Geometry Section 5-2 1112
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SECTION 5-2Medians and Altitudes of Triangles
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ESSENTIAL QUESTIONS
How do you identify and use medians in triangles?
How do you identify and use altitudes in triangles?
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VOCABULARY1. Median:
2. Centroid:
3. Altitude:
4. Orthocenter:
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VOCABULARY1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid:
3. Altitude:
4. Orthocenter:
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VOCABULARY1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside)
3. Altitude:
4. Orthocenter:
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VOCABULARY1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside)
3. Altitude: A segment in a triangle that connects a vertex to the opposite side so that the side and altitude are perpendicular
4. Orthocenter:
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VOCABULARY1. Median: A segment in a triangle that connects a vertex to
the midpoint of the opposite side
2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside)
3. Altitude: A segment in a triangle that connects a vertex to the opposite side so that the side and altitude are perpendicular
4. Orthocenter: The point of concurrency where the altitudes of a triangle intersect
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5.7 - CENTROID THEOREM
The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side
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5.7 - CENTROID THEOREM
The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side
If G is the centroid of ∆ABC, then
AG =
23
AF , BG =23
BD, and CG =23
CE
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Special Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in Triangles
Name Example Point of Concurrency
Special Property Example
Perpendicular Bisector
CircumcenterCircumcenter is equidistant from
each vertex
Angle Bisector
IncenterIn center is
equidistant from each side
Median CentroidCentroid is two-
thirds the distance from vertex to
opposite midpoint
Altitude OrthocenterAltitudes are concurrent at orthocenter
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
PV = YV − YP
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
PV = YV − YP
PV =12 − 8
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EXAMPLE 1
In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.
YP =
23
YV
YP =
23
(12)
YP = 8
PV = YV − YP
PV =12 − 8
PV = 4
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
GE = CE −CG
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
GE = CE −CG
GE = 6 − 4
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EXAMPLE 2
In ∆ABC, CG = 4. Find GE.
CG =
23
CE
4 =
23
CE
CE = 6
GE = CE −CG
GE = 6 − 4
GE = 2
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EXAMPLE 3
An artist is designing a sculpture that balances a triangle on tip of a pole. In the artistʼs design on the coordinate plane, the vertices are located at (1, 4), (3, 0), and (3, 8). What are the coordinates of the point where the artist should place
the pole under the triangle so that is will balance?
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EXAMPLE 3
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y (1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y (1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y (1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
= 3,4( )
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
= 3,4( )
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
We need to find the centroid, so we start by finding the midpoint
of our vertical side.
M =
3 + 32
,0 + 8
2
⎛⎝⎜
⎞⎠⎟ =
62
,82
⎛⎝⎜
⎞⎠⎟
= 3,4( )
(1, 4), (3, 0), (3, 8)
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EXAMPLE 3
x
y
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EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
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EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
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EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
= (−2)2 + 02
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EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
= (−2)2 + 02
= 4
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EXAMPLE 3
x
y
Next, we need the distance from the opposite vertex to this
midpoint.
d = (1− 3)2 + (4 − 4)2
= (−2)2 + 02
= 4 = 2
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EXAMPLE 3
x
y
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EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
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EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
P = 1+
23
(2),4⎛⎝⎜
⎞⎠⎟
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EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
P = 1+
23
(2),4⎛⎝⎜
⎞⎠⎟
P = 1+
43
,4⎛⎝⎜
⎞⎠⎟
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EXAMPLE 3
x
y The centroid P is 2/3 of this distance from the vertex.
P = 1+
23
(2),4⎛⎝⎜
⎞⎠⎟
P = 1+
43
,4⎛⎝⎜
⎞⎠⎟
P =
73
,4⎛⎝⎜
⎞⎠⎟
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2 = −2
Thursday, March 1, 2012
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12 = −2
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12
m HJ( ) = 2 −1
1+ 5
= −2
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12
m HJ( ) = 2 −1
1+ 5 =
16
= −2
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
To find the orthocenter, find the intersection of two altitudes.
H
I
J Let’s find the equations for the altitudes coming from I and H.
m JI( ) = 1+ 3
−5 + 3 =
4−2
⊥ m =12
m HJ( ) = 2 −1
1+ 5 =
16
⊥ m = −6
= −2
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Altitude through H
Thursday, March 1, 2012
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
y − 2 =
12
(x −1)
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
y − 2 =
12
(x −1)
y − 2 =
12
x −12
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J ⊥ m =
12
, (1,2)
Altitude through H
y − 2 =
12
(x −1)
y − 2 =
12
x −12
y =
12
x +32
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
Altitude through I
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
y + 3 = −6(x + 3)
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
y + 3 = −6(x + 3)
y + 3 = −6x −18
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EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).
Find the coordinates of the orthocenter of ∆HIJ.
x
y
H
I
J
⊥ m = −6, (−3,−3)
Altitude through I
y + 3 = −6(x + 3)
y + 3 = −6x −18
y = −6x − 21
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EXAMPLE 4
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EXAMPLE 4
−6x − 21 =
12
x +32
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
Thursday, March 1, 2012
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
−6 −
4513
⎛⎝⎜
⎞⎠⎟− 21 =
12
−4513
⎛⎝⎜
⎞⎠⎟+
32
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
−6 −
4513
⎛⎝⎜
⎞⎠⎟− 21 =
12
−4513
⎛⎝⎜
⎞⎠⎟+
32
−
313
= −3
13
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EXAMPLE 4
−6x − 21 =
12
x +32
+6x +6x
−21 =
132
x +32
−
32
−32
−
452
=132
x
213
−452
⎛⎝⎜
⎞⎠⎟=
132
x⎛⎝⎜
⎞⎠⎟
213
x = −
4513
−6 −
4513
⎛⎝⎜
⎞⎠⎟− 21 =
12
−4513
⎛⎝⎜
⎞⎠⎟+
32
−
313
= −3
13
−
4513
,−3
13
⎛⎝⎜
⎞⎠⎟
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CHECK YOUR UNDERSTANDING
p. 337 #1-4
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PROBLEM SET
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PROBLEM SET
p. 338 #5-31 odd, 49, 53
"Education's purpose is to replace an empty mind with an open one." – Malcolm Forbes
Thursday, March 1, 2012