1.
2.
3.
4.
5.
6.
Algebra II
Twice the cube of a number
The square of a number decreased by
ten
The sum of three times a number and
seven
Seven less than one third a number
Three times the sum of a number and
seven
If n represents an odd integer, what Is
the next consecutive odd integer
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Literal Equations
Algebra II
Algebra II
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Algebra II
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Algebra II
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1. 3x – 2y = 10
3. 10 = 3x – ¾ y
-2y = -3x + 10
-3x + 10 = - ¾y
y = 3/2x – 5
4x – 40/3 = y
2. ½x + 3y = 5
4. 4x + 8y = 17
3y = - ½x + 5
y = - 1/6x + 5/3
Algebra II
8y = -4x + 17
y = -1/2x + 17/8
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5. 6x – 5y = 13
y = 6/5x – 13/5
6.
2/3x + 3/2y = 10
y = -4/9x + 20/3
Algebra II
7. 4/3x – 2y = 12
y = 2/3 x – 6
8. 3x – 8y = 4
y = 3/8x – ½
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Ax + By = C
is solved for C
in terms of A, x, B, and y
½bh = A
Is solved for A
in terms of b and h.
Algebra II
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1. What is P = 2L + 2w solved for? In terms of?
Solved for P in terms of L and W
2. What is C = 2πr solved for? In terms of?
Solved for C in terms of r
3. What are you finding in A = (b)(h)? In terms of?
Finding A in terms of b and h
4. What are you finding in b = A/h ? In terms of?
Finding b in terms A and h
Algebra II
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1. If A = c,
3t
find t in terms of A and
c.
A=c
3t
A = (3t)(c)
A = 3tc
A=t
3c
Algebra II
2. If 3L + 2W = 6,
find w in terms of L.
3L + 2W = 6
2w = -3L + 6
w = -3/2 L + 3
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3. If c = 4a2, find a in
terms of c.
c = 4a 2
c
2
=a
4
c
=a
4
Algebra II
4. If A = πrL + πr2,
Find L in terms of A and
r.
A - πr2 = πrL
A - πr2 = L
πr
A –r=L
πr
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5. If E=mc2, find m in
terms of c and E.
E = mc
E
=m
2
c
Algebra II
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6. If V = 1/3πr2h, find r
in terms of V and h.
1
v = p r 2h
3
3v = p r h
3v
= r2
ph
2
3v
=r
ph
12
p2q
L=
m
7. If
find p in
terms of m, L, and q.
p2q
L=
m
A = ½bh
mL= p2 q
2A = bh
mL 2
=p
q
mL
=p
q
Algebra II
8. If A = 1/2bh, find h in
terms of A and b.
2A = h
b
13
Solve for y: 5x + 2/3 y = 10
2. Solve for x. 3x - 4y = 12
3. Solve for r in terms of V and h:
V = 4/3πr2h
1.
Algebra II
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Literal Equations
Algebra II

What is P in terms of right now?
 L and W

What should you replace?
 L
W=L+3
W–3=L
P = 2L + 2W
P = 2(w – 3) + 2w
P = 4w – 6
Algebra II
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
What is A in terms of right now?
 b and h

What should you replace?
 B
b = 3h
A = ½bh
P = ½ (3h)(h)
P = 3/2 h2
Algebra II
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
What is V in terms of right now?
 r and h

What should you replace?
 h
r = 1/2h
2r = h
V = 1/3πr2h
V = 1/3πr2(2r)
V = 2/3πr3
Algebra II
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
What is the Area in terms of now?
 r

What should you replace?
 R
C = 2πr
R = C/2π
A = π(r)2
A = π(C/2π)2
A = C/(4π)
Algebra II
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
What is the Area in terms of now?
 b and h

What should you replace?
 H
A = ½ bh
A = ½ b(√3/2)b
A = √3b2/4
Algebra II
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

h=d V=πr2h
What is V in terms of now?
 Radius and height

What should we replace?
 Height
h=d
h=2r
V=πr2h
V=πr2(2r)
V=2πr3
Algebra II
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

h=5 + 3r S=2πrh + 2πr2
What is S in terms of now?
 Radius and height

What should we replace?
 Height
h=5+3r
 S= 2πrh + 2πr2
S= 2πr(5 + 3r) + 2πr2
S= 10πr + 6πr2 +2πr2
S= 10πr + 8πr2
Algebra II
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1.
Solve for y: 1/3 x + 2/5 y = 10.
1.
Given 4n + 1/2 m = 12, find m in terms of n.
2.
If the area of a square can be found by A =
s2 and the perimeter of a square can be found
by p = 4s, find area of a square (A) in terms of
its perimeter (p).
Algebra II
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