EE6-7 Equations Involving Fractions

EE6-7 Equations Involving Fractions

Pages 80?81

STANDARDS 6.EE.B.5, 6.EE.B.7

Vocabulary equation expression solving for a variable

Goals

Students will solve equations of the form x + b = c and x - b = c, where b and c are fractions or mixed numbers.

PRIOR KNOWLEDGE REQUIRED

Can add and subtract fractions and mixed numbers Can verify whether a number solves an equation

Review equations with whole numbers. Write several equations on the board and ask students to solve them.

Exercises: Solve each equation by subtracting the same number from both sides.

a) x + 1 = 6 b) x + 13 = 17 c) 34 = x + 5 d) 12 + x = 44

Answers: a) 5, b) 4, c) 29, d) 32

Substituting fractions for variables and evaluating expressions. Write on the board the expression n + 1/4. Tell students that we can replace n with any number and get a numerical expression. For example, if we let n = 2/4, then this expression becomes 2/4 + 1/4, which is 3/4.

Exercises: Replace n with 1/5 in each expression and evaluate the expression.

a) n + 2 5

b) 4 - n 5

c) 2 - n

d) 1 + n

Answers: a) 3 , b) 3 , c) 9 or 1 4 , d) 6 or 1 1

555

5

5

5

Before students attempt the following exercises, review adding mixed

numbers with regrouping and subtracting mix numbers without regrouping.

Exercises: Replace the variable with the given number, then evaluate. Write the answer in lowest terms.

a) h + 1 4 , h = 3

5

5

b) x - 1 1 , x = 2 3

4

4

c) 3 5 + w, w = 1 7

9

9

Answers: a) 2 2 or 12 , b) 1 2 = 1 1 or 3 , c) 5 3 = 5 1 or 16

5 5

4 22

9 33

Now review subtraction with regrouping. Do the first problem below together,

then have students do the others individually.

d) 4 1 - p, p = 1 5

6

6

e) 5 1 - t, t = 3 2

3

3

f) 2 2 - t, t = 1 3

5

5

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Bonus: z - 3 2 , z = 5 1

7

6

Answers: d) 2 2 = 2 1 or 7 , e) 1 2 or 5 , f) 4 , Bonus: 1 37

6 33

33 5

42

Equations with fractions. Start with a word problem: Jennifer has some pizzas. Her friend brings her another pizza. Now she has 3 pizzas altogether. How many pizzas did she have to start? SAY: To answer this question, I can solve an equation. Write the equation x + 1 = 3 on the board. ASK: How do you solve this equation? (by subtracting 1 from both sides) Ask a volunteer to solve the equation by showing the steps:

x+1 = 3 -1 -1 x= 2

Jennifer started with 2 pizzas.

Ask another word problem: Jennifer's mother cut one of the pizzas into 8 pieces and gave some pieces to Jennifer. Jennifer asked for 1 more piece, and her mother gave it to her. Jennifer now has 3 pieces of pizza. How many pieces did she have to start? (2 pieces) ASK: What fraction of the whole pizza did she have at the start? (2/8) SAY: You can find the fraction of the whole pizza that she had at the start by solving an equation with fractions. Write the following on the board beside the previous equation:

x+ 1 = 3 8 8

Tell students that, in this equation, x is the fraction of the whole pizza that Jennifer had at the start. Explain to students that just as we can subtract a whole (1) from both sides of the equation, we can also subtract a piece (1/8) from both sides of the equation.

x+ 1 8

- 1 8

x

= 3 8

- 1 8

= 2 8

Jennifer started with 2 of a whole pizza. 8

Exercises: Solve each equation by subtracting the same number from both sides. Reduce your answer to lowest terms.

a) x + 2 = 5 7 7

b) x + 1 = 3 4 4

c) 7 = x + 5

8

8

Answers: a) 3 , b) 2 = 1 , c) 2 = 1

7

42 84

Make sure all students can solve equations with simple fractions, as above, then proceed to equations with mixed numbers or improper fractions. Finally, solve equations with fractions that have different denominators. (NOTE: Be sure the answers are positive fractions!)

COPYRIGHT ? 2013?2018 JUMP MATH: NOT TO BE COPIED. US EDITION

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(MP.2, MP.4) (MP.2, MP.4)

Exercises: Solve each equation by subtracting or adding the same number to both sides.

a) x + 1 3 = 4 5 7 7

d) x - 3 = 2 1 5 5

b) 2 7 = x + 1 5

8

8

e) 2 2 = x - 1

5

3

c) 3 2 = x + 3

5

5

2 Answers: a) 3

7

, b) 1 2 8

1

4

= 1 4 , c) 2 5

4 , d) 2

5

, e)

41 15

11 = 2 15

NOTE: In the following extensions, encourage students to model the situation using equations with variables (MP.4) and to say what the variables and equations mean in the context of the problems (MP.2).

Extensions

1.Some friends bought pizza and ate 2 7 pizzas. They had 4 of a pizza

with pineapple left and

1

of

a

pizza

10 without

pineapple

left.

5 2

1

of the

2

2

pizzas they ordered were vegetarian. How many pizzas did they buy?

Which fact did you not need?

Solution: Let x be the number of pizzas the friends bought. The amount

of pizza left can be written as an equation: x - 2 7/10 = 4/5 + 1/2.

Adding the fractions on the right side gives: 4/5 + 1/2 = 8/10 + 5/10

= 13/10 = 1 3/10. The equation becomes: x - 2 7/10 = 1 3/10. Adding

2 7/10 to both sides of the equation gives: x = 1 3/10 + 2 7/10 = 4.

So the friends bought 4 pizzas. I did not need to use the fact that

2 1/2 pizzas were vegetarian.

2.Ron bought some ribbon. He used 1 2 meters to wrap a gift for his aunt

and 1 7

5 meters to wrap gifts for his friends.

3

of his friends are boys.

Ron

10 had

2

4

5 meters left. How much ribbon did Ron buy? Which fact

10

did you not need?

Solution: Let x be the length of ribbon Ron bought. I can write the amount of ribbon left as an equation: x - 1 2/5 - 1 7/10 = 2 4/10. The expression on the left side can be simplified: x - 1 2/5 - 1 7/10 = x - (1 2/5 + 1 7/10) = x - (14/10 + 17/10) = x - 31/10 = x - 3 1/10. The equation becomes: x - 3 1/10 = 2 4/10. Adding 3 1/10 to both sides of the equation gives: x = 2 4/10 + 3 1/10 = 5 5/10 = 5 1/2. So Ron bought 5 1/2 meters of ribbon. I did not need to use the fraction of Ron's friends that are boys.

COPYRIGHT ? 2013?2018 JUMP MATH: NOT TO BE COPIED. US EDITION

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