1. z

^{2}- 3z+2 over z^{2 }+ 3z-4 multiplied by z^{2 }+ 6z+8 over z^{2}-4 all of this equals to z2. 2h

^{2 }-2h-4 over (h+1)(h-2) plus h^{2 }+ h-2 over (h+2)(h+1)1. z^{2} - 3z+2 over z^{2 }+ 3z-4 multiplied by z^{2 }+ 6z+8 over z^{2} -4 all of this equals to z

2. 2h^{2 }-2h-4 over (h+1)(h-2) plus h^{2 }+ h-2 over (h+2)(h+1)

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New Milford, CT

1. (z^{2}-3z+2) / (z^{2}+3z-4) * (z^{2}+6z+8) / (z^{2}-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:

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:

(1^{2}-3*1+2)/(1^{2}+3*1-4) * (1^{2}+6*1+8)/(1^{2}-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. (2h^{2}-2h-4)/( (h+1)(h-2) ) + (h^{2}+h-2)/( (h+2)(h+1) )

Factor a two out of the numerator in the first term:

2*(h^{2}-h-2)/( (h+1)(h-2) ) + (h^{2}+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(h^{2}-h-2) + h^{2}-3h+2)/( (h-2)(h+1) )

=(2h^{2}-2h-4+h^{2}-3h+2)/( (h-2)(h+1) )

=(3h^{2}-5h-2)/( (h-2)(h+1) )

factor the numerator:

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

= **(3h+1)/(h+1)**

San Diego, CA

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)

Careful, if z = 1, the first term becomes 0/0

Yes, Pierce

I agree with you.

Thanks

Evan F.

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