Step-by-step explanation:
Move expression to the left side and change its sign
5
y
−
3
+
10
y
2
−
y
−
6
−
y
y
+
2
=
0
Write
−
y
as a sum or difference
5
y
−
3
+
10
y
2
+
2
y
−
3
y
−
6
−
y
y
+
2
=
0
Factor out
y
and
−
3
from the expression
5
y
−
3
+
10
y
(
y
+
2
)
−
3
(
y
+
2
)
−
y
y
+
2
=
0
Factor out
y
+
2
from the expression
5
y
−
3
+
10
(
y
+
2
)
(
y
−
3
)
−
y
y
+
2
=
0
Write all numerators above the least common denominator
5
(
y
+
2
)
+
10
−
y
(
y
−
3
)
(
y
+
2
)
(
y
−
3
)
=
0
Distribute
5
and
−
y
through the parenthesis
5
y
+
10
+
10
−
y
2
+
3
y
(
y
+
2
)
(
y
−
3
)
=
0
Collect the like terms
8
y
+
20
−
y
2
(
y
+
2
)
(
y
−
3
)
=
0
Use the commutative property to reorder the terms
−
y
2
+
8
y
+
20
(
y
+
2
)
(
y
−
3
)
=
0
Write
8
y
as a sum or difference
−
y
2
+
10
y
−
2
y
+
20
(
y
+
2
)
(
y
−
3
)
=
0
Factor out
−
y
and
−
2
from the expression
−
y
(
y
−
10
)
−
2
(
y
−
10
)
(
y
+
2
)
(
y
−
3
)
=
0
Factor out
−
(
y
−
10
)
from the expression
−
(
y
−
10
)
(
y
+
2
)
(
y
+
2
)
(
y
−
3
)
=
0
Reduce the fraction with
y
+
2
−
y
−
10
y
−
3
=
0
Determine the sign of the fraction
−
y
−
10
y
−
3
=
0
Simplify
10
−
y
y
−
3
=
0
When the quotient of expressions equals
0
, the numerator has to be
0
10
−
y
=
0
Move the constant,
10
, to the right side and change its sign
−
y
=
−
10
Change the signs on both sides of the equation
y
=
10
Check if the solution is in the defined range
y
=
10
,
y
≠
3
,
y
≠
−
2
∴
y
=
10
The data for resort A shows more consistency because a larger interquartile range such as the one for resort B, shows more variation. This means that the snowfall for resort A is more likely to be close to the median.
Just did this on edg. :)
I hope this helps you
2a.2a+1.2a-3.2a-3.1
4a^2+2a-6a-3
4a^2-4a-3
