# solve ,show work and explain?

2/a^2-1=6/a^2-2a-3

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• Anonymous

Possible intermediate steps:

2/(a^2-1) = 6/(a^2-2 a-3)

Cross multiply:

2 (a^2-2 a-3) = 6 (a^2-1)

Expand out terms on both sides:

2 a^2-4 a-6 = 6 a^2-6

Subtract 6 a^2-6 from both sides:

-4 a^2-4 a = 0

Solve the quadratic equation by completing the square:

Divide both sides by -4:

a^2+a = 0

a^2+a+1/4 = 1/4

Factor the left hand side:

(a+1/2)^2 = 1/4

Take the square root of both sides:

a+1/2 = -1/2 or a+1/2 = 1/2

Subtract 1/2 from both sides:

a = -1 or a+1/2 = 1/2

Subtract 1/2 from both sides:

a = -1 or a = 0

Now test that these solutions are appropriate by substitution into the original equation:

Check the solution a = 0:

2/(a^2-1) => 2/(-1+0^2) = -2

6/(a^2-2 a-3) => 6/(-3-2 0+0^2) = -2

So the solution is correct.

Thus, the solutions are:

a = -1 or a = 0

• 2/(a^2 - 1) = 6/(a^2 - 2a - 3)

you will need to cross multiply for the denominators to be the same.

2(a^2 - 2a - 3) = 6(a^2 - 1)

2a^2 - 4a - 6 = 6a^2 - 6

-4a^2 -4a = 0

factor

divide both sides by -4 because it is a common factor

a^2 + a = 0

use completing the square method

a^2 + a +(1/2)^2 = 0 + (1/2)^2

a^2 + a + 1/4 = 1/4

now fully factor

(a + 1/2)^2 = 1/4

find the square root of both sides to remove the squared symbol

a + 1/2 = + or - sqrt.(1/4)

= + or - 1/2

a = 0 or -1

• ok, I will just explain, since there is no good way to show work on here. start by cross multiplying.

so 6(a^2 -1) = 2 (a^2-2a-3)

Then distribute the 6 and the 2

then get all numbers on 1 side so that all the a's and stuff =0

Then we use the zero product property to solve

you factor the a's and set the factors equal to 0

then solve for a.

There will be at most 2 solutions.

Source(s): I am a math teacher.
• 2/a^2-1=6/a^2-2a-3 first you need to factor at bottom out

2/(a+1)(a-1)=6/(a-3)(a+1) LCD will be a+1, a-1, a-3 see another miss what LCD time that

2(a-3)=6(a-1)

2a-6=6a-6

6a-2a=-6+6

4a=0