can u solve this?

Question

1/(h+1) = (2h+3)/(h-1)

Answer 1

1/(h+1) = (2h+3)/(h-1)
multiply by (h+1)(h-1) both sides:

h-1 = (2h+3)(h+1)
h-1 = 2h^2 +2h +3h + 3

Answer 2

1/(h+1) = (2h+3)/(h-1)

multiply both sides by (h-1)(h+1) to remove denominators:
(If we get any solution like h=1 or h=-1, then they would be discarded, because denominators don't allow them)

(h-1) = (2h+3)(h+1)

Then solve the binomium:

(h-1) = (2h+3)(h+1)

h-1 = 2*h^2 +5*h +3

Reduce it:

2*h^2 + 4*h+ 4 =0

Simplify dividing by 2:
h^2 + 2*h+ 2 =0

Now we solve it as a quadratic equation- using the formula:
h= -1 -i
h=-1 + i
(roots are complex).

Answer 3

Multiply both sides by (h+1)(h-1) to get rid of the fractions:
h-1=(2h+3)(h+1)
h-1=2h^2+5h+3

Subtract h and add 1 from both sides:
2h^2+4h+4=0

Factor out a 2:
h^2+2h+2=0

Use the quadratic formula:
h=-b±√b^2-4ac/2a
a=1,b=2,c=2
h=-2±√(2)^2-4(1*2)/2(1)
h=(-2±√4-8)/2
h=-2±√-4/2
h=-2±√4i/2
h=-2±2i/2
h=-1±i

Answer 4

You have different denominators. Multiply both sides by (h-1) and by (h+1) and you'll get rid of them:

(h-1) = (2h+3)(h+1)

Now expand out:
(h-1) = (2h^2 + 5h + 3)
0 = 2h^2 + 4h +4
h^2 + 2h + 2 = 0

Now you need solve using the quadratic formula:
a = 1
b = 2
c = 2

h = [ -2 ± sqrt( 2² - 4(1)(2) ) ] / 2
h = -1 ± sqrt(-4)/2
h = -1 ± i

Answer 5

Multiply left side by (h -1)/(h -1) and multiply right side by (h+1)/(h+1( to get common denominator. Leave factored.

Set equation equal to zero.

Factor numerator and eliminate any common factors between numerator and denominator.

Hope that helps!

Answer 6

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