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Simplify the expresions and solve the equations

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2 Answers

 
1. (z2-3z+2) / (z2+3z-4)  *  (z2+6z+8) / (z2-4) = z
 
Factor the left hand side:
 
    (z-2)(z-1) / ( (z+4)(z-1) )  *  (z+4)(z+2) / ( (z-2)(z+2) ) = z
 
Cancel out like terms:
 
    (z-2)/(z+4) * (z+4)/(z-2) = z
 
    (z-2)(z+4) / ((z+4)(z-2) ) = z
 
Cancel like terms again:
 
    1 = z
 
So we find that z = 1. Now we need to check and make sure that when we plug this back into the equality, we do not divide by zero:
 
 (12-3*1+2)/(12+3*1-4)  *  (12+6*1+8)/(12-4) = 1
 
 
 0/0 * 15/-3 = 1
 
So when we plug z = 1 into the original equation, one of our terms turns out to be 0/0. This is impossible, so there is no solution.
 
2. (2h2-2h-4)/( (h+1)(h-2) ) + (h2+h-2)/( (h+2)(h+1) )
 
Factor a two out of the numerator in the first term:
 
    2*(h2-h-2)/( (h+1)(h-2) ) + (h2+h-2)/( (h+2)(h+1) )
 
Factor out our polynomials:
 
    2*(h-2)(h+1)/( (h+1)(h-2) ) + (h+2)(h-1)/( (h+2)(h+1) )
 
Cancel our like terms:
 
   2*(h-2)/(h-2) + (h-1)/(h+1)
 
Get a common denominator:
 
   2*(h-2)(h+1)/( (h-2)(h+1) ) + (h-1)(h-2)/( (h-2)(h+1) )
 
Add
 
   (2*(h-2)(h+1) + (h-1)(h-2) )/( (h-2)(h+1) )
 
Multiply out:
 
  (2(h2-h-2) + h2-3h+2)/( (h-2)(h+1) )
 
 =(2h2-2h-4+h2-3h+2)/( (h-2)(h+1) )
 
 =(3h2-5h-2)/( (h-2)(h+1) )
 
factor the numerator:
 
  = (3h+1)(h-2)/( (h-2)(h+1) )
 
  = (3h+1)/(h+1)
 

Comments

In the 2nd problem when you get 2*(h-2)/(h-2) + (h-1)/(h+1),
it seems the first expression can cancel the (h-2)/(h-2)
to leave 2 + (h-1)/(h+1) then adding to get (h+1)/(h+1) which also reduces to 
1 as the final answer to simplifying of the expression.
z^2-3z+2= (z-1)(z-2)
 
z^2+3z-4= (z-1)(z+4)

z^2+6z+8=(z+2)(z+4)
 
z^2-4=(z-2)(z+2)
 
Putting all this together as per question----
 
((z-1)(z-2)/(z-1)(z+4))*((z+2)(z+4)/(z-2)(z+2))
 
=1=z
 
z=1
 
Second part---------
 
2h^2-2h-4=2(h-2)(h+1)
 
h^2+h-2= (h+2)(h-1)
 
Putting all this together as per question------
 
2(h-2)(h+1)/(h+1)(h-2)=2
 
2+((h-1)(h+1))
 
(3h+1)/(h+1)

Comments

Careful, if z = 1, the first term becomes 0/0
Yes, Pierce
I agree with you.
Thanks