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docker41 [41]
3 years ago
8

The sum of the fractions 2/y-3 and 6/y+3 is equal to their

Mathematics
2 answers:
Serggg [28]3 years ago
7 0

Answer:

I think "Product" is the answer. Not really sure.

Sphinxa [80]3 years ago
6 0

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

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skad [1K]

Answer:

option A and B

x=\frac{-7+\sqrt{17}} {4}

and

x=\frac{-7-\sqrt{17}} {4}

Step-by-step explanation:

we have

2x^2+7x+4=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

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in this problem we have

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so

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substitute in the formula

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x=\frac{-7\pm\sqrt{17}} {4}

so

x=\frac{-7+\sqrt{17}} {4}

and

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