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MAA AMC
AMC 12/AHSME
1970 AMC 12/AHSME
3
Find y in terms of x
Find y in terms of x
Source: 1970 AHSME Problem 3
March 20, 2014
AMC
Problem Statement
If
x
=
1
+
2
p
x=1+2^p
x
=
1
+
2
p
and
y
=
1
+
2
−
p
y=1+2^{-p}
y
=
1
+
2
−
p
, then
y
y
y
in terms of
x
x
x
is:
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
x
+
1
x
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
x
+
2
x
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
x
x
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
2
−
x
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
x
−
1
x
<span class='latex-bold'>(A) </span>\dfrac{x+1}{x-1}\qquad<span class='latex-bold'>(B) </span>\dfrac{x+2}{x-1}\qquad<span class='latex-bold'>(C) </span>\dfrac{x}{x-1}\qquad<span class='latex-bold'>(D) </span>2-x\qquad <span class='latex-bold'>(E) </span>\dfrac{x-1}{x}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
x
−
1
x
+
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
x
−
1
x
+
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
x
−
1
x
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
2
−
x
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
x
x
−
1
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