with the variable. Use zero pairs to remove the chips by adding three positive integer chips to the left side of the mat. Whatever you add on one side of the mat, add on the other side of the mat. This is using the Addition Property of Equality. How many chips are on the right side of the mat? What does this represent?

Step 4 Clear your mat and place chips and variable cubes on the mat to represent the equation 4x = −8. Draw this on your own paper.

Page 6: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Explore! Introduction to Equation Mats

Step 5 Divide the integer chips equally among the variable cubes. This is using the Division Property of Equality. Each variable cube is equal to how many integer chips? Write your answer in the form x = ___.

Step 6 Model an addition equation on your mat. Record the algebraic equation on your paper.

Step 7 Solve your equation. What does your variable equal?

Page 7: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Good to Know!

You can solve equations using inverse operations to keep your equation balanced. Inverse operations are operations that undo each other, such as addition and subtraction.

Even though you may be able to solve many one-step equations mentally, it is important that you show your work. The equations you will be solving in later lessons and in future math classes will become more complex.

Drawing a vertical line through the equals sign can help you stay organized. Whatever is done on one side of the line to cancel out a value must be done on the other side.

Page 8: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Example 1Solve each equation. Show your work and check your solution.

a. x + 13 = 41

The inverse operation ofaddition is subtraction.

Subtract 13 from both sidesof the equation to isolatethe variable.

Check the answer by substitutingthe solution into the original equationfor the variable.

x + 13 = 41 –13 –13 x = 28

(28) + 13 = 41 41 = 41

The vertical line can help you stay organized.

Page 9: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Example 1 Continued…Solve each equation. Show your work and check your solution.

b. 6m = 27

Divide both sides of theequation by 6.

Check the solution.

6m = 27 6 6

m = 4.5

6(4.5) = 27 27 = 27

or m = 412

Page 10: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Example 1 Continued…Solve each equation. Show your work and check your solution.

c.

Multiply both sides ofthe equation by 3.

Check the solution.

3

4y

3

y 12

44

3

)12(4

?

43

y

3

The variable can be on either side of the equation.

Page 11: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Example 2The meteorologist on Channel 3 announced that a recordhigh temperature has been set in Kirkland today. The newrecord is 2.8º F more than the old record. Today’s hightemperature was 98.3º F. What was the old record?

Let x represent the old record temperature.The equation with represents the situation is:

Subtract 2.8 from each side of the equationto isolate the variable.

Check

The old record high temperature in Kirklandwas 95.5º F.

x + 2.8 = 98.3

x + 2.8 = 98.3– 2.8 – 2.8

x = 95.5

95.5 + 2.8 98.398.3 = 98.3

Page 12: Warm-Up 10.28.15 Evaluate each expression for the given value(s). 1.3x – 2 when x = 5 2.7 – 6y when y = –2 3. – 12 when p = 10 4. d + (–2) when d = 12.

Exit ProblemsSolve each equation. Show your work and check your solution.

1. x + 39 = 150

2.

3. 12.9 = y – 14.2

4. –5t = 55

35

d

d = –15

t = –11

x = 111

y = 27.1