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Let’s check out your problem:
What are the
y
y
y
coordinates of the system,
x
2
+
y
2
=
10
x^2+y^2=10
x
2
+
y
2
=
10
and
x
−
2
y
=
5
x-2y=5
x
−
2
y
=
5
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Math Problems
Precalculus
Powers of i
Full solution
Q.
What are the
y
y
y
coordinates of the system,
x
2
+
y
2
=
10
x^2+y^2=10
x
2
+
y
2
=
10
and
x
−
2
y
=
5
x-2y=5
x
−
2
y
=
5
Solve for x:
Step
1
1
1
: Solve the line equation for x.
\newline
x
−
2
y
=
5
x - 2y = 5
x
−
2
y
=
5
\newline
x
=
2
y
+
5
x = 2y + 5
x
=
2
y
+
5
Substitute in circle equation:
Step
2
2
2
: Substitute
x
x
x
in the
circle equation
.
(
2
y
+
5
)
2
+
y
2
=
10
(2y + 5)^2 + y^2 = 10
(
2
y
+
5
)
2
+
y
2
=
10
4
y
2
+
20
y
+
25
+
y
2
=
10
4y^2 + 20y + 25 + y^2 = 10
4
y
2
+
20
y
+
25
+
y
2
=
10
5
y
2
+
20
y
+
25
=
10
5y^2 + 20y + 25 = 10
5
y
2
+
20
y
+
25
=
10
Simplify and solve quadratic:
Step
3
3
3
: Simplify and solve the
quadratic equation
.
\newline
5
y
2
+
20
y
+
15
=
0
5y^2 + 20y + 15 = 0
5
y
2
+
20
y
+
15
=
0
\newline
y
2
+
4
y
+
3
=
0
y^2 + 4y + 3 = 0
y
2
+
4
y
+
3
=
0
\newline
(
y
+
3
)
(
y
+
1
)
=
0
(y + 3)(y + 1) = 0
(
y
+
3
)
(
y
+
1
)
=
0
\newline
y
=
−
3
y = -3
y
=
−
3
,
y
=
−
1
y = -1
y
=
−
1
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)
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\newline
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Question
3
⋅
(
3
+
20
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)
=
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⋅
(
3
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i
)
=
\newline
Your answer should be a complex number in the form
a
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b
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a+b i
a
+
bi
where
a
a
a
and
b
b
b
are real numbers.
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Question
(
35
−
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)
+
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13
+
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i
)
=
(35-23 i)+(13+25 i)=
(
35
−
23
i
)
+
(
13
+
25
i
)
=
\newline
Express your answer in the form
(
a
+
b
i
)
(a+b i)
(
a
+
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