Euclidean m-Space & Linear Equations Systems of Linear Equations.
Lesson 8: Linear Equations in Disguise
Transcript of Lesson 8: Linear Equations in Disguise
Lesson 8: Linear Equations in Disguise Date: 11/8/13
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Lesson 8: Linear Equations in Disguise
Student Outcomes
Students rewrite and solve equations that are not obviously linear equations using properties of equality.
Lesson Notes In this lesson, students learn that some equations that may not look like linear equations are, in fact, linear. This lesson on solving rational equations is included because of the types of equations students will see in later topics of this module related to slope. Students will recognize these equations as proportions. It is not necessary to refer to these types of equations as equations that contain rational expressions. They can be referred to simply as proportions since students are familiar with this terminology. Expressions of this type will be treated carefully in algebra as they involve a discussion about why the denominator of such expressions cannot be equal to zero. That discussion is not included in this lesson.
Classwork Concept Development (3 minutes)
Some linear equations may not look like linear equations upon first glance. A simple example that you should recognize is
π₯5
=6
12
What do we call this kind of problem and how do we solve it?
This is a proportion. We can solve this by multiplying both sides of the equation by 5. We can also solve it by multiplying each numerator with the other fractionβs denominator.
Students may not think of multiplying each numerator with the other fractionβs denominator because multiplying both sides by 5 requires fewer steps and uses the multiplication property of equality that has been used to solve other equations. If necessary, state the theorem and give a brief explanation.
Theorem. Given rational numbers π΄, π΅, πΆ, and π·, so that π΅ β 0 and π· β 0, the property states
If π΄π΅
=πΆπ·
, then π΄π· = π΅πΆ.
To find the value of π₯, we can multiply each numerator by the other fractionβs denominator. π₯5
=6
12
12π₯ = 6(5)
Lesson 8: Linear Equations in Disguise Date: 11/8/13
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
It should be more obvious now that we have a linear equation. We can now solve it as usual using the properties of equality.
12π₯ = 30
π₯ =3012
π₯ =52
In this lesson, our work will be similar, but the numerator and/or the denominator of the fractions may contain more than one term. However, the way we solve these kinds of problems remain the same.
Example 1 (5 minutes)
Given a linear equation in disguise, we will try to solve it. To do so, we must first assume that the following equation is true for some number π₯.
π₯ β 12
=π₯ + 1
34
We want to make this equation look like the linear equations we are used to. For that reason, we will multiply both sides of the equation by 2 and 4, as we normally do with proportions:
2 οΏ½π₯ +13οΏ½ = 4(π₯ β 1)
Is this a linear equation? How do you know?
Yes, this is a linear equation because the expressions on the left and right of the equal sign are linear expressions.
Notice that the expressions that contained more than one term were put in parentheses. We do that so we donβt make a mistake and forget to use the distributive property.
Now that we have a linear equation. We will use the distributive property and solve as usual.
2 οΏ½π₯ +13οΏ½ = 4(π₯ β 1)
2π₯ +23
= 4π₯ β 4
2π₯ β 2π₯ +23
= 4π₯ β 2π₯ β 4 23
= 2π₯ β 4 23
+ 4 = 2π₯ β 4 + 4 143
= 2π₯ 12β
143
=12β 2π₯
73
= π₯
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
How can we verify that 73
is the solution to the equation?
We can replace π₯ with 73
in the original equation:
π₯ β 12
=π₯ + 1
34
73 β 1
2=
73 + 1
34
43 2
= 83
4
4 οΏ½43οΏ½ = 2 οΏ½
83οΏ½
163
=163
Since 73
made the equation true, then we know it is the solution to the equation.
Example 2 (4 minutes)
Can we solve the following equation? Explain. 15 β π₯
7=
2π₯ + 93
We need to multiply each numerator with the other fractionβs denominator.
So, 15 β π₯
7=
2π₯ + 93
7(2π₯ + 9) = 3 οΏ½15β π₯οΏ½
What would be the next step?
Use the distributive property.
Now we have
7(2π₯ + 9) = 3 οΏ½15β π₯οΏ½
14π₯ + 63 =35β 3π₯
Is this a linear equation? How do you know? Yes, this is a linear equation because the left and right side are linear expressions.
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Example 3 (5 minutes)
Can we solve the following equation? If so, go ahead and solve it. If not, explain why not.
Example 3
Can this equation be solved?
π + π
ππ + ππ
=ππ
Give students a few minutes to work. Provide support to students as needed.
Yes, we can solve the equation because we can multiply each numerator with the other fractionβs denominator and then use the distributive property to begin solving it:
π + π
ππ + ππ
=ππ
(π+ π)π = οΏ½ππ +πποΏ½π
ππ + ππ = πππ + π ππ+ ππ β ππ = πππ β ππ + π
ππ = πππ + π ππ β π = πππ + π β π
ππ = πππ ππππ
= π
Example 4 (5 minutes)
Can we solve the following equation? If so, go ahead and solve it. If not, explain why not.
Example 4
Can this equation be solved?
πππ + π
=ππ
Give students a few minutes to work. Provide support to students as needed.
Yes, we can solve the equation because we can multiply each numerator with the other fractions denominator and then use the distributive property to begin solving it.
π
ππ + π=ππ
π(π) = (ππ + π)π ππ = ππ + π
ππ β π = ππ + π β π ππ = ππ πππ
= π
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Example 5 (5 minutes)
Example 5
In the diagram below, β³ π¨π©πͺ~ β³ π¨β²π©β²πͺβ². Using what we know about similar triangles, we can determine the value of π.
Begin by writing the ratios that represent the corresponding sides.
π β ππ.π
=π + πππ
It is possible to write several different proportions in this case. If time, discuss this fact with students.
Now that we have the ratios, solve for π₯ and find the lengths of π΄π΅ and π΄πΆ.
π β ππ.π
=π + πππ
(π β π)ππ = π.π(π + π) πππ β ππ = π.ππ + ππ
πππ β ππ+ ππ = π.ππ + ππ+ ππ πππ = π.ππ + ππ
πππ β π.ππ = π.ππ β π.ππ + ππ π.ππ = ππ
π =πππ.π
π = ππ.π
The length of |π¨π©| = ππ.π cm and the length of |π¨πͺ| = ππ.π cm.
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Exercises 1β4 (8 minutes)
Students complete Exercises 1β4 independently.
Exercises 1β4
Solve the following equations of rational expressions, if possible.
1. ππ+ππ
= πβππ
Sample student work: ππ + ππ
=π β ππ
π(π β π) = (ππ + π)π π β ππ = πππ + π
π β ππ + ππ = πππ + ππ + π π = πππ + π
π β π = πππ + π β π π = πππ πππ
=ππππ
π ππ
= π
2. π+ππππβπ
= ππ
Sample student work: π+ ππππ β π
=ππ
(π+ ππ)π = (ππ β π)π ππ + πππ = πππ β π
ππ β ππ + πππ = πππ β π β ππ πππ = πππ β ππ
πππ β πππ = πππ β πππ β ππ βππ = βππ βπβπ
π =βππβπ
π =πππ
3. π+πππ
=βππβ ππ
π
Sample student work:
π + πππ
=βππ β π
ππ
πποΏ½βππ βπποΏ½ = (π + π)π
βπππ β π = ππ + ππ βπππ + πππ β π = ππ + πππ + ππ
βπ = πππ + ππ βπβ ππ = πππ + ππ β ππ
βππ = πππ βππππ
=ππππ
π
βπππ
= π
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
4. π
πβππ= π
ππ+ππ
Sample student work: π
π β ππ=
π
ππ + ππ
ποΏ½ππ +πποΏ½ = (π β ππ)π
πππ + π = ππ β πππ πππ + π β π = ππ β π β πππ
πππ = ππ β πππ πππ + πππ = ππ β πππ + πππ
πππ = ππ ππππ
π =ππππ
π =ππππ
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
We know that proportions that have more than one term in the numerator and/or denominator can be solved the same way we normally solve a proportion.
When multiplying a fraction with more than one term in the numerator and/or denominator by a number, we should put the expressions with more than one term in parentheses, so we are less likely to forget to use the distributive property.
Exit Ticket (5 minutes)
π + πππ β π
=ππ
π(π + π) = π(ππβ π)
Lesson Summary
Proportions are linear equations in disguise and are solved the same way we normally solve proportions.
When multiplying a fraction with more than one term in the numerator and/or denominator by a number, put the expressions with more than one term in parentheses so you remember to use the distributive property when transforming the equation. For example:
The equation π(π+ π) = π(ππ β π) is now clearly a linear equation and can be solved using the properties of equality.
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Name Date
Lesson 8: Linear Equations in Disguise
Exit Ticket Solve the following equations for π₯.
1. 5π₯β83
=11π₯β95
2. π₯+117
=2π₯+1β8
3. βπ₯β2β4
=3π₯+62
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Exit Ticket Sample Solutions
1. ππβππ
=πππβππ
ππ β ππ
=πππ β π
π
π(ππβ π) = π(πππ β π)
πππ β ππ = πππ β ππ
πππ β πππ β ππ = πππ β πππ β ππ
βππ = ππ β ππ
βππ+ ππ = ππ β ππ+ ππ
βππ = ππ
βπππ
= π
2. π+πππ
=ππ+πβπ
π + πππ
=ππ + πβπ
π(ππ+ π) = βπ(π + ππ) πππ + π = βππ β ππ
πππ + π β π = βππ β ππ β π πππ = βππ β ππ
πππ + ππ = βππ + ππ β ππ πππ = βππ
π = βππππ
3. βπβπβπ
=ππ+ππ
βπ β πβπ
=ππ + ππ
βπ(ππ + π) = π(βπβ π)
βπππ β ππ = βππ β π βπππ β ππ+ ππ = βππ β π + ππ
βπππ = βππ + ππ βπππ + ππ = βππ + ππ + ππ
βπππ = ππ π = βπ
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
Problem Set Sample Solutions Students practice solving equations with rational expressions, if a solution possible.
Solve the following equations of rational expressions, if possible. If it cannot be solved, explain why.
1. π
ππβπ=
βππ+π
Sample student work:
πππ β π
=βππ + π
π(π + π) = βπ(ππβ π) ππ + π = βππ + π
ππ + π β π = βππ + π β π ππ = βππ β π
ππ + ππ = βππ + ππ β π πππ = βπ
π = βπππ
2. πβππ
=ππβππ
Sample student work:
π β ππ
=ππ β ππ
π(ππβ π) = (π β π)π πππ β π = ππ β ππ
πππ β π + π = ππ + π β ππ πππ = ππ β ππ
πππ + ππ = ππ β ππ + ππ πππ = ππ ππππ
π =ππππ
π =ππππ
3. πππ+π
=ππ
Sample student work:
πππ + π
=ππ
π(ππ) = (π + π)π πππ = ππ + ππ
πππ β ππ = ππ β ππ + ππ πππ = ππ ππππ
π =ππππ
π =πππ
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
4. πππ+π
π=πβππ
Sample student work:
πππ + π
π=π β ππ
π(π β π) = ποΏ½πππ + ποΏ½
ππ β π = π + ππ ππ β π + π = π + ππ + π
ππ = π + ππ ππ β π = π β π + ππ
ππ = ππ
π =πππ
5. πβπππ
=πβππ
Sample student work:
π β πππ
=π β ππ
π(π β π) = (π β ππ)π ππ β ππ = π β ππ
ππ β ππ + ππ = π + ππ β ππ ππ = ππ β ππ
ππ + ππ = ππ β ππ + ππ ππ = ππ πππ =
πππ
π =πππ
6. ππ+ππ
=ππβππ
Sample student work:
ππ + ππ
=ππ β ππ
π(ππβ π) = π(ππ + π) ππ β π = πππ + ππ
ππ β π + π = πππ + ππ + π ππ = πππ + ππ
ππ β πππ = πππ β πππ + ππ βππ = ππ
π = βπππ
π = βπππ
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
7. ππ+ππ
=πβππ
Sample student work:
ππ + ππ
=π β ππ
(ππ+ π)π = π(π β π) πππ + π = ππ β ππ
πππ + π β π = ππ β π β ππ πππ = ππ β ππ
πππ + ππ = ππ β ππ + ππ πππ = ππ ππππ
π =ππππ
π =ππ
8. πππβπ
ππ=βπβπππ
Sample student work:
πππ β πππ
=βπβ πππ
ππ(βπβ π) = οΏ½πππ β ποΏ½ππ
βππ β πππ = ππ β πππ βππ β πππ + πππ = ππ + πππ β πππ
βππ = πππ β πππ βππ + πππ = πππ β πππ + πππ
ππ = πππ ππππ
=ππππ
π ππππ
= π
9. πβππβπ
=ππ
Sample student work:
π β ππ β π
=ππ
(π β π)π = (π β π)π π β ππ = π β ππ
π β ππ + ππ = π β ππ + ππ π = π β π
π β π = π β π β π π = βπ
βπ = π
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NYS COMMON CORE MATHEMATICS CURRICULUM 8β’4 Lesson 8
10. In the diagram below, β³ π¨π©πͺ~ β³ π¨β²π©β²πͺβ². Determine the lengths of π¨πͺ and π©πͺ.
Sample student work:
π + ππ.π
=ππ β ππ
π(π+ π) = π.π(ππ β π) ππ + ππ = ππ.ππ β π
ππ + ππ + π = ππ.ππ β π+ π ππ + ππ = ππ.ππ
ππ β ππ + ππ = ππ.ππ β ππ ππ = π.ππ ππ = π
The length of |π¨πͺ| = ππ ππ and the length of |π©πͺ| = ππ ππ.