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Mathematics
Problem
The slope of a curve at any point is given by
d
y
d
x
=
2
y
−
3
x
4
y
+
5
\dfrac{dy}{dx}=\dfrac{2y-3x}{4y+5}
d
x
d
y
=
4
y
+
5
2
y
−
3
x
with
y
≠
−
5
4
y\neq-\dfrac{5}{4}
y
=
−
4
5
. Under which condition on
x
x
x
and
y
y
y
will the tangent to the curve be horizontal? a)
(
2
y
−
3
x
)
(
4
y
+
5
)
=
0
(2y-3x)(4y+5)=0
(
2
y
−
3
x
)
(
4
y
+
5
)
=
0
b)
4
y
+
5
=
0
4y+5=0
4
y
+
5
=
0
c)
2
y
−
3
x
=
0
2y-3x=0
2
y
−
3
x
=
0
d)
2
y
−
3
x
=
4
y
+
5
2y-3x=4y+5
2
y
−
3
x
=
4
y
+
5
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