Lesson 5 - West Ada

Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8β€’4
Lesson 5: Writing and Solving Linear Equations
Lesson Notes
All of the problems in this lesson relate to what students have learned about geometry in recent modules and previous
years. The purpose is two-fold, first, we want to reinforce what students have learned about geometry, and second, we
want to demonstrate a need for writing and solving an equation in a context that is familiar. Throughout the lesson
students solve mathematical problems that relate directly to what students have learned about angle relationships,
congruence, and the triangle sum theorem. Encourage students to draw a diagram to represent the situation presented
in the word problems.
Classwork
Example 1
ο‚§
Solve the following:
Example 1
One angle is five less than three times the size of another angle. Together they have a sum of πŸπŸ’πŸ‘Λš. What are the sizes of
each angle?
MP.1
Provide students with time to make sense of the problem and persevere in solving it. They could begin their work by
guessing and checking, drawing a diagram, or other methods as appropriate. Then move to the algebraic method shown
below.
ο‚§
What do we need to do first to solve this problem?
οƒΊ
ο‚§
If π‘₯ is the size of the first angle, how do you represent the size of the second
angle?
οƒΊ
ο‚§
The second angle is 3π‘₯ βˆ’ 5.
What is the equation that represents this situation?
οƒΊ
ο‚§
First we need to define our variable (symbol). Let π‘₯ be the size of the first
angle.
The equation is: π‘₯ + 3π‘₯ βˆ’ 5 = 143.
Scaffolding:
Model for students how to use
diagrams to help make sense of
the problems throughout this
lesson. Encourage students to
use diagrams to help them
understand the situation.
The equation that represents this situation is: π‘₯ + 3π‘₯ βˆ’ 5 = 143. Solve for π‘₯,
then determine the size of each angle.
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8β€’4
Student work:
𝒙 + πŸ‘π’™ βˆ’ πŸ“ = πŸπŸ’πŸ‘
(𝟏 + πŸ‘)𝒙 βˆ’ πŸ“ = πŸπŸ’πŸ‘
πŸ’π’™ βˆ’ πŸ“ = πŸπŸ’πŸ‘
πŸ’π’™ βˆ’ πŸ“ + πŸ“ = πŸπŸ’πŸ‘ + πŸ“
πŸ’π’™ = πŸπŸ’πŸ–
𝒙 = πŸ‘πŸ•
The size of the first angle is πŸ‘πŸ•Λš. The second angle is πŸ‘(πŸ‘πŸ•) βˆ’ πŸ“ = 𝟏𝟏𝟏 βˆ’ πŸ“ = πŸπŸŽπŸ”Λš.
ο‚§
Compare the method you tried at the beginning of the problem with the algebraic method. What advantage
does writing and solving an equation have?
οƒΊ
ο‚§
Could we have defined π‘₯ to be the size of the second angle? If so, what, if anything, would change?
οƒΊ
ο‚§
ο‚§
If we let π‘₯ be the size of the second angle, then the equation would change, but the answers for the
sizes of the angles should remain the same.
If π‘₯ is the size of the second angle, how would we write the size of the first angle?
οƒΊ
ο‚§
Writing and solving an equation is a more direct method than the one I tried before. It allows me to
find the answer more quickly.
The first angle would be
π‘₯+5
3
.
The equation that represents the situation is π‘₯ +
π‘₯+5
= 143. How should we solve this equation?
3
οƒΊ
We could add the fractions together, then solve for π‘₯.
οƒΊ
We could multiply every term by 3 to change the fraction to a whole number.
Using either method, solve the equation. Verify that the sizes of the angles are the same as before.
οƒΊ
π‘₯+5
= 143
3
3π‘₯ π‘₯ + 5
+
= 143
3
3
3π‘₯ + π‘₯ + 5
= 143
3
3π‘₯ + π‘₯ + 5 = 143(3)
(3 + 1)π‘₯ + 5 = 429
4π‘₯ + 5 = 429
4π‘₯ + 5 βˆ’ 5 = 429 βˆ’ 5
4π‘₯ = 424
π‘₯ = 106
π‘₯+
Lesson 5:
Date:
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Scaffolding:
You may need to remind
students how to add fractions
by rewriting term(s) as
equivalent fractions then
adding the numerators.
Provide support as needed.
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
οƒΊ
The size of the second angle is 106˚. The measure of the first angle is
π‘₯+5
3
=
106+5
3
=
111
3
8β€’4
= 37°.
OR
π‘₯+5
= 143
3
π‘₯+5
3 (π‘₯ +
= 143)
3
3π‘₯ + π‘₯ + 5 = 429
4π‘₯ + 5 = 429
4π‘₯ + 5 βˆ’ 5 = 429 βˆ’ 5
4π‘₯ = 424
π‘₯ = 106
π‘₯+
The size of the second angle is 106˚. The measure of the first angle is
ο‚§
π‘₯+5
3
=
106+5
3
=
111
3
= 37˚.
Whether we let π‘₯ represent the first angle or the second angle does not change our answers. Whether we
solve the equation using the first or second method does not change our answers. What matters is that we
accurately write the information in the problem and correctly use the properties of equality. You may solve a
problem differently than your classmates or teachers. Again, what matters most is that what you do is
accurate and correct.
Example 2
ο‚§
Solve the following:
Example 2
Given a right triangle, find the size of the angles if one angle is ten more than four times the other angle and the third
angle is the right angle.
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8β€’4
Solution One
Let 𝒙 be the size of the first angle. Then the second angle is πŸ’π’™ + 𝟏𝟎. The sum of the measures for the angles for this
right triangle is as follows: 𝒙 + πŸ’π’™ + 𝟏𝟎 + πŸ—πŸŽ = πŸπŸ–πŸŽ.
𝒙 + πŸ’π’™ + 𝟏𝟎 + πŸ—πŸŽ = πŸπŸ–πŸŽ
(𝟏 + πŸ’)𝒙 + 𝟏𝟎𝟎 = πŸπŸ–πŸŽ
πŸ“π’™ + 𝟏𝟎𝟎 = πŸπŸ–πŸŽ
πŸ“π’™ + 𝟏𝟎𝟎 βˆ’ 𝟏𝟎𝟎 = πŸπŸ–πŸŽ βˆ’ 𝟏𝟎𝟎
πŸ“π’™ = πŸ–πŸŽ
𝒙 = πŸπŸ”
The measure of the first angle is πŸπŸ”Λš, the measure of the second angle is πŸ’(πŸπŸ”) + 𝟏𝟎 = πŸ”πŸ’ + 𝟏𝟎 = πŸ•πŸ’Λš, and the third
angle is πŸ—πŸŽΛš.
Solution Two
Let 𝒙 be the size of the first angle. Then the second angle is πŸ’π’™ + 𝟏𝟎. Since we have a right triangle we already know that
one angle is πŸ—πŸŽΛš which means that the sum of the other two angles is πŸ—πŸŽ: 𝒙 + πŸ’π’™ + 𝟏𝟎 = πŸ—πŸŽ.
𝒙 + πŸ’π’™ + 𝟏𝟎 = πŸ—πŸŽ
(𝟏 + πŸ’)𝒙 + 𝟏𝟎 = πŸ—πŸŽ
πŸ“π’™ + 𝟏𝟎 = πŸ—πŸŽ
πŸ“π’™ + 𝟏𝟎 βˆ’ 𝟏𝟎 = πŸ—πŸŽ βˆ’ 𝟏𝟎
πŸ“π’™ = πŸ–πŸŽ
𝒙 = πŸπŸ”
The measure of the first angle is πŸπŸ”Λš, the measure of the second angle is πŸ’(πŸπŸ”) + 𝟏𝟎 = πŸ”πŸ’ + 𝟏𝟎 = πŸ•πŸ’Λš, and the third
angle is πŸ—πŸŽΛš.
Solution Three
Let 𝒙 be the size of the second angle. Then the first angle is
π’™βˆ’πŸπŸŽ
πŸ’
. Since we have a right triangle we already know that
one angle is πŸ—πŸŽΛš which means that the sum of the other two angles is πŸ—πŸŽ: 𝒙 +
π’™βˆ’πŸπŸŽ
= πŸ—πŸŽ.
πŸ’
𝒙 βˆ’ 𝟏𝟎
= πŸ—πŸŽ
πŸ’
𝒙 βˆ’ 𝟏𝟎
πŸ’ (𝒙 +
= πŸ—πŸŽ)
πŸ’
πŸ’π’™ + 𝒙 βˆ’ 𝟏𝟎 = πŸ‘πŸ”πŸŽ
(πŸ’ + 𝟏)𝒙 βˆ’ 𝟏𝟎 = πŸ‘πŸ”πŸŽ
πŸ“π’™ βˆ’ 𝟏𝟎 = πŸ‘πŸ”πŸŽ
πŸ“π’™ βˆ’ 𝟏𝟎 + 𝟏𝟎 = πŸ‘πŸ”πŸŽ + 𝟏𝟎
πŸ“π’™ = πŸ‘πŸ•πŸŽ
𝒙 = πŸ•πŸ’
𝒙+
The measure of the second angle is πŸ•πŸ’Λš, the measure of the first angle is
Lesson 5:
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πŸ•πŸ’βˆ’πŸπŸŽ
πŸ’
=
πŸ”πŸ’
πŸ’
= πŸπŸ”Λš, and the third angle is πŸ—πŸŽΛš.
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8β€’4
Solution Four
Let 𝒙 be the size of the second angle. Then the first angle is
π’™βˆ’πŸπŸŽ
πŸ’
. The sum of the three angles is as follows: 𝒙 +
π’™βˆ’πŸπŸŽ
+
πŸ’
πŸ—πŸŽ = πŸπŸ–πŸŽ.
𝒙 βˆ’ 𝟏𝟎
+ πŸ—πŸŽ = πŸπŸ–πŸŽ
πŸ’
𝒙 βˆ’ 𝟏𝟎
𝒙+
+ πŸ—πŸŽ βˆ’ πŸ—πŸŽ = πŸπŸ–πŸŽ βˆ’ πŸ—πŸŽ
πŸ’
𝒙 βˆ’ 𝟏𝟎
𝒙+
= πŸ—πŸŽ
πŸ’
πŸ’π’™ 𝒙 βˆ’ 𝟏𝟎
+
= πŸ—πŸŽ
πŸ’
πŸ’
πŸ’π’™ + 𝒙 βˆ’ 𝟏𝟎
= πŸ—πŸŽ
πŸ’
πŸ’π’™ + 𝒙 βˆ’ 𝟏𝟎 = πŸ‘πŸ”πŸŽ
πŸ“π’™ βˆ’ 𝟏𝟎 + 𝟏𝟎 = πŸ‘πŸ”πŸŽ + 𝟏𝟎
πŸ“π’™ = πŸ‘πŸ•πŸŽ
𝒙 = πŸ•πŸ’
𝒙+
The second angle is πŸ•πŸ’Λš, the first angle is
πŸ•πŸ’βˆ’πŸπŸŽ
πŸ’
=
πŸ”πŸ’
πŸ’
= πŸπŸ”Λš, and the third angle is πŸ—πŸŽΛš.
Make sure students see at least four different methods of solving the problem. Conclude this example with the
statements below.
ο‚§
Each method was slightly different either in terms of how the variable was defined or how the properties of
equality were used to solve the equation. The way you find the answer may be different than your classmates
or your teacher.
ο‚§
As long as you write accurately and do what is mathematically correct, you will find the correct answer.
Example 3
ο‚§
A pair of alternate interior angles are described as follows. One angle is fourteen more than half a number.
The other angle is six less than half a number. Are the angles congruent?
ο‚§
We will begin by assuming that the angles are congruent. If the angles are congruent, what does that mean
about their measures?
οƒΊ
ο‚§
Write an expression that describes each angle.
οƒΊ
ο‚§
It means that they are equal in measure.
One angle is
π‘₯
2
+ 14 and the other angle is
π‘₯
2
βˆ’ 6.
If the angles are congruent, we can write the equation is
π‘₯
2
+ 14 =
π‘₯
2
βˆ’ 6. We know that our properties of
equality allow us to transform equations while making sure that they remain true.
π‘₯
π‘₯
+ 14 = βˆ’ 6
2
2
π‘₯ π‘₯
π‘₯ π‘₯
βˆ’ + 14 = βˆ’ βˆ’ 6,
2 2
2 2
14 β‰  βˆ’6.
Therefore, our assumption was not correct and the angles are not congruent.
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8β€’4
Exercises 1–6
Exercises 1–6
For each of the following problems, write an equation and solve.
1.
A pair of congruent angles are described as follows: the measure of one angle is three more than twice a number
and the other angle’s measure is πŸ“πŸ’. πŸ“ less than three times the number. Determine the size of the angles.
Let 𝒙 be the number. Then the measure of one angle is πŸ‘ + πŸπ’™ and the measure of the other angle is πŸ‘π’™ βˆ’ πŸ“πŸ’. πŸ“.
Because the angles are congruent, their measures are equal. Therefore,
πŸ‘ + πŸπ’™ = πŸ‘π’™ βˆ’ πŸ“πŸ’. πŸ“
πŸ‘ + πŸπ’™ βˆ’ πŸπ’™ = πŸ‘π’™ βˆ’ πŸπ’™ βˆ’ πŸ“πŸ’. πŸ“
πŸ‘ = 𝒙 βˆ’ πŸ“πŸ’. πŸ“
πŸ‘ + πŸ“πŸ’. πŸ“ = 𝒙 βˆ’ πŸ“πŸ’. πŸ“ + πŸ“πŸ’. πŸ“
πŸ“πŸ•. πŸ“ = 𝒙
Then each angle is πŸ‘ + 𝟐(πŸ“πŸ•. πŸ“) = πŸ‘ + πŸπŸπŸ“ = πŸπŸπŸ–Λš.
2.
The measure of one angle is described as twelve more than four times a number. Its supplement is twice as large.
Find the measure of each angle.
Let 𝒙 be the number. Then the measure of one angle is πŸ’π’™ + 𝟏𝟐. The other angle is 𝟐(πŸ’π’™ + 𝟏𝟐) = πŸ–π’™ + πŸπŸ’. Since
the angles are supplementary, their sum must be πŸπŸ–πŸŽ:
πŸ’π’™ + 𝟏𝟐 + πŸ–π’™ + πŸπŸ’ = πŸπŸ–πŸŽ
πŸπŸπ’™ + πŸ‘πŸ” = πŸπŸ–πŸŽ
πŸπŸπ’™ + πŸ‘πŸ” βˆ’ πŸ‘πŸ” = πŸπŸ–πŸŽ βˆ’ πŸ‘πŸ”
πŸπŸπ’™ = πŸπŸ’πŸ’
𝒙 = 𝟏𝟐
One angle is πŸ’(𝟏𝟐) + 𝟏𝟐 = πŸ’πŸ– + 𝟏𝟐 = πŸ”πŸŽΛš and the other angle is πŸπŸ–πŸŽ βˆ’ πŸ”πŸŽ = 𝟏𝟐𝟎˚.
3.
A triangle has angles described as follows: the first angle is four more than seven times a number, another angle is
four less than the first and the third angle is twice as large as the first. What are the sizes of each of the angles?
Let 𝒙 be the number. The measure of the first angle is πŸ•π’™ + πŸ’. The measure of the second angle is πŸ•π’™ + πŸ’ βˆ’ πŸ’ = πŸ•π’™.
The measure of the third angle is 𝟐(πŸ•π’™ + πŸ’) = πŸπŸ’π’™ + πŸ–. The sum of the angles of a triangle must be πŸπŸ–πŸŽΛš.
πŸ•π’™ + πŸ’ + πŸ•π’™ + πŸπŸ’π’™ + πŸ– = πŸπŸ–πŸŽ
πŸπŸ–π’™ + 𝟏𝟐 = πŸπŸ–πŸŽ
πŸπŸ–π’™ + 𝟏𝟐 βˆ’ 𝟏𝟐 = πŸπŸ–πŸŽ βˆ’ 𝟏𝟐
πŸπŸ–π’™ = πŸπŸ”πŸ–
𝒙=πŸ”
The measure of the first angle is πŸ•(πŸ”) + πŸ’ = πŸ’πŸ + πŸ’ = πŸ’πŸ”Λš. The measure of the second angle is πŸ•(πŸ”) = πŸ’πŸΛš. The
measure of the third angle is 𝟐(πŸ’πŸ”) = πŸ—πŸΛš.
4.
One angle measures nine more than six times a number. A sequence of rigid motions maps the angle onto another
angle that is described as being thirty less than nine times the number. What is the measure of the angles?
Let 𝒙 be the number. Then the measure of one angle is πŸ”π’™ + πŸ—. The measure of the other angle is πŸ—π’™ βˆ’ πŸ‘πŸŽ. Since
rigid motions preserve the measures of angles, then the measure of these angles is equal.
πŸ”π’™ + πŸ— = πŸ—π’™ βˆ’ πŸ‘πŸŽ
πŸ”π’™ + πŸ— βˆ’ πŸ— = πŸ—π’™ βˆ’ πŸ‘πŸŽ βˆ’ πŸ—
πŸ”π’™ = πŸ—π’™ βˆ’ πŸ‘πŸ—
πŸ”π’™ βˆ’ πŸ—π’™ = πŸ—π’™ βˆ’ πŸ—π’™ βˆ’ πŸ‘πŸ—
βˆ’πŸ‘π’™ = βˆ’πŸ‘πŸ—
𝒙 = πŸπŸ‘
The angles measure πŸ”(πŸπŸ‘) + πŸ— = πŸ•πŸ– + πŸ— = πŸ–πŸ•Λš.
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
8β€’4
A right triangle is described as having an angle of size β€œsix less than negative two times a number,” another angle
that is β€œthree less than negative one-fourth the number”, and a right angle. What are the measures of the angles?
𝒙
Let 𝒙 be a number. Then one angle is βˆ’πŸπ’™ βˆ’ πŸ”. The other angle is βˆ’ βˆ’ πŸ‘. The sum of the two angles must be πŸ—πŸŽΛš.
πŸ’
𝒙
βˆ’πŸπ’™ βˆ’ πŸ” + (βˆ’ ) βˆ’ πŸ‘ = πŸ—πŸŽ
πŸ’
πŸ–π’™
𝒙
(βˆ’ ) + (βˆ’ ) βˆ’ πŸ— = πŸ—πŸŽ
πŸ’
πŸ’
πŸ—π’™
(βˆ’ ) βˆ’ πŸ— + πŸ— = πŸ—πŸŽ + πŸ—
πŸ’
πŸ—π’™
βˆ’
= πŸ—πŸ—
πŸ’
βˆ’πŸ—π’™ = πŸ‘πŸ—πŸ”
𝒙 = βˆ’πŸ’πŸ’
One of the angles is βˆ’πŸ(βˆ’πŸ’πŸ’) βˆ’ πŸ” = πŸ–πŸ– βˆ’ πŸ” = πŸ–πŸΛš. The other angle is πŸ—πŸŽ βˆ’ πŸ–πŸ = πŸ–Λš.
6.
One angle is one less than six times the size of another. The two angles are complementary angles. Find the size of
each angle.
Let 𝒙 be the measure of the first angle. Then the measure of the second angle is πŸ”π’™ βˆ’ 𝟏. The sum of the measures
will be πŸ—πŸŽ because the angles are complementary.
𝒙 + πŸ”π’™ βˆ’ 𝟏 = πŸ—πŸŽ
πŸ•π’™ βˆ’ 𝟏 = πŸ—πŸŽ
πŸ•π’™ βˆ’ 𝟏 + 𝟏 = πŸ—πŸŽ + 𝟏
πŸ•π’™ = πŸ—πŸ
𝒙 = πŸπŸ‘
One angle is πŸπŸ‘Λš and the other angle is πŸ”(πŸπŸ‘) βˆ’ 𝟏 = πŸ•πŸ– βˆ’ 𝟏 = πŸ•πŸ•Λš.
Lesson 5:
Date:
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Writing and Solving Linear Equations
11/16/14
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8β€’4
Problem Set Sample Solutions
Students practice writing and solving linear equations.
For each of the following problems, write an equation and solve.
1.
The measure of one angle is thirteen less than five times the measure of another angle. The sum of the measures of
the two angles is πŸπŸ’πŸŽΛš. Determine the measures of each of the angles.
Let 𝒙 be the size of the one angle. Then the measure of the other angle is πŸ“π’™ βˆ’ πŸπŸ‘.
𝒙 + πŸ“π’™ βˆ’ πŸπŸ‘ = πŸπŸ’πŸŽ
πŸ”π’™ βˆ’ πŸπŸ‘ = πŸπŸ’πŸŽ
πŸ”π’™ βˆ’ πŸπŸ‘ + πŸπŸ‘ = πŸπŸ’πŸŽ + πŸπŸ‘
πŸ”π’™ = πŸπŸ“πŸ‘
𝒙 = πŸπŸ“. πŸ“
The measure of one angle is πŸπŸ“. πŸ“Λš and the measure of the other angle is πŸπŸ’πŸŽ βˆ’ πŸπŸ“. πŸ“ = πŸπŸπŸ’. πŸ“Λš.
2.
An angle measures seventeen more than three times a number. Its supplement is three more than seven times the
number. What is the measure of each angle?
Let 𝒙 be the number. Then one angle is πŸ‘π’™ + πŸπŸ•. The other angle is πŸ•π’™ + πŸ‘. Since the angles are supplementary,
the sum of their measures will be πŸπŸ–πŸŽ.
πŸ‘π’™ + πŸπŸ• + πŸ•π’™ + πŸ‘ = πŸπŸ–πŸŽ
πŸπŸŽπ’™ + 𝟐𝟎 = πŸπŸ–πŸŽ
πŸπŸŽπ’™ + 𝟐𝟎 βˆ’ 𝟐𝟎 = πŸπŸ–πŸŽ βˆ’ 𝟐𝟎
πŸπŸŽπ’™ = πŸπŸ”πŸŽ
𝒙 = πŸπŸ”
One angle is πŸ‘(πŸπŸ”) + πŸπŸ• = πŸ”πŸ“Λš. The other angle is πŸπŸ–πŸŽ βˆ’ πŸ”πŸ“ = πŸπŸπŸ“Λš.
3.
The angles of a triangle are described as follows: βˆ π‘¨ is the largest angle, its measure is twice the measure of βˆ π‘©.
The measure of ∠π‘ͺ is 𝟐 less than half the measure of βˆ π‘©. Find the measures of the three angles.
Let 𝒙 be the measure of βˆ π‘©. Then the measure of βˆ π‘¨ = πŸπ’™ and ∠π‘ͺ =
𝒙
βˆ’ 𝟐. The sum of the measures of the angles
𝟐
must be πŸπŸ–πŸŽ.
𝒙
𝒙 + πŸπ’™ + βˆ’ 𝟐 = πŸπŸ–πŸŽ
𝟐
𝒙
πŸ‘π’™ + βˆ’ 𝟐 + 𝟐 = πŸπŸ–πŸŽ + 𝟐
𝟐
𝒙
πŸ‘π’™ + = πŸπŸ–πŸ
𝟐
πŸ”π’™ 𝒙
+ = πŸπŸ–πŸ
𝟐 𝟐
πŸ•π’™
= πŸπŸ–πŸ
𝟐
πŸ•π’™ = πŸ‘πŸ”πŸ’
𝒙 = πŸ“πŸ
The measures of the angles are as follows: βˆ π‘¨ = πŸπŸŽπŸ’Λš, βˆ π‘© = πŸ“πŸΛš, 𝒂𝒏𝒅 ∠π‘ͺ =
Lesson 5:
Date:
© 2013 Common Core, Inc. Some rights reserved. commoncore.org
πŸ“πŸ
βˆ’ 𝟐 = πŸπŸ” βˆ’ 𝟐 = πŸπŸ’Λš.
𝟐
Writing and Solving Linear Equations
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
4.
8β€’4
A pair of corresponding angles are described as follows: the measure of one angle is five less than seven times a
number and the measure of the other angle is eight more than seven times the number. Are the angles congruent?
Why or why not?
Let 𝒙 be the number. Then measure of one angle is πŸ•π’™ βˆ’ πŸ“ and the measure of the other angle is πŸ•π’™ + πŸ–. Assume
they are congruent, which means their measures are equal.
πŸ•π’™ βˆ’ πŸ“ = πŸ•π’™ + πŸ–
πŸ•π’™ βˆ’ πŸ•π’™ βˆ’ πŸ“ = πŸ•π’™ βˆ’ πŸ•π’™ + πŸ–
βˆ’πŸ“ β‰  πŸ–
Since – πŸ“ β‰  πŸ–, the angles are not congruent.
5.
The measure of one angle is eleven more than four times a number. Another angle is twice the first angle’s
measure. The sum of the measures of the angles is πŸπŸ—πŸ“Λš. What is the measure of each angle?
Let 𝒙 be the number. The measure of one angle can be represented with πŸ’π’™ + 𝟏𝟏 and the other angle’s measure
can be represented as 𝟐(πŸ’π’™ + 𝟏𝟏) = πŸ–π’™ + 𝟐𝟐.
πŸ’π’™ + 𝟏𝟏 + πŸ–π’™ + 𝟐𝟐 = πŸπŸ—πŸ“
πŸπŸπ’™ + πŸ‘πŸ‘ = πŸπŸ—πŸ“
πŸπŸπ’™ + πŸ‘πŸ‘ βˆ’ πŸ‘πŸ‘ = πŸπŸ—πŸ“ βˆ’ πŸ‘πŸ‘
πŸπŸπ’™ = πŸπŸ”πŸ
𝒙 = πŸπŸ‘. πŸ“
The measure of one angle is πŸ’(πŸπŸ‘. πŸ“) + 𝟏𝟏 = πŸ“πŸ’ + 𝟏𝟏 = πŸ”πŸ“Λš and the measure of the other angle is πŸπŸ‘πŸŽΛš.
6.
Three angles are described as follows: βˆ π‘© is half the size of βˆ π‘¨. The measure of ∠π‘ͺ is equal to one less than
𝟐 times the measure of βˆ π‘©. The sum of βˆ π‘¨ and βˆ π‘© is πŸπŸπŸ’. Can the three angles form a triangle? Why or why not?
Let 𝒙 represent the measure of βˆ π‘¨. Then the measure of βˆ π‘© =
𝒙
𝒙
and the measure of ∠π‘ͺ = 𝟐 ( ) βˆ’ 𝟏 = 𝒙 βˆ’ 𝟏 .
𝟐
𝟐
The sum of βˆ π‘¨ + βˆ π‘© = πŸπŸπŸ’.
𝒙
= πŸπŸπŸ’
𝟐
πŸ‘π’™
= πŸπŸπŸ’
𝟐
πŸ‘π’™ = πŸπŸπŸ–
𝒙 = πŸ•πŸ”
𝒙+
The measure of βˆ π‘¨ = πŸ•πŸ”Λš, βˆ π‘© =
πŸ•πŸ”
= πŸ‘πŸ–Λš, and ∠π‘ͺ = πŸ•πŸ“Λš. The sum of the three angles is πŸ•πŸ” + πŸ‘πŸ– + πŸ•πŸ“ = πŸπŸ–πŸ—.
𝟐
Since the sum of the measures of the interior angles of a triangle must sum to πŸπŸ–πŸŽΛš, these angles do not form a
triangle. Their sum is too large.
Lesson 5:
Date:
© 2013 Common Core, Inc. Some rights reserved. commoncore.org
Writing and Solving Linear Equations
11/16/14
58
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.