Solving Double Absolute Value Equations
Case 1: Both Abs are “+”
Case 2: Abs are “+ & -”
Case 3: Both Abs are “-”
(𝒙 − 𝟏) + (𝟑𝒙 − 𝟐) = 2 (𝒙 − 𝟏) + −(𝟑𝒙 − 𝟐) = 2
4x − 3 = 2
−2x + 1 = 2
−2x = 1
4x = 5
1
5
x =−
x =
2
4
−(𝒙 − 𝟏) + (𝟑𝒙 − 𝟐) = 2
2x − 1 = 2
2x = 3
3
x =
2
−(𝒙 − 𝟏) −
(𝟑𝒙 − 𝟐) = 2
−4x + 3 = 2
−4x = −1
1
x =
4
Solving Double Absolute Value Equations
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Ans:
𝟓 𝟏
,
𝟒 𝟒
Warm – Up: Page 28
Left 2, down 3
Slope  𝟏𝒂 𝟑𝒂 𝟓𝒂
,
,
,…
𝟏 𝟏 𝟏
𝟏𝒂
𝟏
𝟏𝒂 𝟕𝒂
, ,…
𝟏 𝟏
Inverses
(reciprocal)
𝒂 𝒂 𝒂
, , ,…
𝟏 𝟑 𝟓
Translations
Reflections
Dilations
Shift “c” up
Shift “c” down
Shift “c” left
Shift “c” right
Reflection over y-axis
Reflection over x-axis
Vertical Shrink (narrow)
Vertical Stretch (wide)
Horizontal Stretch (wide)
Horizontal Shrink (narrow)
Shift “6” left
Shift “2” down
Shift “3” right and 7 up
Refl. x-axis, 7 left and 10 down
Vert. Shrink of
1
,
2
2 up
Right 4, up 2
left 3, down 5
Right 8, down 4
Vertex = (-1, -4)
𝟐
y = 𝒙+𝟏 −𝟒
y =(x + 1)(x + 1) −𝟒
y = 𝒙𝟐 + 𝟏𝒙 + 𝟏𝒙 + 𝟏 − 𝟒
y = 𝒙𝟐 + 𝟐𝒙 − 𝟑
b =𝟐
c = -3
Slope  𝟏𝒂 𝟑𝒂 𝟓𝒂
,
,
,…
𝟏 𝟏 𝟏
𝟏𝒂
𝟏
𝟏𝒂 𝟕𝒂
, ,…
𝟏 𝟏
Inverses
(reciprocal)
𝒂 𝒂 𝒂
, , ,…
𝟏 𝟑 𝟓
𝒚 = 𝒙; v=(0,0)
𝟏 𝟏 𝟏
𝒔𝒍𝒐𝒑𝒆 = , , , … .
𝟏 𝟑 𝟓
𝒚=
𝒙
𝒚= 𝒙+𝟑−𝟐
𝒚=
𝒙 + 𝟑 − 𝟐;v=(-3,-2)
𝒍𝒆𝒇𝒕 𝟑, 𝒅𝒐𝒘𝒏 𝟐
v=(-4,0); 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒗𝒆𝒓 𝒙 − 𝒂𝒙𝒊𝒔, 𝒍𝒆𝒇𝒕 𝟒
Right 5, up 7
v = (5, 7)
v=(-4,1); 𝒍𝒆𝒇𝒕 𝟒, 𝒖𝒑 𝟏
v=(2,5);
𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒗𝒆𝒓 𝒙 − 𝒂𝒙𝒊𝒔, 𝒓𝒊𝒈𝒉𝒕 𝟐, 𝒖𝒑 𝟓
𝒚 = −𝒙𝟐 + 𝟏𝟎
𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒗𝒆𝒓 𝒙 − 𝒂𝒙𝒊𝒔,
𝒖𝒑 𝟏𝟎